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Compare nested objects in JavaScript and return keys equality
JavaScript: Deep comparison recursively: Objects and propertiesGet the property of the difference between two objects in javascriptLength of a JavaScript objectWhat is the most efficient way to deep clone an object in JavaScript?How do I remove a property from a JavaScript object?How do I check if an array includes an object in JavaScript?Which equals operator (== vs ===) should be used in JavaScript comparisons?Compare two dates with JavaScriptHow do I test for an empty JavaScript object?How do I loop through or enumerate a JavaScript object?How do I correctly clone a JavaScript object?Checking if a key exists in a JavaScript object?
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I have two nested objects obj1
and obj2
and I want to compare them and the recursively return an object that for each nested key has a equality-like boolean flag
So for a given obj1
like
obj1 =
prop1: 1,
prop2: "foo",
prop3:
prop4: 2,
prop5: "bar"
and the obj2
like
obj2 =
prop1: 3,
prop2: "foo",
prop3:
prop4: 2,
prop5: "foobar"
,
prop6: "new"
it should return
equality =
prop1: false,
prop2: true,
prop3 :
prop4: true,
prop5: false
,
prop6: false
If an object has a new property, like obj2.prop6
, then the equality will be equality.prop6 = false
.
For non-nested object a simple keys comparison solutions is here Get the property of the difference between two objects in javascript
While to recursively compare nested objects it is showed here JavaScript: Deep comparison recursively: Objects and properties
javascript
add a comment |
I have two nested objects obj1
and obj2
and I want to compare them and the recursively return an object that for each nested key has a equality-like boolean flag
So for a given obj1
like
obj1 =
prop1: 1,
prop2: "foo",
prop3:
prop4: 2,
prop5: "bar"
and the obj2
like
obj2 =
prop1: 3,
prop2: "foo",
prop3:
prop4: 2,
prop5: "foobar"
,
prop6: "new"
it should return
equality =
prop1: false,
prop2: true,
prop3 :
prop4: true,
prop5: false
,
prop6: false
If an object has a new property, like obj2.prop6
, then the equality will be equality.prop6 = false
.
For non-nested object a simple keys comparison solutions is here Get the property of the difference between two objects in javascript
While to recursively compare nested objects it is showed here JavaScript: Deep comparison recursively: Objects and properties
javascript
1
Will both objects always have exact match properties?
– holydragon
2 hours ago
good point. Nope, so the equality could have a new key set tofalse
. Updating with this point. Thank you.
– loretoparisi
2 hours ago
add a comment |
I have two nested objects obj1
and obj2
and I want to compare them and the recursively return an object that for each nested key has a equality-like boolean flag
So for a given obj1
like
obj1 =
prop1: 1,
prop2: "foo",
prop3:
prop4: 2,
prop5: "bar"
and the obj2
like
obj2 =
prop1: 3,
prop2: "foo",
prop3:
prop4: 2,
prop5: "foobar"
,
prop6: "new"
it should return
equality =
prop1: false,
prop2: true,
prop3 :
prop4: true,
prop5: false
,
prop6: false
If an object has a new property, like obj2.prop6
, then the equality will be equality.prop6 = false
.
For non-nested object a simple keys comparison solutions is here Get the property of the difference between two objects in javascript
While to recursively compare nested objects it is showed here JavaScript: Deep comparison recursively: Objects and properties
javascript
I have two nested objects obj1
and obj2
and I want to compare them and the recursively return an object that for each nested key has a equality-like boolean flag
So for a given obj1
like
obj1 =
prop1: 1,
prop2: "foo",
prop3:
prop4: 2,
prop5: "bar"
and the obj2
like
obj2 =
prop1: 3,
prop2: "foo",
prop3:
prop4: 2,
prop5: "foobar"
,
prop6: "new"
it should return
equality =
prop1: false,
prop2: true,
prop3 :
prop4: true,
prop5: false
,
prop6: false
If an object has a new property, like obj2.prop6
, then the equality will be equality.prop6 = false
.
For non-nested object a simple keys comparison solutions is here Get the property of the difference between two objects in javascript
While to recursively compare nested objects it is showed here JavaScript: Deep comparison recursively: Objects and properties
javascript
javascript
edited 2 hours ago
loretoparisi
asked 3 hours ago
loretoparisiloretoparisi
8,06654973
8,06654973
1
Will both objects always have exact match properties?
– holydragon
2 hours ago
good point. Nope, so the equality could have a new key set tofalse
. Updating with this point. Thank you.
– loretoparisi
2 hours ago
add a comment |
1
Will both objects always have exact match properties?
– holydragon
2 hours ago
good point. Nope, so the equality could have a new key set tofalse
. Updating with this point. Thank you.
– loretoparisi
2 hours ago
1
1
Will both objects always have exact match properties?
– holydragon
2 hours ago
Will both objects always have exact match properties?
– holydragon
2 hours ago
good point. Nope, so the equality could have a new key set to
false
. Updating with this point. Thank you.– loretoparisi
2 hours ago
good point. Nope, so the equality could have a new key set to
false
. Updating with this point. Thank you.– loretoparisi
2 hours ago
add a comment |
4 Answers
4
active
oldest
votes
You could use reduce
to build new object and another get
method to get nested props from other object by string
and compare it to current prop value in first object.
const obj1 = prop1: 1, prop2: "foo", prop3: prop4: 2, prop5: "bar"
const obj2 = prop1: 3, prop2: "foo", prop3: prop4: 2, prop5: "foobar"
function get(obj, path)
return path.split('.').reduce((r, e) => , obj)
function compare(a, b, prev = "")
return Object.keys(a).reduce((r, e) =>
const path = prev + (prev ? '.' + e : e);
const value = a[e] === get(b, path);
r[e] = typeof a[e] === 'object' ? compare(a[e], b, path) : value
return r;
, )
const result = compare(obj1, obj2);
console.log(result)
To compare all properties of both objects you could create extra function that will perform loop by both objects.
const obj1 = "prop1":1,"prop2":"foo","prop3":"prop4":2,"prop5":"bar","prop7":"prop9":"prop10":"foo"
const obj2 = "prop1":3,"prop2":"foo","prop3":"prop4":2,"prop5":"foobar","prop6":"new","prop7":"foo":"foo","bar":"baz":"baz"
function get(obj, path)
return path.split('.').reduce((r, e) => undefined;
, obj);
function isEmpty(o)
if (typeof o !== 'object') return true;
else return !Object.keys(o).length;
function build(a, b, o = null, prev = '')
return Object.keys(a).reduce(
(r, e) =>
const path = prev + (prev ? '.' + e : e);
const bObj = get(b, path);
const value = a[e] === bObj;
if (typeof a[e] === 'object')
if (isEmpty(a[e]) && isEmpty(bObj))
if (e in r) r[e] = r[e];
else r[e] = true;
else if (!bObj && isEmpty(a[e]))
r[e] = value;
else
r[e] = build(a[e], b, r[e], path);
else
r[e] = value;
return r;
,
o ? o :
);
function compare(a, b)
const o = build(a, b);
return build(b, a, o);
const result = compare(obj1, obj2);
console.log(result)
1
I thinks this is the best solution since it supports all recent version of ECMAScript.
– loretoparisi
1 hour ago
Just a thing, supposed that the property is a void object like, it does not make the comparison.
– loretoparisi
1 hour ago
1
In that case you could use this jsfiddle.net/vqwn3zLf
– Nenad Vracar
1 hour ago
Ok, now it handles a nested lonelyprop11=
object, but if you have like the sameprop12=
in both, you will get theas result for the
prop12
keyword, instead ofbool
. See here jsfiddle.net/gpu20nwy
– loretoparisi
1 hour ago
1
@loretoparisi Try this jsfiddle.net/r0y8nd3q/1
– Nenad Vracar
39 mins ago
|
show 2 more comments
You could iterate all keys and check the nested objects if both values are objects.
const isObject = v => v && typeof v === 'object';
function getDifference(a, b)
return Object.assign(...Array.from(
new Set([...Object.keys(a), ...Object.keys(b)]),
k => ( [k]: isObject(a[k]) && isObject(b[k])
? getDifference(a[k], b[k])
: a[k] === b[k]
)
));
var obj1 = prop1: 1, prop2: "foo", prop3: prop4: 2, prop5: "bar" ,
obj2 = prop1: 3, prop2: "foo", prop3: prop4: 2, prop5: "foobar" , prop6: "new" ;
console.log(getDifference(obj1, obj2));
Thanks! This solution works in every condition even if the property is a void object likeprop7 =
. Maybe avoiding the ECMA6Spread
notation for more compatibility could help.
– loretoparisi
1 hour ago
1
without spread notSet
.
– Nina Scholz
1 hour ago
maybe likeObject.assign(Array.from(new Set( [].concat(Object.keys(a)).concat(Object.keys(b)))
– loretoparisi
1 hour ago
w h a t e v e r ...
– Nina Scholz
1 hour ago
add a comment |
Loop through each key and compare the properties. If the property is an object, recursively compare the properties. This will work for any level of nesting. Since the properties could be missing from either of the objects value ||
check is added.
const obj1=prop1:1,prop2:"foo",prop3:prop4:2,prop5:"bar",prop7:pro8:"only in 1",
obj2=prop1:3,prop2:"foo",prop3:prop4:2,prop5:"foobar", prop6: "only in 2";
const isObject = val => val && typeof val === 'object'; // required for "null" comparison
function compare(obj1, obj2)
let equality = ,
merged = ...obj1, ...obj2 ; // has properties of both
for (let key in merged)
let value1 = obj1[key], value2 = obj2[key];
if (isObject(value1)
return equality;
console.log(compare(obj1, obj2))
Thanks, just updated the code, assuming you can have new properties as well.
– loretoparisi
2 hours ago
what if obj1 have some key with a child obj as its value and obj2 does not have the same key, then the method will be called with compare(obj1, undefined) which will throw an error at obj2[key]
– AZ_
2 hours ago
@loretoparisi updated
– adiga
2 hours ago
@adiga not sure, but the comments in the question do mention that, and the updated question as well, it cant be assumed that extra ket always have a string value.
– AZ_
2 hours ago
@AZ_ updated. Not sure if it will fail for any scenario
– adiga
2 hours ago
add a comment |
A recursive example,
var obj1 =
prop1: 1,
prop2: "foo",
prop3:
prop4: 2,
prop5: "bar"
,
prop7: ,
var obj2 =
prop1: 3,
prop2: "foo",
prop3:
prop4: 2,
prop5: "foobar"
,
prop6: "new",
prop7: ,
prop8: ,
var result = ;
function compare(obj1, obj2, obj_)
for (let k in obj1)
var type = typeof obj1[k];
if (type === 'object')
obj_[k] = ;
if (!obj2[k])
obj_[k] = false;
else if ((Object.entries(obj1[k]).length === 0 && obj1[k].constructor === Object) && (Object.entries(obj2[k]).length === 0 && obj2[k].constructor === Object))
obj_[k] = true;
else
compare(obj1[k], obj2[k], obj_[k]);
else
obj_[k] = (obj1[k] === obj2[k]);
if (Object.keys(obj1).length < Object.keys(obj2).length) //check if both objects vary in length.
var tmp = obj1;
obj1 = obj2;
obj2 = tmp;
compare(obj1, obj2, result);
console.log(result);
Thank you. If you try this case - jsfiddle.net/47fcqtom you will get ain the results for properties of type
object
that belong to both objects and that are.
– loretoparisi
57 mins ago
1
@loretoparisi I made a change, try now.
– Shoyeb Sheikh
18 mins ago
add a comment |
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4 Answers
4
active
oldest
votes
4 Answers
4
active
oldest
votes
active
oldest
votes
active
oldest
votes
You could use reduce
to build new object and another get
method to get nested props from other object by string
and compare it to current prop value in first object.
const obj1 = prop1: 1, prop2: "foo", prop3: prop4: 2, prop5: "bar"
const obj2 = prop1: 3, prop2: "foo", prop3: prop4: 2, prop5: "foobar"
function get(obj, path)
return path.split('.').reduce((r, e) => , obj)
function compare(a, b, prev = "")
return Object.keys(a).reduce((r, e) =>
const path = prev + (prev ? '.' + e : e);
const value = a[e] === get(b, path);
r[e] = typeof a[e] === 'object' ? compare(a[e], b, path) : value
return r;
, )
const result = compare(obj1, obj2);
console.log(result)
To compare all properties of both objects you could create extra function that will perform loop by both objects.
const obj1 = "prop1":1,"prop2":"foo","prop3":"prop4":2,"prop5":"bar","prop7":"prop9":"prop10":"foo"
const obj2 = "prop1":3,"prop2":"foo","prop3":"prop4":2,"prop5":"foobar","prop6":"new","prop7":"foo":"foo","bar":"baz":"baz"
function get(obj, path)
return path.split('.').reduce((r, e) => undefined;
, obj);
function isEmpty(o)
if (typeof o !== 'object') return true;
else return !Object.keys(o).length;
function build(a, b, o = null, prev = '')
return Object.keys(a).reduce(
(r, e) =>
const path = prev + (prev ? '.' + e : e);
const bObj = get(b, path);
const value = a[e] === bObj;
if (typeof a[e] === 'object')
if (isEmpty(a[e]) && isEmpty(bObj))
if (e in r) r[e] = r[e];
else r[e] = true;
else if (!bObj && isEmpty(a[e]))
r[e] = value;
else
r[e] = build(a[e], b, r[e], path);
else
r[e] = value;
return r;
,
o ? o :
);
function compare(a, b)
const o = build(a, b);
return build(b, a, o);
const result = compare(obj1, obj2);
console.log(result)
1
I thinks this is the best solution since it supports all recent version of ECMAScript.
– loretoparisi
1 hour ago
Just a thing, supposed that the property is a void object like, it does not make the comparison.
– loretoparisi
1 hour ago
1
In that case you could use this jsfiddle.net/vqwn3zLf
– Nenad Vracar
1 hour ago
Ok, now it handles a nested lonelyprop11=
object, but if you have like the sameprop12=
in both, you will get theas result for the
prop12
keyword, instead ofbool
. See here jsfiddle.net/gpu20nwy
– loretoparisi
1 hour ago
1
@loretoparisi Try this jsfiddle.net/r0y8nd3q/1
– Nenad Vracar
39 mins ago
|
show 2 more comments
You could use reduce
to build new object and another get
method to get nested props from other object by string
and compare it to current prop value in first object.
const obj1 = prop1: 1, prop2: "foo", prop3: prop4: 2, prop5: "bar"
const obj2 = prop1: 3, prop2: "foo", prop3: prop4: 2, prop5: "foobar"
function get(obj, path)
return path.split('.').reduce((r, e) => , obj)
function compare(a, b, prev = "")
return Object.keys(a).reduce((r, e) =>
const path = prev + (prev ? '.' + e : e);
const value = a[e] === get(b, path);
r[e] = typeof a[e] === 'object' ? compare(a[e], b, path) : value
return r;
, )
const result = compare(obj1, obj2);
console.log(result)
To compare all properties of both objects you could create extra function that will perform loop by both objects.
const obj1 = "prop1":1,"prop2":"foo","prop3":"prop4":2,"prop5":"bar","prop7":"prop9":"prop10":"foo"
const obj2 = "prop1":3,"prop2":"foo","prop3":"prop4":2,"prop5":"foobar","prop6":"new","prop7":"foo":"foo","bar":"baz":"baz"
function get(obj, path)
return path.split('.').reduce((r, e) => undefined;
, obj);
function isEmpty(o)
if (typeof o !== 'object') return true;
else return !Object.keys(o).length;
function build(a, b, o = null, prev = '')
return Object.keys(a).reduce(
(r, e) =>
const path = prev + (prev ? '.' + e : e);
const bObj = get(b, path);
const value = a[e] === bObj;
if (typeof a[e] === 'object')
if (isEmpty(a[e]) && isEmpty(bObj))
if (e in r) r[e] = r[e];
else r[e] = true;
else if (!bObj && isEmpty(a[e]))
r[e] = value;
else
r[e] = build(a[e], b, r[e], path);
else
r[e] = value;
return r;
,
o ? o :
);
function compare(a, b)
const o = build(a, b);
return build(b, a, o);
const result = compare(obj1, obj2);
console.log(result)
1
I thinks this is the best solution since it supports all recent version of ECMAScript.
– loretoparisi
1 hour ago
Just a thing, supposed that the property is a void object like, it does not make the comparison.
– loretoparisi
1 hour ago
1
In that case you could use this jsfiddle.net/vqwn3zLf
– Nenad Vracar
1 hour ago
Ok, now it handles a nested lonelyprop11=
object, but if you have like the sameprop12=
in both, you will get theas result for the
prop12
keyword, instead ofbool
. See here jsfiddle.net/gpu20nwy
– loretoparisi
1 hour ago
1
@loretoparisi Try this jsfiddle.net/r0y8nd3q/1
– Nenad Vracar
39 mins ago
|
show 2 more comments
You could use reduce
to build new object and another get
method to get nested props from other object by string
and compare it to current prop value in first object.
const obj1 = prop1: 1, prop2: "foo", prop3: prop4: 2, prop5: "bar"
const obj2 = prop1: 3, prop2: "foo", prop3: prop4: 2, prop5: "foobar"
function get(obj, path)
return path.split('.').reduce((r, e) => , obj)
function compare(a, b, prev = "")
return Object.keys(a).reduce((r, e) =>
const path = prev + (prev ? '.' + e : e);
const value = a[e] === get(b, path);
r[e] = typeof a[e] === 'object' ? compare(a[e], b, path) : value
return r;
, )
const result = compare(obj1, obj2);
console.log(result)
To compare all properties of both objects you could create extra function that will perform loop by both objects.
const obj1 = "prop1":1,"prop2":"foo","prop3":"prop4":2,"prop5":"bar","prop7":"prop9":"prop10":"foo"
const obj2 = "prop1":3,"prop2":"foo","prop3":"prop4":2,"prop5":"foobar","prop6":"new","prop7":"foo":"foo","bar":"baz":"baz"
function get(obj, path)
return path.split('.').reduce((r, e) => undefined;
, obj);
function isEmpty(o)
if (typeof o !== 'object') return true;
else return !Object.keys(o).length;
function build(a, b, o = null, prev = '')
return Object.keys(a).reduce(
(r, e) =>
const path = prev + (prev ? '.' + e : e);
const bObj = get(b, path);
const value = a[e] === bObj;
if (typeof a[e] === 'object')
if (isEmpty(a[e]) && isEmpty(bObj))
if (e in r) r[e] = r[e];
else r[e] = true;
else if (!bObj && isEmpty(a[e]))
r[e] = value;
else
r[e] = build(a[e], b, r[e], path);
else
r[e] = value;
return r;
,
o ? o :
);
function compare(a, b)
const o = build(a, b);
return build(b, a, o);
const result = compare(obj1, obj2);
console.log(result)
You could use reduce
to build new object and another get
method to get nested props from other object by string
and compare it to current prop value in first object.
const obj1 = prop1: 1, prop2: "foo", prop3: prop4: 2, prop5: "bar"
const obj2 = prop1: 3, prop2: "foo", prop3: prop4: 2, prop5: "foobar"
function get(obj, path)
return path.split('.').reduce((r, e) => , obj)
function compare(a, b, prev = "")
return Object.keys(a).reduce((r, e) =>
const path = prev + (prev ? '.' + e : e);
const value = a[e] === get(b, path);
r[e] = typeof a[e] === 'object' ? compare(a[e], b, path) : value
return r;
, )
const result = compare(obj1, obj2);
console.log(result)
To compare all properties of both objects you could create extra function that will perform loop by both objects.
const obj1 = "prop1":1,"prop2":"foo","prop3":"prop4":2,"prop5":"bar","prop7":"prop9":"prop10":"foo"
const obj2 = "prop1":3,"prop2":"foo","prop3":"prop4":2,"prop5":"foobar","prop6":"new","prop7":"foo":"foo","bar":"baz":"baz"
function get(obj, path)
return path.split('.').reduce((r, e) => undefined;
, obj);
function isEmpty(o)
if (typeof o !== 'object') return true;
else return !Object.keys(o).length;
function build(a, b, o = null, prev = '')
return Object.keys(a).reduce(
(r, e) =>
const path = prev + (prev ? '.' + e : e);
const bObj = get(b, path);
const value = a[e] === bObj;
if (typeof a[e] === 'object')
if (isEmpty(a[e]) && isEmpty(bObj))
if (e in r) r[e] = r[e];
else r[e] = true;
else if (!bObj && isEmpty(a[e]))
r[e] = value;
else
r[e] = build(a[e], b, r[e], path);
else
r[e] = value;
return r;
,
o ? o :
);
function compare(a, b)
const o = build(a, b);
return build(b, a, o);
const result = compare(obj1, obj2);
console.log(result)
const obj1 = prop1: 1, prop2: "foo", prop3: prop4: 2, prop5: "bar"
const obj2 = prop1: 3, prop2: "foo", prop3: prop4: 2, prop5: "foobar"
function get(obj, path)
return path.split('.').reduce((r, e) => , obj)
function compare(a, b, prev = "")
return Object.keys(a).reduce((r, e) =>
const path = prev + (prev ? '.' + e : e);
const value = a[e] === get(b, path);
r[e] = typeof a[e] === 'object' ? compare(a[e], b, path) : value
return r;
, )
const result = compare(obj1, obj2);
console.log(result)
const obj1 = prop1: 1, prop2: "foo", prop3: prop4: 2, prop5: "bar"
const obj2 = prop1: 3, prop2: "foo", prop3: prop4: 2, prop5: "foobar"
function get(obj, path)
return path.split('.').reduce((r, e) => , obj)
function compare(a, b, prev = "")
return Object.keys(a).reduce((r, e) =>
const path = prev + (prev ? '.' + e : e);
const value = a[e] === get(b, path);
r[e] = typeof a[e] === 'object' ? compare(a[e], b, path) : value
return r;
, )
const result = compare(obj1, obj2);
console.log(result)
const obj1 = "prop1":1,"prop2":"foo","prop3":"prop4":2,"prop5":"bar","prop7":"prop9":"prop10":"foo"
const obj2 = "prop1":3,"prop2":"foo","prop3":"prop4":2,"prop5":"foobar","prop6":"new","prop7":"foo":"foo","bar":"baz":"baz"
function get(obj, path)
return path.split('.').reduce((r, e) => undefined;
, obj);
function isEmpty(o)
if (typeof o !== 'object') return true;
else return !Object.keys(o).length;
function build(a, b, o = null, prev = '')
return Object.keys(a).reduce(
(r, e) =>
const path = prev + (prev ? '.' + e : e);
const bObj = get(b, path);
const value = a[e] === bObj;
if (typeof a[e] === 'object')
if (isEmpty(a[e]) && isEmpty(bObj))
if (e in r) r[e] = r[e];
else r[e] = true;
else if (!bObj && isEmpty(a[e]))
r[e] = value;
else
r[e] = build(a[e], b, r[e], path);
else
r[e] = value;
return r;
,
o ? o :
);
function compare(a, b)
const o = build(a, b);
return build(b, a, o);
const result = compare(obj1, obj2);
console.log(result)
const obj1 = "prop1":1,"prop2":"foo","prop3":"prop4":2,"prop5":"bar","prop7":"prop9":"prop10":"foo"
const obj2 = "prop1":3,"prop2":"foo","prop3":"prop4":2,"prop5":"foobar","prop6":"new","prop7":"foo":"foo","bar":"baz":"baz"
function get(obj, path)
return path.split('.').reduce((r, e) => undefined;
, obj);
function isEmpty(o)
if (typeof o !== 'object') return true;
else return !Object.keys(o).length;
function build(a, b, o = null, prev = '')
return Object.keys(a).reduce(
(r, e) =>
const path = prev + (prev ? '.' + e : e);
const bObj = get(b, path);
const value = a[e] === bObj;
if (typeof a[e] === 'object')
if (isEmpty(a[e]) && isEmpty(bObj))
if (e in r) r[e] = r[e];
else r[e] = true;
else if (!bObj && isEmpty(a[e]))
r[e] = value;
else
r[e] = build(a[e], b, r[e], path);
else
r[e] = value;
return r;
,
o ? o :
);
function compare(a, b)
const o = build(a, b);
return build(b, a, o);
const result = compare(obj1, obj2);
console.log(result)
edited 25 mins ago
answered 2 hours ago
Nenad VracarNenad Vracar
73.3k126085
73.3k126085
1
I thinks this is the best solution since it supports all recent version of ECMAScript.
– loretoparisi
1 hour ago
Just a thing, supposed that the property is a void object like, it does not make the comparison.
– loretoparisi
1 hour ago
1
In that case you could use this jsfiddle.net/vqwn3zLf
– Nenad Vracar
1 hour ago
Ok, now it handles a nested lonelyprop11=
object, but if you have like the sameprop12=
in both, you will get theas result for the
prop12
keyword, instead ofbool
. See here jsfiddle.net/gpu20nwy
– loretoparisi
1 hour ago
1
@loretoparisi Try this jsfiddle.net/r0y8nd3q/1
– Nenad Vracar
39 mins ago
|
show 2 more comments
1
I thinks this is the best solution since it supports all recent version of ECMAScript.
– loretoparisi
1 hour ago
Just a thing, supposed that the property is a void object like, it does not make the comparison.
– loretoparisi
1 hour ago
1
In that case you could use this jsfiddle.net/vqwn3zLf
– Nenad Vracar
1 hour ago
Ok, now it handles a nested lonelyprop11=
object, but if you have like the sameprop12=
in both, you will get theas result for the
prop12
keyword, instead ofbool
. See here jsfiddle.net/gpu20nwy
– loretoparisi
1 hour ago
1
@loretoparisi Try this jsfiddle.net/r0y8nd3q/1
– Nenad Vracar
39 mins ago
1
1
I thinks this is the best solution since it supports all recent version of ECMAScript.
– loretoparisi
1 hour ago
I thinks this is the best solution since it supports all recent version of ECMAScript.
– loretoparisi
1 hour ago
Just a thing, supposed that the property is a void object like
, it does not make the comparison.– loretoparisi
1 hour ago
Just a thing, supposed that the property is a void object like
, it does not make the comparison.– loretoparisi
1 hour ago
1
1
In that case you could use this jsfiddle.net/vqwn3zLf
– Nenad Vracar
1 hour ago
In that case you could use this jsfiddle.net/vqwn3zLf
– Nenad Vracar
1 hour ago
Ok, now it handles a nested lonely
prop11=
object, but if you have like the same prop12=
in both, you will get the
as result for the prop12
keyword, instead of bool
. See here jsfiddle.net/gpu20nwy– loretoparisi
1 hour ago
Ok, now it handles a nested lonely
prop11=
object, but if you have like the same prop12=
in both, you will get the
as result for the prop12
keyword, instead of bool
. See here jsfiddle.net/gpu20nwy– loretoparisi
1 hour ago
1
1
@loretoparisi Try this jsfiddle.net/r0y8nd3q/1
– Nenad Vracar
39 mins ago
@loretoparisi Try this jsfiddle.net/r0y8nd3q/1
– Nenad Vracar
39 mins ago
|
show 2 more comments
You could iterate all keys and check the nested objects if both values are objects.
const isObject = v => v && typeof v === 'object';
function getDifference(a, b)
return Object.assign(...Array.from(
new Set([...Object.keys(a), ...Object.keys(b)]),
k => ( [k]: isObject(a[k]) && isObject(b[k])
? getDifference(a[k], b[k])
: a[k] === b[k]
)
));
var obj1 = prop1: 1, prop2: "foo", prop3: prop4: 2, prop5: "bar" ,
obj2 = prop1: 3, prop2: "foo", prop3: prop4: 2, prop5: "foobar" , prop6: "new" ;
console.log(getDifference(obj1, obj2));
Thanks! This solution works in every condition even if the property is a void object likeprop7 =
. Maybe avoiding the ECMA6Spread
notation for more compatibility could help.
– loretoparisi
1 hour ago
1
without spread notSet
.
– Nina Scholz
1 hour ago
maybe likeObject.assign(Array.from(new Set( [].concat(Object.keys(a)).concat(Object.keys(b)))
– loretoparisi
1 hour ago
w h a t e v e r ...
– Nina Scholz
1 hour ago
add a comment |
You could iterate all keys and check the nested objects if both values are objects.
const isObject = v => v && typeof v === 'object';
function getDifference(a, b)
return Object.assign(...Array.from(
new Set([...Object.keys(a), ...Object.keys(b)]),
k => ( [k]: isObject(a[k]) && isObject(b[k])
? getDifference(a[k], b[k])
: a[k] === b[k]
)
));
var obj1 = prop1: 1, prop2: "foo", prop3: prop4: 2, prop5: "bar" ,
obj2 = prop1: 3, prop2: "foo", prop3: prop4: 2, prop5: "foobar" , prop6: "new" ;
console.log(getDifference(obj1, obj2));
Thanks! This solution works in every condition even if the property is a void object likeprop7 =
. Maybe avoiding the ECMA6Spread
notation for more compatibility could help.
– loretoparisi
1 hour ago
1
without spread notSet
.
– Nina Scholz
1 hour ago
maybe likeObject.assign(Array.from(new Set( [].concat(Object.keys(a)).concat(Object.keys(b)))
– loretoparisi
1 hour ago
w h a t e v e r ...
– Nina Scholz
1 hour ago
add a comment |
You could iterate all keys and check the nested objects if both values are objects.
const isObject = v => v && typeof v === 'object';
function getDifference(a, b)
return Object.assign(...Array.from(
new Set([...Object.keys(a), ...Object.keys(b)]),
k => ( [k]: isObject(a[k]) && isObject(b[k])
? getDifference(a[k], b[k])
: a[k] === b[k]
)
));
var obj1 = prop1: 1, prop2: "foo", prop3: prop4: 2, prop5: "bar" ,
obj2 = prop1: 3, prop2: "foo", prop3: prop4: 2, prop5: "foobar" , prop6: "new" ;
console.log(getDifference(obj1, obj2));
You could iterate all keys and check the nested objects if both values are objects.
const isObject = v => v && typeof v === 'object';
function getDifference(a, b)
return Object.assign(...Array.from(
new Set([...Object.keys(a), ...Object.keys(b)]),
k => ( [k]: isObject(a[k]) && isObject(b[k])
? getDifference(a[k], b[k])
: a[k] === b[k]
)
));
var obj1 = prop1: 1, prop2: "foo", prop3: prop4: 2, prop5: "bar" ,
obj2 = prop1: 3, prop2: "foo", prop3: prop4: 2, prop5: "foobar" , prop6: "new" ;
console.log(getDifference(obj1, obj2));
const isObject = v => v && typeof v === 'object';
function getDifference(a, b)
return Object.assign(...Array.from(
new Set([...Object.keys(a), ...Object.keys(b)]),
k => ( [k]: isObject(a[k]) && isObject(b[k])
? getDifference(a[k], b[k])
: a[k] === b[k]
)
));
var obj1 = prop1: 1, prop2: "foo", prop3: prop4: 2, prop5: "bar" ,
obj2 = prop1: 3, prop2: "foo", prop3: prop4: 2, prop5: "foobar" , prop6: "new" ;
console.log(getDifference(obj1, obj2));
const isObject = v => v && typeof v === 'object';
function getDifference(a, b)
return Object.assign(...Array.from(
new Set([...Object.keys(a), ...Object.keys(b)]),
k => ( [k]: isObject(a[k]) && isObject(b[k])
? getDifference(a[k], b[k])
: a[k] === b[k]
)
));
var obj1 = prop1: 1, prop2: "foo", prop3: prop4: 2, prop5: "bar" ,
obj2 = prop1: 3, prop2: "foo", prop3: prop4: 2, prop5: "foobar" , prop6: "new" ;
console.log(getDifference(obj1, obj2));
edited 2 hours ago
answered 2 hours ago
Nina ScholzNina Scholz
197k15109179
197k15109179
Thanks! This solution works in every condition even if the property is a void object likeprop7 =
. Maybe avoiding the ECMA6Spread
notation for more compatibility could help.
– loretoparisi
1 hour ago
1
without spread notSet
.
– Nina Scholz
1 hour ago
maybe likeObject.assign(Array.from(new Set( [].concat(Object.keys(a)).concat(Object.keys(b)))
– loretoparisi
1 hour ago
w h a t e v e r ...
– Nina Scholz
1 hour ago
add a comment |
Thanks! This solution works in every condition even if the property is a void object likeprop7 =
. Maybe avoiding the ECMA6Spread
notation for more compatibility could help.
– loretoparisi
1 hour ago
1
without spread notSet
.
– Nina Scholz
1 hour ago
maybe likeObject.assign(Array.from(new Set( [].concat(Object.keys(a)).concat(Object.keys(b)))
– loretoparisi
1 hour ago
w h a t e v e r ...
– Nina Scholz
1 hour ago
Thanks! This solution works in every condition even if the property is a void object like
prop7 =
. Maybe avoiding the ECMA6 Spread
notation for more compatibility could help.– loretoparisi
1 hour ago
Thanks! This solution works in every condition even if the property is a void object like
prop7 =
. Maybe avoiding the ECMA6 Spread
notation for more compatibility could help.– loretoparisi
1 hour ago
1
1
without spread not
Set
.– Nina Scholz
1 hour ago
without spread not
Set
.– Nina Scholz
1 hour ago
maybe like
Object.assign(Array.from(new Set( [].concat(Object.keys(a)).concat(Object.keys(b)))
– loretoparisi
1 hour ago
maybe like
Object.assign(Array.from(new Set( [].concat(Object.keys(a)).concat(Object.keys(b)))
– loretoparisi
1 hour ago
w h a t e v e r ...
– Nina Scholz
1 hour ago
w h a t e v e r ...
– Nina Scholz
1 hour ago
add a comment |
Loop through each key and compare the properties. If the property is an object, recursively compare the properties. This will work for any level of nesting. Since the properties could be missing from either of the objects value ||
check is added.
const obj1=prop1:1,prop2:"foo",prop3:prop4:2,prop5:"bar",prop7:pro8:"only in 1",
obj2=prop1:3,prop2:"foo",prop3:prop4:2,prop5:"foobar", prop6: "only in 2";
const isObject = val => val && typeof val === 'object'; // required for "null" comparison
function compare(obj1, obj2)
let equality = ,
merged = ...obj1, ...obj2 ; // has properties of both
for (let key in merged)
let value1 = obj1[key], value2 = obj2[key];
if (isObject(value1)
return equality;
console.log(compare(obj1, obj2))
Thanks, just updated the code, assuming you can have new properties as well.
– loretoparisi
2 hours ago
what if obj1 have some key with a child obj as its value and obj2 does not have the same key, then the method will be called with compare(obj1, undefined) which will throw an error at obj2[key]
– AZ_
2 hours ago
@loretoparisi updated
– adiga
2 hours ago
@adiga not sure, but the comments in the question do mention that, and the updated question as well, it cant be assumed that extra ket always have a string value.
– AZ_
2 hours ago
@AZ_ updated. Not sure if it will fail for any scenario
– adiga
2 hours ago
add a comment |
Loop through each key and compare the properties. If the property is an object, recursively compare the properties. This will work for any level of nesting. Since the properties could be missing from either of the objects value ||
check is added.
const obj1=prop1:1,prop2:"foo",prop3:prop4:2,prop5:"bar",prop7:pro8:"only in 1",
obj2=prop1:3,prop2:"foo",prop3:prop4:2,prop5:"foobar", prop6: "only in 2";
const isObject = val => val && typeof val === 'object'; // required for "null" comparison
function compare(obj1, obj2)
let equality = ,
merged = ...obj1, ...obj2 ; // has properties of both
for (let key in merged)
let value1 = obj1[key], value2 = obj2[key];
if (isObject(value1)
return equality;
console.log(compare(obj1, obj2))
Thanks, just updated the code, assuming you can have new properties as well.
– loretoparisi
2 hours ago
what if obj1 have some key with a child obj as its value and obj2 does not have the same key, then the method will be called with compare(obj1, undefined) which will throw an error at obj2[key]
– AZ_
2 hours ago
@loretoparisi updated
– adiga
2 hours ago
@adiga not sure, but the comments in the question do mention that, and the updated question as well, it cant be assumed that extra ket always have a string value.
– AZ_
2 hours ago
@AZ_ updated. Not sure if it will fail for any scenario
– adiga
2 hours ago
add a comment |
Loop through each key and compare the properties. If the property is an object, recursively compare the properties. This will work for any level of nesting. Since the properties could be missing from either of the objects value ||
check is added.
const obj1=prop1:1,prop2:"foo",prop3:prop4:2,prop5:"bar",prop7:pro8:"only in 1",
obj2=prop1:3,prop2:"foo",prop3:prop4:2,prop5:"foobar", prop6: "only in 2";
const isObject = val => val && typeof val === 'object'; // required for "null" comparison
function compare(obj1, obj2)
let equality = ,
merged = ...obj1, ...obj2 ; // has properties of both
for (let key in merged)
let value1 = obj1[key], value2 = obj2[key];
if (isObject(value1)
return equality;
console.log(compare(obj1, obj2))
Loop through each key and compare the properties. If the property is an object, recursively compare the properties. This will work for any level of nesting. Since the properties could be missing from either of the objects value ||
check is added.
const obj1=prop1:1,prop2:"foo",prop3:prop4:2,prop5:"bar",prop7:pro8:"only in 1",
obj2=prop1:3,prop2:"foo",prop3:prop4:2,prop5:"foobar", prop6: "only in 2";
const isObject = val => val && typeof val === 'object'; // required for "null" comparison
function compare(obj1, obj2)
let equality = ,
merged = ...obj1, ...obj2 ; // has properties of both
for (let key in merged)
let value1 = obj1[key], value2 = obj2[key];
if (isObject(value1)
return equality;
console.log(compare(obj1, obj2))
const obj1=prop1:1,prop2:"foo",prop3:prop4:2,prop5:"bar",prop7:pro8:"only in 1",
obj2=prop1:3,prop2:"foo",prop3:prop4:2,prop5:"foobar", prop6: "only in 2";
const isObject = val => val && typeof val === 'object'; // required for "null" comparison
function compare(obj1, obj2)
let equality = ,
merged = ...obj1, ...obj2 ; // has properties of both
for (let key in merged)
let value1 = obj1[key], value2 = obj2[key];
if (isObject(value1)
return equality;
console.log(compare(obj1, obj2))
const obj1=prop1:1,prop2:"foo",prop3:prop4:2,prop5:"bar",prop7:pro8:"only in 1",
obj2=prop1:3,prop2:"foo",prop3:prop4:2,prop5:"foobar", prop6: "only in 2";
const isObject = val => val && typeof val === 'object'; // required for "null" comparison
function compare(obj1, obj2)
let equality = ,
merged = ...obj1, ...obj2 ; // has properties of both
for (let key in merged)
let value1 = obj1[key], value2 = obj2[key];
if (isObject(value1)
return equality;
console.log(compare(obj1, obj2))
edited 1 hour ago
answered 2 hours ago
adigaadiga
12.3k62645
12.3k62645
Thanks, just updated the code, assuming you can have new properties as well.
– loretoparisi
2 hours ago
what if obj1 have some key with a child obj as its value and obj2 does not have the same key, then the method will be called with compare(obj1, undefined) which will throw an error at obj2[key]
– AZ_
2 hours ago
@loretoparisi updated
– adiga
2 hours ago
@adiga not sure, but the comments in the question do mention that, and the updated question as well, it cant be assumed that extra ket always have a string value.
– AZ_
2 hours ago
@AZ_ updated. Not sure if it will fail for any scenario
– adiga
2 hours ago
add a comment |
Thanks, just updated the code, assuming you can have new properties as well.
– loretoparisi
2 hours ago
what if obj1 have some key with a child obj as its value and obj2 does not have the same key, then the method will be called with compare(obj1, undefined) which will throw an error at obj2[key]
– AZ_
2 hours ago
@loretoparisi updated
– adiga
2 hours ago
@adiga not sure, but the comments in the question do mention that, and the updated question as well, it cant be assumed that extra ket always have a string value.
– AZ_
2 hours ago
@AZ_ updated. Not sure if it will fail for any scenario
– adiga
2 hours ago
Thanks, just updated the code, assuming you can have new properties as well.
– loretoparisi
2 hours ago
Thanks, just updated the code, assuming you can have new properties as well.
– loretoparisi
2 hours ago
what if obj1 have some key with a child obj as its value and obj2 does not have the same key, then the method will be called with compare(obj1, undefined) which will throw an error at obj2[key]
– AZ_
2 hours ago
what if obj1 have some key with a child obj as its value and obj2 does not have the same key, then the method will be called with compare(obj1, undefined) which will throw an error at obj2[key]
– AZ_
2 hours ago
@loretoparisi updated
– adiga
2 hours ago
@loretoparisi updated
– adiga
2 hours ago
@adiga not sure, but the comments in the question do mention that, and the updated question as well, it cant be assumed that extra ket always have a string value.
– AZ_
2 hours ago
@adiga not sure, but the comments in the question do mention that, and the updated question as well, it cant be assumed that extra ket always have a string value.
– AZ_
2 hours ago
@AZ_ updated. Not sure if it will fail for any scenario
– adiga
2 hours ago
@AZ_ updated. Not sure if it will fail for any scenario
– adiga
2 hours ago
add a comment |
A recursive example,
var obj1 =
prop1: 1,
prop2: "foo",
prop3:
prop4: 2,
prop5: "bar"
,
prop7: ,
var obj2 =
prop1: 3,
prop2: "foo",
prop3:
prop4: 2,
prop5: "foobar"
,
prop6: "new",
prop7: ,
prop8: ,
var result = ;
function compare(obj1, obj2, obj_)
for (let k in obj1)
var type = typeof obj1[k];
if (type === 'object')
obj_[k] = ;
if (!obj2[k])
obj_[k] = false;
else if ((Object.entries(obj1[k]).length === 0 && obj1[k].constructor === Object) && (Object.entries(obj2[k]).length === 0 && obj2[k].constructor === Object))
obj_[k] = true;
else
compare(obj1[k], obj2[k], obj_[k]);
else
obj_[k] = (obj1[k] === obj2[k]);
if (Object.keys(obj1).length < Object.keys(obj2).length) //check if both objects vary in length.
var tmp = obj1;
obj1 = obj2;
obj2 = tmp;
compare(obj1, obj2, result);
console.log(result);
Thank you. If you try this case - jsfiddle.net/47fcqtom you will get ain the results for properties of type
object
that belong to both objects and that are.
– loretoparisi
57 mins ago
1
@loretoparisi I made a change, try now.
– Shoyeb Sheikh
18 mins ago
add a comment |
A recursive example,
var obj1 =
prop1: 1,
prop2: "foo",
prop3:
prop4: 2,
prop5: "bar"
,
prop7: ,
var obj2 =
prop1: 3,
prop2: "foo",
prop3:
prop4: 2,
prop5: "foobar"
,
prop6: "new",
prop7: ,
prop8: ,
var result = ;
function compare(obj1, obj2, obj_)
for (let k in obj1)
var type = typeof obj1[k];
if (type === 'object')
obj_[k] = ;
if (!obj2[k])
obj_[k] = false;
else if ((Object.entries(obj1[k]).length === 0 && obj1[k].constructor === Object) && (Object.entries(obj2[k]).length === 0 && obj2[k].constructor === Object))
obj_[k] = true;
else
compare(obj1[k], obj2[k], obj_[k]);
else
obj_[k] = (obj1[k] === obj2[k]);
if (Object.keys(obj1).length < Object.keys(obj2).length) //check if both objects vary in length.
var tmp = obj1;
obj1 = obj2;
obj2 = tmp;
compare(obj1, obj2, result);
console.log(result);
Thank you. If you try this case - jsfiddle.net/47fcqtom you will get ain the results for properties of type
object
that belong to both objects and that are.
– loretoparisi
57 mins ago
1
@loretoparisi I made a change, try now.
– Shoyeb Sheikh
18 mins ago
add a comment |
A recursive example,
var obj1 =
prop1: 1,
prop2: "foo",
prop3:
prop4: 2,
prop5: "bar"
,
prop7: ,
var obj2 =
prop1: 3,
prop2: "foo",
prop3:
prop4: 2,
prop5: "foobar"
,
prop6: "new",
prop7: ,
prop8: ,
var result = ;
function compare(obj1, obj2, obj_)
for (let k in obj1)
var type = typeof obj1[k];
if (type === 'object')
obj_[k] = ;
if (!obj2[k])
obj_[k] = false;
else if ((Object.entries(obj1[k]).length === 0 && obj1[k].constructor === Object) && (Object.entries(obj2[k]).length === 0 && obj2[k].constructor === Object))
obj_[k] = true;
else
compare(obj1[k], obj2[k], obj_[k]);
else
obj_[k] = (obj1[k] === obj2[k]);
if (Object.keys(obj1).length < Object.keys(obj2).length) //check if both objects vary in length.
var tmp = obj1;
obj1 = obj2;
obj2 = tmp;
compare(obj1, obj2, result);
console.log(result);
A recursive example,
var obj1 =
prop1: 1,
prop2: "foo",
prop3:
prop4: 2,
prop5: "bar"
,
prop7: ,
var obj2 =
prop1: 3,
prop2: "foo",
prop3:
prop4: 2,
prop5: "foobar"
,
prop6: "new",
prop7: ,
prop8: ,
var result = ;
function compare(obj1, obj2, obj_)
for (let k in obj1)
var type = typeof obj1[k];
if (type === 'object')
obj_[k] = ;
if (!obj2[k])
obj_[k] = false;
else if ((Object.entries(obj1[k]).length === 0 && obj1[k].constructor === Object) && (Object.entries(obj2[k]).length === 0 && obj2[k].constructor === Object))
obj_[k] = true;
else
compare(obj1[k], obj2[k], obj_[k]);
else
obj_[k] = (obj1[k] === obj2[k]);
if (Object.keys(obj1).length < Object.keys(obj2).length) //check if both objects vary in length.
var tmp = obj1;
obj1 = obj2;
obj2 = tmp;
compare(obj1, obj2, result);
console.log(result);
var obj1 =
prop1: 1,
prop2: "foo",
prop3:
prop4: 2,
prop5: "bar"
,
prop7: ,
var obj2 =
prop1: 3,
prop2: "foo",
prop3:
prop4: 2,
prop5: "foobar"
,
prop6: "new",
prop7: ,
prop8: ,
var result = ;
function compare(obj1, obj2, obj_)
for (let k in obj1)
var type = typeof obj1[k];
if (type === 'object')
obj_[k] = ;
if (!obj2[k])
obj_[k] = false;
else if ((Object.entries(obj1[k]).length === 0 && obj1[k].constructor === Object) && (Object.entries(obj2[k]).length === 0 && obj2[k].constructor === Object))
obj_[k] = true;
else
compare(obj1[k], obj2[k], obj_[k]);
else
obj_[k] = (obj1[k] === obj2[k]);
if (Object.keys(obj1).length < Object.keys(obj2).length) //check if both objects vary in length.
var tmp = obj1;
obj1 = obj2;
obj2 = tmp;
compare(obj1, obj2, result);
console.log(result);
var obj1 =
prop1: 1,
prop2: "foo",
prop3:
prop4: 2,
prop5: "bar"
,
prop7: ,
var obj2 =
prop1: 3,
prop2: "foo",
prop3:
prop4: 2,
prop5: "foobar"
,
prop6: "new",
prop7: ,
prop8: ,
var result = ;
function compare(obj1, obj2, obj_)
for (let k in obj1)
var type = typeof obj1[k];
if (type === 'object')
obj_[k] = ;
if (!obj2[k])
obj_[k] = false;
else if ((Object.entries(obj1[k]).length === 0 && obj1[k].constructor === Object) && (Object.entries(obj2[k]).length === 0 && obj2[k].constructor === Object))
obj_[k] = true;
else
compare(obj1[k], obj2[k], obj_[k]);
else
obj_[k] = (obj1[k] === obj2[k]);
if (Object.keys(obj1).length < Object.keys(obj2).length) //check if both objects vary in length.
var tmp = obj1;
obj1 = obj2;
obj2 = tmp;
compare(obj1, obj2, result);
console.log(result);
edited 19 mins ago
answered 1 hour ago
Shoyeb SheikhShoyeb Sheikh
652211
652211
Thank you. If you try this case - jsfiddle.net/47fcqtom you will get ain the results for properties of type
object
that belong to both objects and that are.
– loretoparisi
57 mins ago
1
@loretoparisi I made a change, try now.
– Shoyeb Sheikh
18 mins ago
add a comment |
Thank you. If you try this case - jsfiddle.net/47fcqtom you will get ain the results for properties of type
object
that belong to both objects and that are.
– loretoparisi
57 mins ago
1
@loretoparisi I made a change, try now.
– Shoyeb Sheikh
18 mins ago
Thank you. If you try this case - jsfiddle.net/47fcqtom you will get a
in the results for properties of type object
that belong to both objects and that are
.– loretoparisi
57 mins ago
Thank you. If you try this case - jsfiddle.net/47fcqtom you will get a
in the results for properties of type object
that belong to both objects and that are
.– loretoparisi
57 mins ago
1
1
@loretoparisi I made a change, try now.
– Shoyeb Sheikh
18 mins ago
@loretoparisi I made a change, try now.
– Shoyeb Sheikh
18 mins ago
add a comment |
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1
Will both objects always have exact match properties?
– holydragon
2 hours ago
good point. Nope, so the equality could have a new key set to
false
. Updating with this point. Thank you.– loretoparisi
2 hours ago