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Argument list too long when zipping large list of certain files in a folder
2019 Community Moderator ElectionSolving “mv: Argument list too long”?How to print a range of IP addresses with Linux seq commandbash: /usr/bin/perl: Argument list too long/usr/bin/awk: Argument list too longArgument list too long with just 5000 filesAWK Using a Variable in an Equality Expressionbash array with variable in the nameReplace a long string with the sed command: Argument list too long errorAdd text to each value while looping thru and printing them in a array?Moving random files using shuf and mv - Argument list too long
I'm trying to zip a certain large number of files by creating an array of them. Then I tried to run the zip command but met with "Argument list too long" error.
declare -a arr=()
fixed=5
for i in `seq 10 1 200`; do
for j in `seq $((i+fixed)) 1 200`; do
arr+=("$i_$j.xxx")
done
done
new_arr=$(printf ",%s" "$arr[@]")
new_arr=$new_arr:1
zip all_data.zip $new_arr
bash
asked 49 mins ago
ZackZack
63
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Zack is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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I'm trying to zip a certain large number of files by creating an array of them. Then I tried to run the zip command but met with "Argument list too long" error.
declare -a arr=()
fixed=5
for i in `seq 10 1 200`; do
for j in `seq $((i+fixed)) 1 200`; do
arr+=("$i_$j.xxx")
done
done
new_arr=$(printf ",%s" "$arr[@]")
new_arr=$new_arr:1
zip all_data.zip $new_arr
bash
asked 49 mins ago
ZackZack
63
New contributor
Zack is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
add a comment |
I'm trying to zip a certain large number of files by creating an array of them. Then I tried to run the zip command but met with "Argument list too long" error.
declare -a arr=()
fixed=5
for i in `seq 10 1 200`; do
for j in `seq $((i+fixed)) 1 200`; do
arr+=("$i_$j.xxx")
done
done
new_arr=$(printf ",%s" "$arr[@]")
new_arr=$new_arr:1
zip all_data.zip $new_arr
bash
asked 49 mins ago
ZackZack
63
New contributor
Zack is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
I'm trying to zip a certain large number of files by creating an array of them. Then I tried to run the zip command but met with "Argument list too long" error.
declare -a arr=()
fixed=5
for i in `seq 10 1 200`; do
for j in `seq $((i+fixed)) 1 200`; do
arr+=("$i_$j.xxx")
done
done
new_arr=$(printf ",%s" "$arr[@]")
new_arr=$new_arr:1
zip all_data.zip $new_arr
bash
bash
asked 49 mins ago
ZackZack
63
New contributor
Zack is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 49 mins ago
ZackZack
63
New contributor
Zack is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited 45 mins ago
Zack
asked 49 mins ago
ZackZack
63
New contributor
Zack is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 49 mins ago
ZackZack
63
asked 49 mins ago
ZackZack
63
63
New contributor
Zack is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
Zack is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Zack is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
add a comment |
add a comment |
1 Answer
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votes
extract from man zip ( linux version )
zip -@ foo
will store the files listed one per line on stdin in foo.zip.
example from the same man page
find . -name "*.[ch]" -print | zip source -@
So steps will be :
build a list off all files to be archive , format must one file name by line
run
zipcommandcat BIG_FILENAME_LIST.txt | zip thebigziparchive -@
answered 28 mins ago
EchoMike444EchoMike444
9525
add a comment |
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1 Answer
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1 Answer
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active
oldest
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active
oldest
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active
oldest
votes
extract from man zip ( linux version )
zip -@ foo
will store the files listed one per line on stdin in foo.zip.
example from the same man page
find . -name "*.[ch]" -print | zip source -@
So steps will be :
build a list off all files to be archive , format must one file name by line
run
zipcommandcat BIG_FILENAME_LIST.txt | zip thebigziparchive -@
answered 28 mins ago
EchoMike444EchoMike444
9525
add a comment |
extract from man zip ( linux version )
zip -@ foo
will store the files listed one per line on stdin in foo.zip.
example from the same man page
find . -name "*.[ch]" -print | zip source -@
So steps will be :
build a list off all files to be archive , format must one file name by line
run
zipcommandcat BIG_FILENAME_LIST.txt | zip thebigziparchive -@
answered 28 mins ago
EchoMike444EchoMike444
9525
add a comment |
extract from man zip ( linux version )
zip -@ foo
will store the files listed one per line on stdin in foo.zip.
example from the same man page
find . -name "*.[ch]" -print | zip source -@
So steps will be :
build a list off all files to be archive , format must one file name by line
run
zipcommandcat BIG_FILENAME_LIST.txt | zip thebigziparchive -@
answered 28 mins ago
EchoMike444EchoMike444
9525
extract from man zip ( linux version )
zip -@ foo
will store the files listed one per line on stdin in foo.zip.
example from the same man page
find . -name "*.[ch]" -print | zip source -@
So steps will be :
build a list off all files to be archive , format must one file name by line
run
zipcommandcat BIG_FILENAME_LIST.txt | zip thebigziparchive -@
answered 28 mins ago
EchoMike444EchoMike444
9525
answered 28 mins ago
EchoMike444EchoMike444
9525
answered 28 mins ago
EchoMike444EchoMike444
9525
answered 28 mins ago
EchoMike444EchoMike444
9525
9525
add a comment |
add a comment |
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