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count occurances of text in file with bash
The Next CEO of Stack OverflowCount duplicated words in a text fileHelp with bash script with textuniq --count command is yields incorrect result?bash - wrong answer by word count commandSort file/folder output in BashHow long it will take to sort uniq a 62GB file?How to sort a file with bashHow to count number of process instances with bash?Wrong sorting a text fileBash incrementing bash string - from text file
2
I have a log file sorted by IP addresses,
I want to find the number of occurrences of each unique IP address.
how can I do this with bash? possibly listing the number of occurrences next to an ip, such as:
5.135.134.16 count: 5
13.57.220.172: count 30
18.206.226 count:2
....and so on.
heres a sample of the log:
5.135.134.16 - - [23/Mar/2019:08:42:54 -0400] "GET /wp-login.php HTTP/1.1" 200 2988 "-" "Mozilla/5.0 (X11; Ubuntu; Linux x86_64; rv:62.0) Gecko/20100101 Firefox/62.0"
5.135.134.16 - - [23/Mar/2019:08:42:55 -0400] "GET /wp-login.php HTTP/1.1" 200 2988 "-" "Mozilla/5.0 (X11; Ubuntu; Linux x86_64; rv:62.0) Gecko/20100101 Firefox/62.0"
5.135.134.16 - - [23/Mar/2019:08:42:55 -0400] "POST /wp-login.php HTTP/1.1" 200 3836 "-" "Mozilla/5.0 (X11; Ubuntu; Linux x86_64; rv:62.0) Gecko/20100101 Firefox/62.0"
5.135.134.16 - - [23/Mar/2019:08:42:55 -0400] "POST /wp-login.php HTTP/1.1" 200 3988 "-" "Mozilla/5.0 (X11; Ubuntu; Linux x86_64; rv:62.0) Gecko/20100101 Firefox/62.0"
5.135.134.16 - - [23/Mar/2019:08:42:56 -0400] "POST /xmlrpc.php HTTP/1.1" 200 413 "-" "Mozilla/5.0 (X11; Ubuntu; Linux x86_64; rv:62.0) Gecko/20100101 Firefox/62.0"
13.57.220.172 - - [23/Mar/2019:11:01:05 -0400] "GET /wp-login.php HTTP/1.1" 200 2988 "-" "Mozilla/5.0 (X11; Ubuntu; Linux x86_64; rv:62.0) Gecko/20100101 Firefox/62.0"
13.57.220.172 - - [23/Mar/2019:11:01:06 -0400] "POST /wp-login.php HTTP/1.1" 200 3985 "-" "Mozilla/5.0 (X11; Ubuntu; Linux x86_64; rv:62.0) Gecko/20100101 Firefox/62.0"
13.57.220.172 - - [23/Mar/2019:11:01:07 -0400] "GET /wp-login.php HTTP/1.1" 200 2988 "-" "Mozilla/5.0 (X11; Ubuntu; Linux x86_64; rv:62.0) Gecko/20100101 Firefox/62.0"
13.57.220.172 - - [23/Mar/2019:11:01:08 -0400] "POST /wp-login.php HTTP/1.1" 200 3833 "-" "Mozilla/5.0 (X11; Ubuntu; Linux x86_64; rv:62.0) Gecko/20100101 Firefox/62.0"
13.57.220.172 - - [23/Mar/2019:11:01:09 -0400] "GET /wp-login.php HTTP/1.1" 200 2988 "-" "Mozilla/5.0 (X11; Ubuntu; Linux x86_64; rv:62.0) Gecko/20100101 Firefox/62.0"
13.57.220.172 - - [23/Mar/2019:11:01:11 -0400] "POST /wp-login.php HTTP/1.1" 200 3836 "-" "Mozilla/5.0 (X11; Ubuntu; Linux x86_64; rv:62.0) Gecko/20100101 Firefox/62.0"
13.57.220.172 - - [23/Mar/2019:11:01:12 -0400] "GET /wp-login.php HTTP/1.1" 200 2988 "-" "Mozilla/5.0 (X11; Ubuntu; Linux x86_64; rv:62.0) Gecko/20100101 Firefox/62.0"
13.57.220.172 - - [23/Mar/2019:11:01:15 -0400] "POST /wp-login.php HTTP/1.1" 200 3837 "-" "Mozilla/5.0 (X11; Ubuntu; Linux x86_64; rv:62.0) Gecko/20100101 Firefox/62.0"
13.57.220.172 - - [23/Mar/2019:11:01:17 -0400] "POST /xmlrpc.php HTTP/1.1" 200 413 "-" "Mozilla/5.0 (X11; Ubuntu; Linux x86_64; rv:62.0) Gecko/20100101 Firefox/62.0"
13.57.233.99 - - [23/Mar/2019:04:17:45 -0400] "GET / HTTP/1.1" 200 25160 "-" "Mozilla/5.0 (Macintosh; Intel Mac OS X 10_12_6) AppleWebKit/537.36 (KHTML, like Gecko) Chrome/61.0.3163.100 Safari/537.36"
18.206.226.75 - - [23/Mar/2019:21:58:07 -0400] "GET /wp-login.php HTTP/1.1" 200 2988 "https://www.google.com/url?3a622303df89920683e4421b2cf28977" "Mozilla/5.0 (Windows NT 6.2; rv:33.0) Gecko/20100101 Firefox/33.0"
18.206.226.75 - - [23/Mar/2019:21:58:07 -0400] "POST /wp-login.php HTTP/1.1" 200 3988 "https://www.google.com/url?3a622303df89920683e4421b2cf28977" "Mozilla/5.0 (Windows NT 6.2; rv:33.0) Gecko/20100101 Firefox/33.0"
18.213.10.181 - - [23/Mar/2019:14:45:42 -0400] "GET /wp-login.php HTTP/1.1" 200 2988 "-" "Mozilla/5.0 (X11; Ubuntu; Linux x86_64; rv:62.0) Gecko/20100101 Firefox/62.0"
18.213.10.181 - - [23/Mar/2019:14:45:42 -0400] "GET /wp-login.php HTTP/1.1" 200 2988 "-" "Mozilla/5.0 (X11; Ubuntu; Linux x86_64; rv:62.0) Gecko/20100101 Firefox/62.0"
18.213.10.181 - - [23/Mar/2019:14:45:42 -0400] "GET /wp-login.php HTTP/1.1" 200 2988 "-" "Mozilla/5.0 (X11; Ubuntu; Linux x86_64; rv:62.0) Gecko/20100101 Firefox/62.0"
command-line bash sort uniq
asked 27 mins ago
j0hj0h
6,4801657119
add a comment |
2
I have a log file sorted by IP addresses,
I want to find the number of occurrences of each unique IP address.
how can I do this with bash? possibly listing the number of occurrences next to an ip, such as:
5.135.134.16 count: 5
13.57.220.172: count 30
18.206.226 count:2
....and so on.
heres a sample of the log:
5.135.134.16 - - [23/Mar/2019:08:42:54 -0400] "GET /wp-login.php HTTP/1.1" 200 2988 "-" "Mozilla/5.0 (X11; Ubuntu; Linux x86_64; rv:62.0) Gecko/20100101 Firefox/62.0"
5.135.134.16 - - [23/Mar/2019:08:42:55 -0400] "GET /wp-login.php HTTP/1.1" 200 2988 "-" "Mozilla/5.0 (X11; Ubuntu; Linux x86_64; rv:62.0) Gecko/20100101 Firefox/62.0"
5.135.134.16 - - [23/Mar/2019:08:42:55 -0400] "POST /wp-login.php HTTP/1.1" 200 3836 "-" "Mozilla/5.0 (X11; Ubuntu; Linux x86_64; rv:62.0) Gecko/20100101 Firefox/62.0"
5.135.134.16 - - [23/Mar/2019:08:42:55 -0400] "POST /wp-login.php HTTP/1.1" 200 3988 "-" "Mozilla/5.0 (X11; Ubuntu; Linux x86_64; rv:62.0) Gecko/20100101 Firefox/62.0"
5.135.134.16 - - [23/Mar/2019:08:42:56 -0400] "POST /xmlrpc.php HTTP/1.1" 200 413 "-" "Mozilla/5.0 (X11; Ubuntu; Linux x86_64; rv:62.0) Gecko/20100101 Firefox/62.0"
13.57.220.172 - - [23/Mar/2019:11:01:05 -0400] "GET /wp-login.php HTTP/1.1" 200 2988 "-" "Mozilla/5.0 (X11; Ubuntu; Linux x86_64; rv:62.0) Gecko/20100101 Firefox/62.0"
13.57.220.172 - - [23/Mar/2019:11:01:06 -0400] "POST /wp-login.php HTTP/1.1" 200 3985 "-" "Mozilla/5.0 (X11; Ubuntu; Linux x86_64; rv:62.0) Gecko/20100101 Firefox/62.0"
13.57.220.172 - - [23/Mar/2019:11:01:07 -0400] "GET /wp-login.php HTTP/1.1" 200 2988 "-" "Mozilla/5.0 (X11; Ubuntu; Linux x86_64; rv:62.0) Gecko/20100101 Firefox/62.0"
13.57.220.172 - - [23/Mar/2019:11:01:08 -0400] "POST /wp-login.php HTTP/1.1" 200 3833 "-" "Mozilla/5.0 (X11; Ubuntu; Linux x86_64; rv:62.0) Gecko/20100101 Firefox/62.0"
13.57.220.172 - - [23/Mar/2019:11:01:09 -0400] "GET /wp-login.php HTTP/1.1" 200 2988 "-" "Mozilla/5.0 (X11; Ubuntu; Linux x86_64; rv:62.0) Gecko/20100101 Firefox/62.0"
13.57.220.172 - - [23/Mar/2019:11:01:11 -0400] "POST /wp-login.php HTTP/1.1" 200 3836 "-" "Mozilla/5.0 (X11; Ubuntu; Linux x86_64; rv:62.0) Gecko/20100101 Firefox/62.0"
13.57.220.172 - - [23/Mar/2019:11:01:12 -0400] "GET /wp-login.php HTTP/1.1" 200 2988 "-" "Mozilla/5.0 (X11; Ubuntu; Linux x86_64; rv:62.0) Gecko/20100101 Firefox/62.0"
13.57.220.172 - - [23/Mar/2019:11:01:15 -0400] "POST /wp-login.php HTTP/1.1" 200 3837 "-" "Mozilla/5.0 (X11; Ubuntu; Linux x86_64; rv:62.0) Gecko/20100101 Firefox/62.0"
13.57.220.172 - - [23/Mar/2019:11:01:17 -0400] "POST /xmlrpc.php HTTP/1.1" 200 413 "-" "Mozilla/5.0 (X11; Ubuntu; Linux x86_64; rv:62.0) Gecko/20100101 Firefox/62.0"
13.57.233.99 - - [23/Mar/2019:04:17:45 -0400] "GET / HTTP/1.1" 200 25160 "-" "Mozilla/5.0 (Macintosh; Intel Mac OS X 10_12_6) AppleWebKit/537.36 (KHTML, like Gecko) Chrome/61.0.3163.100 Safari/537.36"
18.206.226.75 - - [23/Mar/2019:21:58:07 -0400] "GET /wp-login.php HTTP/1.1" 200 2988 "https://www.google.com/url?3a622303df89920683e4421b2cf28977" "Mozilla/5.0 (Windows NT 6.2; rv:33.0) Gecko/20100101 Firefox/33.0"
18.206.226.75 - - [23/Mar/2019:21:58:07 -0400] "POST /wp-login.php HTTP/1.1" 200 3988 "https://www.google.com/url?3a622303df89920683e4421b2cf28977" "Mozilla/5.0 (Windows NT 6.2; rv:33.0) Gecko/20100101 Firefox/33.0"
18.213.10.181 - - [23/Mar/2019:14:45:42 -0400] "GET /wp-login.php HTTP/1.1" 200 2988 "-" "Mozilla/5.0 (X11; Ubuntu; Linux x86_64; rv:62.0) Gecko/20100101 Firefox/62.0"
18.213.10.181 - - [23/Mar/2019:14:45:42 -0400] "GET /wp-login.php HTTP/1.1" 200 2988 "-" "Mozilla/5.0 (X11; Ubuntu; Linux x86_64; rv:62.0) Gecko/20100101 Firefox/62.0"
18.213.10.181 - - [23/Mar/2019:14:45:42 -0400] "GET /wp-login.php HTTP/1.1" 200 2988 "-" "Mozilla/5.0 (X11; Ubuntu; Linux x86_64; rv:62.0) Gecko/20100101 Firefox/62.0"
command-line bash sort uniq
asked 27 mins ago
j0hj0h
6,4801657119
With “bash”, do you mean the plain shell or the command line in general?
– dessert
23 mins ago
add a comment |
2
2
2
2
I have a log file sorted by IP addresses,
I want to find the number of occurrences of each unique IP address.
how can I do this with bash? possibly listing the number of occurrences next to an ip, such as:
5.135.134.16 count: 5
13.57.220.172: count 30
18.206.226 count:2
....and so on.
heres a sample of the log:
5.135.134.16 - - [23/Mar/2019:08:42:54 -0400] "GET /wp-login.php HTTP/1.1" 200 2988 "-" "Mozilla/5.0 (X11; Ubuntu; Linux x86_64; rv:62.0) Gecko/20100101 Firefox/62.0"
5.135.134.16 - - [23/Mar/2019:08:42:55 -0400] "GET /wp-login.php HTTP/1.1" 200 2988 "-" "Mozilla/5.0 (X11; Ubuntu; Linux x86_64; rv:62.0) Gecko/20100101 Firefox/62.0"
5.135.134.16 - - [23/Mar/2019:08:42:55 -0400] "POST /wp-login.php HTTP/1.1" 200 3836 "-" "Mozilla/5.0 (X11; Ubuntu; Linux x86_64; rv:62.0) Gecko/20100101 Firefox/62.0"
5.135.134.16 - - [23/Mar/2019:08:42:55 -0400] "POST /wp-login.php HTTP/1.1" 200 3988 "-" "Mozilla/5.0 (X11; Ubuntu; Linux x86_64; rv:62.0) Gecko/20100101 Firefox/62.0"
5.135.134.16 - - [23/Mar/2019:08:42:56 -0400] "POST /xmlrpc.php HTTP/1.1" 200 413 "-" "Mozilla/5.0 (X11; Ubuntu; Linux x86_64; rv:62.0) Gecko/20100101 Firefox/62.0"
13.57.220.172 - - [23/Mar/2019:11:01:05 -0400] "GET /wp-login.php HTTP/1.1" 200 2988 "-" "Mozilla/5.0 (X11; Ubuntu; Linux x86_64; rv:62.0) Gecko/20100101 Firefox/62.0"
13.57.220.172 - - [23/Mar/2019:11:01:06 -0400] "POST /wp-login.php HTTP/1.1" 200 3985 "-" "Mozilla/5.0 (X11; Ubuntu; Linux x86_64; rv:62.0) Gecko/20100101 Firefox/62.0"
13.57.220.172 - - [23/Mar/2019:11:01:07 -0400] "GET /wp-login.php HTTP/1.1" 200 2988 "-" "Mozilla/5.0 (X11; Ubuntu; Linux x86_64; rv:62.0) Gecko/20100101 Firefox/62.0"
13.57.220.172 - - [23/Mar/2019:11:01:08 -0400] "POST /wp-login.php HTTP/1.1" 200 3833 "-" "Mozilla/5.0 (X11; Ubuntu; Linux x86_64; rv:62.0) Gecko/20100101 Firefox/62.0"
13.57.220.172 - - [23/Mar/2019:11:01:09 -0400] "GET /wp-login.php HTTP/1.1" 200 2988 "-" "Mozilla/5.0 (X11; Ubuntu; Linux x86_64; rv:62.0) Gecko/20100101 Firefox/62.0"
13.57.220.172 - - [23/Mar/2019:11:01:11 -0400] "POST /wp-login.php HTTP/1.1" 200 3836 "-" "Mozilla/5.0 (X11; Ubuntu; Linux x86_64; rv:62.0) Gecko/20100101 Firefox/62.0"
13.57.220.172 - - [23/Mar/2019:11:01:12 -0400] "GET /wp-login.php HTTP/1.1" 200 2988 "-" "Mozilla/5.0 (X11; Ubuntu; Linux x86_64; rv:62.0) Gecko/20100101 Firefox/62.0"
13.57.220.172 - - [23/Mar/2019:11:01:15 -0400] "POST /wp-login.php HTTP/1.1" 200 3837 "-" "Mozilla/5.0 (X11; Ubuntu; Linux x86_64; rv:62.0) Gecko/20100101 Firefox/62.0"
13.57.220.172 - - [23/Mar/2019:11:01:17 -0400] "POST /xmlrpc.php HTTP/1.1" 200 413 "-" "Mozilla/5.0 (X11; Ubuntu; Linux x86_64; rv:62.0) Gecko/20100101 Firefox/62.0"
13.57.233.99 - - [23/Mar/2019:04:17:45 -0400] "GET / HTTP/1.1" 200 25160 "-" "Mozilla/5.0 (Macintosh; Intel Mac OS X 10_12_6) AppleWebKit/537.36 (KHTML, like Gecko) Chrome/61.0.3163.100 Safari/537.36"
18.206.226.75 - - [23/Mar/2019:21:58:07 -0400] "GET /wp-login.php HTTP/1.1" 200 2988 "https://www.google.com/url?3a622303df89920683e4421b2cf28977" "Mozilla/5.0 (Windows NT 6.2; rv:33.0) Gecko/20100101 Firefox/33.0"
18.206.226.75 - - [23/Mar/2019:21:58:07 -0400] "POST /wp-login.php HTTP/1.1" 200 3988 "https://www.google.com/url?3a622303df89920683e4421b2cf28977" "Mozilla/5.0 (Windows NT 6.2; rv:33.0) Gecko/20100101 Firefox/33.0"
18.213.10.181 - - [23/Mar/2019:14:45:42 -0400] "GET /wp-login.php HTTP/1.1" 200 2988 "-" "Mozilla/5.0 (X11; Ubuntu; Linux x86_64; rv:62.0) Gecko/20100101 Firefox/62.0"
18.213.10.181 - - [23/Mar/2019:14:45:42 -0400] "GET /wp-login.php HTTP/1.1" 200 2988 "-" "Mozilla/5.0 (X11; Ubuntu; Linux x86_64; rv:62.0) Gecko/20100101 Firefox/62.0"
18.213.10.181 - - [23/Mar/2019:14:45:42 -0400] "GET /wp-login.php HTTP/1.1" 200 2988 "-" "Mozilla/5.0 (X11; Ubuntu; Linux x86_64; rv:62.0) Gecko/20100101 Firefox/62.0"
command-line bash sort uniq
asked 27 mins ago
j0hj0h
6,4801657119
I have a log file sorted by IP addresses,
I want to find the number of occurrences of each unique IP address.
how can I do this with bash? possibly listing the number of occurrences next to an ip, such as:
5.135.134.16 count: 5
13.57.220.172: count 30
18.206.226 count:2
....and so on.
heres a sample of the log:
5.135.134.16 - - [23/Mar/2019:08:42:54 -0400] "GET /wp-login.php HTTP/1.1" 200 2988 "-" "Mozilla/5.0 (X11; Ubuntu; Linux x86_64; rv:62.0) Gecko/20100101 Firefox/62.0"
5.135.134.16 - - [23/Mar/2019:08:42:55 -0400] "GET /wp-login.php HTTP/1.1" 200 2988 "-" "Mozilla/5.0 (X11; Ubuntu; Linux x86_64; rv:62.0) Gecko/20100101 Firefox/62.0"
5.135.134.16 - - [23/Mar/2019:08:42:55 -0400] "POST /wp-login.php HTTP/1.1" 200 3836 "-" "Mozilla/5.0 (X11; Ubuntu; Linux x86_64; rv:62.0) Gecko/20100101 Firefox/62.0"
5.135.134.16 - - [23/Mar/2019:08:42:55 -0400] "POST /wp-login.php HTTP/1.1" 200 3988 "-" "Mozilla/5.0 (X11; Ubuntu; Linux x86_64; rv:62.0) Gecko/20100101 Firefox/62.0"
5.135.134.16 - - [23/Mar/2019:08:42:56 -0400] "POST /xmlrpc.php HTTP/1.1" 200 413 "-" "Mozilla/5.0 (X11; Ubuntu; Linux x86_64; rv:62.0) Gecko/20100101 Firefox/62.0"
13.57.220.172 - - [23/Mar/2019:11:01:05 -0400] "GET /wp-login.php HTTP/1.1" 200 2988 "-" "Mozilla/5.0 (X11; Ubuntu; Linux x86_64; rv:62.0) Gecko/20100101 Firefox/62.0"
13.57.220.172 - - [23/Mar/2019:11:01:06 -0400] "POST /wp-login.php HTTP/1.1" 200 3985 "-" "Mozilla/5.0 (X11; Ubuntu; Linux x86_64; rv:62.0) Gecko/20100101 Firefox/62.0"
13.57.220.172 - - [23/Mar/2019:11:01:07 -0400] "GET /wp-login.php HTTP/1.1" 200 2988 "-" "Mozilla/5.0 (X11; Ubuntu; Linux x86_64; rv:62.0) Gecko/20100101 Firefox/62.0"
13.57.220.172 - - [23/Mar/2019:11:01:08 -0400] "POST /wp-login.php HTTP/1.1" 200 3833 "-" "Mozilla/5.0 (X11; Ubuntu; Linux x86_64; rv:62.0) Gecko/20100101 Firefox/62.0"
13.57.220.172 - - [23/Mar/2019:11:01:09 -0400] "GET /wp-login.php HTTP/1.1" 200 2988 "-" "Mozilla/5.0 (X11; Ubuntu; Linux x86_64; rv:62.0) Gecko/20100101 Firefox/62.0"
13.57.220.172 - - [23/Mar/2019:11:01:11 -0400] "POST /wp-login.php HTTP/1.1" 200 3836 "-" "Mozilla/5.0 (X11; Ubuntu; Linux x86_64; rv:62.0) Gecko/20100101 Firefox/62.0"
13.57.220.172 - - [23/Mar/2019:11:01:12 -0400] "GET /wp-login.php HTTP/1.1" 200 2988 "-" "Mozilla/5.0 (X11; Ubuntu; Linux x86_64; rv:62.0) Gecko/20100101 Firefox/62.0"
13.57.220.172 - - [23/Mar/2019:11:01:15 -0400] "POST /wp-login.php HTTP/1.1" 200 3837 "-" "Mozilla/5.0 (X11; Ubuntu; Linux x86_64; rv:62.0) Gecko/20100101 Firefox/62.0"
13.57.220.172 - - [23/Mar/2019:11:01:17 -0400] "POST /xmlrpc.php HTTP/1.1" 200 413 "-" "Mozilla/5.0 (X11; Ubuntu; Linux x86_64; rv:62.0) Gecko/20100101 Firefox/62.0"
13.57.233.99 - - [23/Mar/2019:04:17:45 -0400] "GET / HTTP/1.1" 200 25160 "-" "Mozilla/5.0 (Macintosh; Intel Mac OS X 10_12_6) AppleWebKit/537.36 (KHTML, like Gecko) Chrome/61.0.3163.100 Safari/537.36"
18.206.226.75 - - [23/Mar/2019:21:58:07 -0400] "GET /wp-login.php HTTP/1.1" 200 2988 "https://www.google.com/url?3a622303df89920683e4421b2cf28977" "Mozilla/5.0 (Windows NT 6.2; rv:33.0) Gecko/20100101 Firefox/33.0"
18.206.226.75 - - [23/Mar/2019:21:58:07 -0400] "POST /wp-login.php HTTP/1.1" 200 3988 "https://www.google.com/url?3a622303df89920683e4421b2cf28977" "Mozilla/5.0 (Windows NT 6.2; rv:33.0) Gecko/20100101 Firefox/33.0"
18.213.10.181 - - [23/Mar/2019:14:45:42 -0400] "GET /wp-login.php HTTP/1.1" 200 2988 "-" "Mozilla/5.0 (X11; Ubuntu; Linux x86_64; rv:62.0) Gecko/20100101 Firefox/62.0"
18.213.10.181 - - [23/Mar/2019:14:45:42 -0400] "GET /wp-login.php HTTP/1.1" 200 2988 "-" "Mozilla/5.0 (X11; Ubuntu; Linux x86_64; rv:62.0) Gecko/20100101 Firefox/62.0"
18.213.10.181 - - [23/Mar/2019:14:45:42 -0400] "GET /wp-login.php HTTP/1.1" 200 2988 "-" "Mozilla/5.0 (X11; Ubuntu; Linux x86_64; rv:62.0) Gecko/20100101 Firefox/62.0"
command-line bash sort uniq
command-line bash sort uniq
asked 27 mins ago
j0hj0h
6,4801657119
asked 27 mins ago
j0hj0h
6,4801657119
edited 22 mins ago
j0h
asked 27 mins ago
j0hj0h
6,4801657119
asked 27 mins ago
j0hj0h
6,4801657119
asked 27 mins ago
j0hj0h
6,4801657119
6,4801657119
With “bash”, do you mean the plain shell or the command line in general?
– dessert
23 mins ago
add a comment |
With “bash”, do you mean the plain shell or the command line in general?
– dessert
23 mins ago
With “bash”, do you mean the plain shell or the command line in general?
– dessert
23 mins ago
With “bash”, do you mean the plain shell or the command line in general?
– dessert
23 mins ago
add a comment |
4 Answers
4
active
oldest
votes
5
You can use cut and uniq tools:
cut -d ' ' -f1 test.txt | uniq -c
5 5.135.134.16
9 13.57.220.172
1 13.57.233.99
2 18.206.226.75
3 18.213.10.181
Explanation :
cut -d ' ' -f1: extract first field (ip address)uniq -c: report repeated lines and display the number of occurences
answered 15 mins ago
Mikael FloraMikael Flora
512
New contributor
Mikael Flora 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 |
3
Here is one possible solution:
IN_FILE="file.log"
for IP in $(awk 'print $1' $IN_FILE | sort -u)
do
echo -en "$IPtcount: "
grep -c "$IP" "$IN_FILE"
done
- replace
file.logwith the actual file name. - the command substitution expression
$(awk 'print $1' $IN_FILE | sort -u)will provide a list of the unique values of the first column. - then
grep -cwill count each of these value within the file.
$ IN_FILE="file.log"; for IP in $(awk 'print $1' $IN_FILE | sort -u); do echo -en "$IPtcount: "; grep -c "$IP" "$IN_FILE"; done
13.57.220.172 count: 9
13.57.233.99 count: 1
18.206.226.75 count: 2
18.213.10.181 count: 3
5.135.134.16 count: 5
answered 12 mins ago
pa4080pa4080
14.7k52872
add a comment |
3
You can use grep and uniq for the list of addresses, loop over them and grep again for the count:
for i in $(<log grep -o '^[^ ]*' | uniq); do
printf '%s count %dn' "$i" $(<log grep -c "$i")
done
Example run
$ for i in $(<log grep -o '^[^ ]*'|uniq);do printf '%s count %dn' "$i" $(<log grep -c "$i");done
5.135.134.16 count 5
13.57.220.172 count 9
13.57.233.99 count 1
18.206.226.75 count 2
18.213.10.181 count 3
answered 10 mins ago
dessertdessert
25.2k673106
add a comment |
3
If you really need the given output format, then since your input is already sorted, a single-pass way to do it in Awk would be
awk '
NR==1 last=$1
$1 != last print last, "count: " c[last]; last = $1
c[$1]++
END print last, "count: " c[last]
'
Ex.
$ awk 'NR==1 last=$1 $1 != last print last, "count: " c[last]; last = $1 c[$1]++ ENDprint last, "count: " c[last]' log
5.135.134.16 count: 5
13.57.220.172 count: 9
13.57.233.99 count: 1
18.206.226.75 count: 2
18.213.10.181 count: 3
Otherwise, I would recommend this cut + uniq based answer
If not already sorted, then
awk 'c[$1]++ ENDfor(i in c) print i, "count: " c[i]' log
(This works on sorted input as well, however unnecessarily reads all the IPs into memory.)
answered 7 mins ago
steeldriversteeldriver
70.3k11114186
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
5
You can use cut and uniq tools:
cut -d ' ' -f1 test.txt | uniq -c
5 5.135.134.16
9 13.57.220.172
1 13.57.233.99
2 18.206.226.75
3 18.213.10.181
Explanation :
cut -d ' ' -f1: extract first field (ip address)uniq -c: report repeated lines and display the number of occurences
answered 15 mins ago
Mikael FloraMikael Flora
512
New contributor
Mikael Flora 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 |
5
You can use cut and uniq tools:
cut -d ' ' -f1 test.txt | uniq -c
5 5.135.134.16
9 13.57.220.172
1 13.57.233.99
2 18.206.226.75
3 18.213.10.181
Explanation :
cut -d ' ' -f1: extract first field (ip address)uniq -c: report repeated lines and display the number of occurences
answered 15 mins ago
Mikael FloraMikael Flora
512
New contributor
Mikael Flora 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 |
5
5
5
You can use cut and uniq tools:
cut -d ' ' -f1 test.txt | uniq -c
5 5.135.134.16
9 13.57.220.172
1 13.57.233.99
2 18.206.226.75
3 18.213.10.181
Explanation :
cut -d ' ' -f1: extract first field (ip address)uniq -c: report repeated lines and display the number of occurences
answered 15 mins ago
Mikael FloraMikael Flora
512
New contributor
Mikael Flora is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
You can use cut and uniq tools:
cut -d ' ' -f1 test.txt | uniq -c
5 5.135.134.16
9 13.57.220.172
1 13.57.233.99
2 18.206.226.75
3 18.213.10.181
Explanation :
cut -d ' ' -f1: extract first field (ip address)uniq -c: report repeated lines and display the number of occurences
answered 15 mins ago
Mikael FloraMikael Flora
512
New contributor
Mikael Flora is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
answered 15 mins ago
Mikael FloraMikael Flora
512
New contributor
Mikael Flora is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
answered 15 mins ago
Mikael FloraMikael Flora
512
answered 15 mins ago
Mikael FloraMikael Flora
512
512
New contributor
Mikael Flora is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
Mikael Flora is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Mikael Flora 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 |
3
Here is one possible solution:
IN_FILE="file.log"
for IP in $(awk 'print $1' $IN_FILE | sort -u)
do
echo -en "$IPtcount: "
grep -c "$IP" "$IN_FILE"
done
- replace
file.logwith the actual file name. - the command substitution expression
$(awk 'print $1' $IN_FILE | sort -u)will provide a list of the unique values of the first column. - then
grep -cwill count each of these value within the file.
$ IN_FILE="file.log"; for IP in $(awk 'print $1' $IN_FILE | sort -u); do echo -en "$IPtcount: "; grep -c "$IP" "$IN_FILE"; done
13.57.220.172 count: 9
13.57.233.99 count: 1
18.206.226.75 count: 2
18.213.10.181 count: 3
5.135.134.16 count: 5
answered 12 mins ago
pa4080pa4080
14.7k52872
add a comment |
3
Here is one possible solution:
IN_FILE="file.log"
for IP in $(awk 'print $1' $IN_FILE | sort -u)
do
echo -en "$IPtcount: "
grep -c "$IP" "$IN_FILE"
done
- replace
file.logwith the actual file name. - the command substitution expression
$(awk 'print $1' $IN_FILE | sort -u)will provide a list of the unique values of the first column. - then
grep -cwill count each of these value within the file.
$ IN_FILE="file.log"; for IP in $(awk 'print $1' $IN_FILE | sort -u); do echo -en "$IPtcount: "; grep -c "$IP" "$IN_FILE"; done
13.57.220.172 count: 9
13.57.233.99 count: 1
18.206.226.75 count: 2
18.213.10.181 count: 3
5.135.134.16 count: 5
answered 12 mins ago
pa4080pa4080
14.7k52872
add a comment |
3
3
3
Here is one possible solution:
IN_FILE="file.log"
for IP in $(awk 'print $1' $IN_FILE | sort -u)
do
echo -en "$IPtcount: "
grep -c "$IP" "$IN_FILE"
done
- replace
file.logwith the actual file name. - the command substitution expression
$(awk 'print $1' $IN_FILE | sort -u)will provide a list of the unique values of the first column. - then
grep -cwill count each of these value within the file.
$ IN_FILE="file.log"; for IP in $(awk 'print $1' $IN_FILE | sort -u); do echo -en "$IPtcount: "; grep -c "$IP" "$IN_FILE"; done
13.57.220.172 count: 9
13.57.233.99 count: 1
18.206.226.75 count: 2
18.213.10.181 count: 3
5.135.134.16 count: 5
answered 12 mins ago
pa4080pa4080
14.7k52872
Here is one possible solution:
IN_FILE="file.log"
for IP in $(awk 'print $1' $IN_FILE | sort -u)
do
echo -en "$IPtcount: "
grep -c "$IP" "$IN_FILE"
done
- replace
file.logwith the actual file name. - the command substitution expression
$(awk 'print $1' $IN_FILE | sort -u)will provide a list of the unique values of the first column. - then
grep -cwill count each of these value within the file.
$ IN_FILE="file.log"; for IP in $(awk 'print $1' $IN_FILE | sort -u); do echo -en "$IPtcount: "; grep -c "$IP" "$IN_FILE"; done
13.57.220.172 count: 9
13.57.233.99 count: 1
18.206.226.75 count: 2
18.213.10.181 count: 3
5.135.134.16 count: 5
answered 12 mins ago
pa4080pa4080
14.7k52872
answered 12 mins ago
pa4080pa4080
14.7k52872
answered 12 mins ago
pa4080pa4080
14.7k52872
answered 12 mins ago
pa4080pa4080
14.7k52872
14.7k52872
add a comment |
add a comment |
3
You can use grep and uniq for the list of addresses, loop over them and grep again for the count:
for i in $(<log grep -o '^[^ ]*' | uniq); do
printf '%s count %dn' "$i" $(<log grep -c "$i")
done
Example run
$ for i in $(<log grep -o '^[^ ]*'|uniq);do printf '%s count %dn' "$i" $(<log grep -c "$i");done
5.135.134.16 count 5
13.57.220.172 count 9
13.57.233.99 count 1
18.206.226.75 count 2
18.213.10.181 count 3
answered 10 mins ago
dessertdessert
25.2k673106
add a comment |
3
You can use grep and uniq for the list of addresses, loop over them and grep again for the count:
for i in $(<log grep -o '^[^ ]*' | uniq); do
printf '%s count %dn' "$i" $(<log grep -c "$i")
done
Example run
$ for i in $(<log grep -o '^[^ ]*'|uniq);do printf '%s count %dn' "$i" $(<log grep -c "$i");done
5.135.134.16 count 5
13.57.220.172 count 9
13.57.233.99 count 1
18.206.226.75 count 2
18.213.10.181 count 3
answered 10 mins ago
dessertdessert
25.2k673106
add a comment |
3
3
3
You can use grep and uniq for the list of addresses, loop over them and grep again for the count:
for i in $(<log grep -o '^[^ ]*' | uniq); do
printf '%s count %dn' "$i" $(<log grep -c "$i")
done
Example run
$ for i in $(<log grep -o '^[^ ]*'|uniq);do printf '%s count %dn' "$i" $(<log grep -c "$i");done
5.135.134.16 count 5
13.57.220.172 count 9
13.57.233.99 count 1
18.206.226.75 count 2
18.213.10.181 count 3
answered 10 mins ago
dessertdessert
25.2k673106
You can use grep and uniq for the list of addresses, loop over them and grep again for the count:
for i in $(<log grep -o '^[^ ]*' | uniq); do
printf '%s count %dn' "$i" $(<log grep -c "$i")
done
Example run
$ for i in $(<log grep -o '^[^ ]*'|uniq);do printf '%s count %dn' "$i" $(<log grep -c "$i");done
5.135.134.16 count 5
13.57.220.172 count 9
13.57.233.99 count 1
18.206.226.75 count 2
18.213.10.181 count 3
answered 10 mins ago
dessertdessert
25.2k673106
answered 10 mins ago
dessertdessert
25.2k673106
answered 10 mins ago
dessertdessert
25.2k673106
answered 10 mins ago
dessertdessert
25.2k673106
25.2k673106
add a comment |
add a comment |
3
If you really need the given output format, then since your input is already sorted, a single-pass way to do it in Awk would be
awk '
NR==1 last=$1
$1 != last print last, "count: " c[last]; last = $1
c[$1]++
END print last, "count: " c[last]
'
Ex.
$ awk 'NR==1 last=$1 $1 != last print last, "count: " c[last]; last = $1 c[$1]++ ENDprint last, "count: " c[last]' log
5.135.134.16 count: 5
13.57.220.172 count: 9
13.57.233.99 count: 1
18.206.226.75 count: 2
18.213.10.181 count: 3
Otherwise, I would recommend this cut + uniq based answer
If not already sorted, then
awk 'c[$1]++ ENDfor(i in c) print i, "count: " c[i]' log
(This works on sorted input as well, however unnecessarily reads all the IPs into memory.)
answered 7 mins ago
steeldriversteeldriver
70.3k11114186
add a comment |
3
If you really need the given output format, then since your input is already sorted, a single-pass way to do it in Awk would be
awk '
NR==1 last=$1
$1 != last print last, "count: " c[last]; last = $1
c[$1]++
END print last, "count: " c[last]
'
Ex.
$ awk 'NR==1 last=$1 $1 != last print last, "count: " c[last]; last = $1 c[$1]++ ENDprint last, "count: " c[last]' log
5.135.134.16 count: 5
13.57.220.172 count: 9
13.57.233.99 count: 1
18.206.226.75 count: 2
18.213.10.181 count: 3
Otherwise, I would recommend this cut + uniq based answer
If not already sorted, then
awk 'c[$1]++ ENDfor(i in c) print i, "count: " c[i]' log
(This works on sorted input as well, however unnecessarily reads all the IPs into memory.)
answered 7 mins ago
steeldriversteeldriver
70.3k11114186
add a comment |
3
3
3
If you really need the given output format, then since your input is already sorted, a single-pass way to do it in Awk would be
awk '
NR==1 last=$1
$1 != last print last, "count: " c[last]; last = $1
c[$1]++
END print last, "count: " c[last]
'
Ex.
$ awk 'NR==1 last=$1 $1 != last print last, "count: " c[last]; last = $1 c[$1]++ ENDprint last, "count: " c[last]' log
5.135.134.16 count: 5
13.57.220.172 count: 9
13.57.233.99 count: 1
18.206.226.75 count: 2
18.213.10.181 count: 3
Otherwise, I would recommend this cut + uniq based answer
If not already sorted, then
awk 'c[$1]++ ENDfor(i in c) print i, "count: " c[i]' log
(This works on sorted input as well, however unnecessarily reads all the IPs into memory.)
answered 7 mins ago
steeldriversteeldriver
70.3k11114186
If you really need the given output format, then since your input is already sorted, a single-pass way to do it in Awk would be
awk '
NR==1 last=$1
$1 != last print last, "count: " c[last]; last = $1
c[$1]++
END print last, "count: " c[last]
'
Ex.
$ awk 'NR==1 last=$1 $1 != last print last, "count: " c[last]; last = $1 c[$1]++ ENDprint last, "count: " c[last]' log
5.135.134.16 count: 5
13.57.220.172 count: 9
13.57.233.99 count: 1
18.206.226.75 count: 2
18.213.10.181 count: 3
Otherwise, I would recommend this cut + uniq based answer
If not already sorted, then
awk 'c[$1]++ ENDfor(i in c) print i, "count: " c[i]' log
(This works on sorted input as well, however unnecessarily reads all the IPs into memory.)
answered 7 mins ago
steeldriversteeldriver
70.3k11114186
edited 1 min ago
answered 7 mins ago
steeldriversteeldriver
70.3k11114186
answered 7 mins ago
steeldriversteeldriver
70.3k11114186
answered 7 mins ago
steeldriversteeldriver
70.3k11114186
70.3k11114186
add a comment |
add a comment |
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With “bash”, do you mean the plain shell or the command line in general?
– dessert
23 mins ago