How to count occurrences of text in a file? The Next CEO of Stack OverflowCount duplicated...
What can we do to stop prior company from asking us questions?
WOW air has ceased operation, can I get my tickets refunded?
Why didn't Theresa May consult with Parliament before negotiating a deal with the EU?
Does it take more energy to get to Venus or to Mars?
What makes a siege story/plot interesting?
Anatomically Correct Mesopelagic Aves
Describing a person. What needs to be mentioned?
Is HostGator storing my password in plaintext?
How to start emacs in "nothing" mode (`fundamental-mode`)
Fastest way to shutdown Ubuntu Mate 18.10
What is meant by a M next to a roman numeral?
Is it my responsibility to learn a new technology in my own time my employer wants to implement?
Removing read access from a file
Apart from "berlinern", do any other German dialects have a corresponding verb?
Unreliable Magic - Is it worth it?
Can a single photon have an energy density?
Are there languages with no euphemisms?
Implement the Thanos sorting algorithm
How should I support this large drywall patch?
What does "Its cash flow is deeply negative" mean?
Does the Brexit deal have to be agreed by both Houses?
Why did we only see the N-1 starfighters in one film?
When did Lisp start using symbols for arithmetic?
What does this shorthand mean?
How to count occurrences of text in a file?
The Next CEO of Stack OverflowCount duplicated words in a text fileuniq --count command is yields incorrect result?How to compare two (vague) file lists and print the duplicates?How to count occurrences of each character?How do I count text lines?How long it will take to sort uniq a 62GB file?How does uniq work?Count number of data points in fileWrong sorting a text fileHow to count number of partial occurrences of a string in a file
5
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.
Here’s 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
edited 21 mins ago
dessert
25.2k673106
asked 55 mins ago
j0hj0h
6,4951657119
add a comment |
5
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.
Here’s 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
edited 21 mins ago
dessert
25.2k673106
asked 55 mins ago
j0hj0h
6,4951657119
With “bash”, do you mean the plain shell or the command line in general?
– dessert
51 mins ago
add a comment |
5
5
5
3
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.
Here’s 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
edited 21 mins ago
dessert
25.2k673106
asked 55 mins ago
j0hj0h
6,4951657119
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.
Here’s 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
edited 21 mins ago
dessert
25.2k673106
asked 55 mins ago
j0hj0h
6,4951657119
edited 21 mins ago
dessert
25.2k673106
asked 55 mins ago
j0hj0h
6,4951657119
edited 21 mins ago
dessert
25.2k673106
edited 21 mins ago
dessert
25.2k673106
edited 21 mins ago
dessert
25.2k673106
25.2k673106
asked 55 mins ago
j0hj0h
6,4951657119
asked 55 mins ago
j0hj0h
6,4951657119
asked 55 mins ago
j0hj0h
6,4951657119
6,4951657119
With “bash”, do you mean the plain shell or the command line in general?
– dessert
51 mins ago
add a comment |
With “bash”, do you mean the plain shell or the command line in general?
– dessert
51 mins ago
With “bash”, do you mean the plain shell or the command line in general?
– dessert
51 mins ago
With “bash”, do you mean the plain shell or the command line in general?
– dessert
51 mins ago
add a comment |
4 Answers
4
active
oldest
votes
6
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 42 mins ago
Mikael FloraMikael Flora
615
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.
1
One could usesed, e.g.sed -E 's/ *(S*) *(S*)/2 count: 1/'to get the output exactly like OP wanted.
– dessert
24 mins ago
add a comment |
5
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 38 mins ago
dessertdessert
25.2k673106
add a comment |
4
Here is one possible solution:
IN_FILE="file.log"
for IP in $(awk '{print $1}' "$IN_FILE" | sort -u)
do
echo -en "${IP}tcount: "
grep -c "$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 values within the file.
$ IN_FILE="file.log"; for IP in $(awk '{print $1}' "$IN_FILE" | sort -u); do echo -en "${IP}tcount: "; grep -c "$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 40 mins ago
pa4080pa4080
14.7k52872
add a comment |
4
If you don't specifically require the given output format, then I would recommend the already posted cut + uniq based answer
If you really need the given output format, a single-pass way to do it in Awk would be
awk '{c[$1]++} END{for(i in c) print i, "count: " c[i]}' log
This is somewhat non-ideal when the input is already sorted since it unnecessarily stores all the IPs into memory - a better, though more complicated, way to do it in the pre-sorted case (more directly equivalent to uniq -c) 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]++} END{print 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
answered 35 mins ago
steeldriversteeldriver
70.3k11114186
add a comment |
Your Answer
StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "89"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});
function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});
}
});
draft saved
draft discarded
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2faskubuntu.com%2fquestions%2f1129521%2fhow-to-count-occurrences-of-text-in-a-file%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
4 Answers
4
active
oldest
votes
4 Answers
4
active
oldest
votes
active
oldest
votes
active
oldest
votes
6
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 42 mins ago
Mikael FloraMikael Flora
615
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.
1
One could usesed, e.g.sed -E 's/ *(S*) *(S*)/2 count: 1/'to get the output exactly like OP wanted.
– dessert
24 mins ago
add a comment |
6
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 42 mins ago
Mikael FloraMikael Flora
615
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.
1
One could usesed, e.g.sed -E 's/ *(S*) *(S*)/2 count: 1/'to get the output exactly like OP wanted.
– dessert
24 mins ago
add a comment |
6
6
6
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 42 mins ago
Mikael FloraMikael Flora
615
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 42 mins ago
Mikael FloraMikael Flora
615
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.
edited 12 mins ago
answered 42 mins ago
Mikael FloraMikael Flora
615
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 42 mins ago
Mikael FloraMikael Flora
615
answered 42 mins ago
Mikael FloraMikael Flora
615
615
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.
1
One could usesed, e.g.sed -E 's/ *(S*) *(S*)/2 count: 1/'to get the output exactly like OP wanted.
– dessert
24 mins ago
add a comment |
1
One could usesed, e.g.sed -E 's/ *(S*) *(S*)/2 count: 1/'to get the output exactly like OP wanted.
– dessert
24 mins ago
1
1
One could use
sed, e.g. sed -E 's/ *(S*) *(S*)/2 count: 1/' to get the output exactly like OP wanted.– dessert
24 mins ago
One could use
sed, e.g. sed -E 's/ *(S*) *(S*)/2 count: 1/' to get the output exactly like OP wanted.– dessert
24 mins ago
add a comment |
5
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 38 mins ago
dessertdessert
25.2k673106
add a comment |
5
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 38 mins ago
dessertdessert
25.2k673106
add a comment |
5
5
5
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 38 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 38 mins ago
dessertdessert
25.2k673106
answered 38 mins ago
dessertdessert
25.2k673106
answered 38 mins ago
dessertdessert
25.2k673106
answered 38 mins ago
dessertdessert
25.2k673106
25.2k673106
add a comment |
add a comment |
4
Here is one possible solution:
IN_FILE="file.log"
for IP in $(awk '{print $1}' "$IN_FILE" | sort -u)
do
echo -en "${IP}tcount: "
grep -c "$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 values within the file.
$ IN_FILE="file.log"; for IP in $(awk '{print $1}' "$IN_FILE" | sort -u); do echo -en "${IP}tcount: "; grep -c "$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 40 mins ago
pa4080pa4080
14.7k52872
add a comment |
4
Here is one possible solution:
IN_FILE="file.log"
for IP in $(awk '{print $1}' "$IN_FILE" | sort -u)
do
echo -en "${IP}tcount: "
grep -c "$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 values within the file.
$ IN_FILE="file.log"; for IP in $(awk '{print $1}' "$IN_FILE" | sort -u); do echo -en "${IP}tcount: "; grep -c "$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 40 mins ago
pa4080pa4080
14.7k52872
add a comment |
4
4
4
Here is one possible solution:
IN_FILE="file.log"
for IP in $(awk '{print $1}' "$IN_FILE" | sort -u)
do
echo -en "${IP}tcount: "
grep -c "$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 values within the file.
$ IN_FILE="file.log"; for IP in $(awk '{print $1}' "$IN_FILE" | sort -u); do echo -en "${IP}tcount: "; grep -c "$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 40 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 "${IP}tcount: "
grep -c "$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 values within the file.
$ IN_FILE="file.log"; for IP in $(awk '{print $1}' "$IN_FILE" | sort -u); do echo -en "${IP}tcount: "; grep -c "$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 40 mins ago
pa4080pa4080
14.7k52872
edited 27 mins ago
answered 40 mins ago
pa4080pa4080
14.7k52872
answered 40 mins ago
pa4080pa4080
14.7k52872
answered 40 mins ago
pa4080pa4080
14.7k52872
14.7k52872
add a comment |
add a comment |
4
If you don't specifically require the given output format, then I would recommend the already posted cut + uniq based answer
If you really need the given output format, a single-pass way to do it in Awk would be
awk '{c[$1]++} END{for(i in c) print i, "count: " c[i]}' log
This is somewhat non-ideal when the input is already sorted since it unnecessarily stores all the IPs into memory - a better, though more complicated, way to do it in the pre-sorted case (more directly equivalent to uniq -c) 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]++} END{print 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
answered 35 mins ago
steeldriversteeldriver
70.3k11114186
add a comment |
4
If you don't specifically require the given output format, then I would recommend the already posted cut + uniq based answer
If you really need the given output format, a single-pass way to do it in Awk would be
awk '{c[$1]++} END{for(i in c) print i, "count: " c[i]}' log
This is somewhat non-ideal when the input is already sorted since it unnecessarily stores all the IPs into memory - a better, though more complicated, way to do it in the pre-sorted case (more directly equivalent to uniq -c) 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]++} END{print 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
answered 35 mins ago
steeldriversteeldriver
70.3k11114186
add a comment |
4
4
4
If you don't specifically require the given output format, then I would recommend the already posted cut + uniq based answer
If you really need the given output format, a single-pass way to do it in Awk would be
awk '{c[$1]++} END{for(i in c) print i, "count: " c[i]}' log
This is somewhat non-ideal when the input is already sorted since it unnecessarily stores all the IPs into memory - a better, though more complicated, way to do it in the pre-sorted case (more directly equivalent to uniq -c) 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]++} END{print 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
answered 35 mins ago
steeldriversteeldriver
70.3k11114186
If you don't specifically require the given output format, then I would recommend the already posted cut + uniq based answer
If you really need the given output format, a single-pass way to do it in Awk would be
awk '{c[$1]++} END{for(i in c) print i, "count: " c[i]}' log
This is somewhat non-ideal when the input is already sorted since it unnecessarily stores all the IPs into memory - a better, though more complicated, way to do it in the pre-sorted case (more directly equivalent to uniq -c) 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]++} END{print 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
answered 35 mins ago
steeldriversteeldriver
70.3k11114186
edited 11 mins ago
answered 35 mins ago
steeldriversteeldriver
70.3k11114186
answered 35 mins ago
steeldriversteeldriver
70.3k11114186
answered 35 mins ago
steeldriversteeldriver
70.3k11114186
70.3k11114186
add a comment |
add a comment |
draft saved
draft discarded
Thanks for contributing an answer to Ask Ubuntu!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
To learn more, see our tips on writing great answers.
draft saved
draft discarded
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2faskubuntu.com%2fquestions%2f1129521%2fhow-to-count-occurrences-of-text-in-a-file%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
With “bash”, do you mean the plain shell or the command line in general?
– dessert
51 mins ago