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
Why does standard notation not preserve intervals (visually)
What does "Its cash flow is deeply negative" mean?
ls Ordering[Ordering[list]] optimal?
Why is Miller's case titled R (Miller)?
Trouble understanding the speech of overseas colleagues
How to write the block matrix in LaTex?
Grabbing quick drinks
Can the Reverse Gravity spell affect the Meteor Swarm spell?
How to start emacs in "nothing" mode (`fundamental-mode`)
Should I tutor a student who I know has cheated on their homework?
How long to clear the 'suck zone' of a turbofan after start is initiated?
Why here is plural "We went to the movies last night."
How to be diplomatic in refusing to write code that breaches the privacy of our users
When did Lisp start using symbols for arithmetic?
Does the Brexit deal have to be agreed by both Houses?
How do spells that require an ability check vs. the caster's spell save DC work?
Would this house-rule that treats advantage as a +1 to the roll instead (and disadvantage as -1) and allows them to stack be balanced?
A pseudo-riley?
How do I solve this limit?
Customer Requests (Sometimes) Drive Me Bonkers!
Under what conditions does the function C = f(A,B) satisfy H(C|A) = H(B)?
How easy is it to start Magic from scratch?
I believe this to be a fraud - hired, then asked to cash check and send cash as Bitcoin
Anatomically Correct Strange Women In Ponds Distributing Swords
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
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.
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 32 mins ago
j0hj0h
6,4851657119
add a comment |
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.
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 32 mins ago
j0hj0h
6,4851657119
With “bash”, do you mean the plain shell or the command line in general?
– dessert
28 mins ago
add a comment |
3
3
3
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 32 mins ago
j0hj0h
6,4851657119
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 32 mins ago
j0hj0h
6,4851657119
asked 32 mins ago
j0hj0h
6,4851657119
edited 26 mins ago
j0h
asked 32 mins ago
j0hj0h
6,4851657119
asked 32 mins ago
j0hj0h
6,4851657119
asked 32 mins ago
j0hj0h
6,4851657119
6,4851657119
With “bash”, do you mean the plain shell or the command line in general?
– dessert
28 mins ago
add a comment |
With “bash”, do you mean the plain shell or the command line in general?
– dessert
28 mins ago
With “bash”, do you mean the plain shell or the command line in general?
– dessert
28 mins ago
With “bash”, do you mean the plain shell or the command line in general?
– dessert
28 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 19 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.
One could usesed, e.g.sed -E 's/ *(S*) *(S*)/2 count: 1/'to get the output exactly like OP wanted.
– dessert
1 min ago
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 15 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 11 mins ago
steeldriversteeldriver
70.3k11114186
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 values 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 16 mins ago
pa4080pa4080
14.7k52872
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%2fcount-occurances-of-text-in-file-with-bash%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
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 19 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.
One could usesed, e.g.sed -E 's/ *(S*) *(S*)/2 count: 1/'to get the output exactly like OP wanted.
– dessert
1 min ago
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 19 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.
One could usesed, e.g.sed -E 's/ *(S*) *(S*)/2 count: 1/'to get the output exactly like OP wanted.
– dessert
1 min ago
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 19 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 19 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 19 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 19 mins ago
Mikael FloraMikael Flora
512
answered 19 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.
One could usesed, e.g.sed -E 's/ *(S*) *(S*)/2 count: 1/'to get the output exactly like OP wanted.
– dessert
1 min ago
add a comment |
One could usesed, e.g.sed -E 's/ *(S*) *(S*)/2 count: 1/'to get the output exactly like OP wanted.
– dessert
1 min ago
One could use
sed, e.g. sed -E 's/ *(S*) *(S*)/2 count: 1/' to get the output exactly like OP wanted.– dessert
1 min ago
One could use
sed, e.g. sed -E 's/ *(S*) *(S*)/2 count: 1/' to get the output exactly like OP wanted.– dessert
1 min ago
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 15 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 15 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 15 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 15 mins ago
dessertdessert
25.2k673106
answered 15 mins ago
dessertdessert
25.2k673106
answered 15 mins ago
dessertdessert
25.2k673106
answered 15 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 11 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 11 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 11 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 11 mins ago
steeldriversteeldriver
70.3k11114186
edited 5 mins ago
answered 11 mins ago
steeldriversteeldriver
70.3k11114186
answered 11 mins ago
steeldriversteeldriver
70.3k11114186
answered 11 mins ago
steeldriversteeldriver
70.3k11114186
70.3k11114186
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 values 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 16 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 values 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 16 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 values 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 16 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 values 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 16 mins ago
pa4080pa4080
14.7k52872
edited 3 mins ago
answered 16 mins ago
pa4080pa4080
14.7k52872
answered 16 mins ago
pa4080pa4080
14.7k52872
answered 16 mins ago
pa4080pa4080
14.7k52872
14.7k52872
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%2fcount-occurances-of-text-in-file-with-bash%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
28 mins ago