How to trim a log file using grep command

Question

For example: If below is the content of my log file, I want to keep only the contents between the first occurrence of value '2018' and last occurrence of '2018'

sssssssssssssssssssss
ddddddddddddddddddddd
2018fffffffffffffffff
dddddd2018ddddddddddd
wwww2018wwwwwwwwwwwww
aaaaaaaaaaaa2018aaaaa

Output should be:

2018fffffffffffffffff
dddddd2018ddddddddddd
wwww2018wwwwwwwwwwwww
aaaaaaaaaaaa2018

What have you tried? Please show your attempt so far, so that we can help you with your code. And what does this have to do with winscp? — ghoti, Oct 03 '18 at 11:22
I have created a bat file which runs WinSCP script to download a log file (response log) from remote server. The log file obviously downloads as-is (responses from all the requests). Once the log file is downloaded, I want to cut only the response specific to my unique ID.Script within my bat file is as below: `winscp.com/command ^ "open sftp://xxx.com/ -hostkey=*" "get /var/log/jboss_sit/suFile.log" "exit" ^` — Shawn, Oct 03 '18 at 14:25
You'll need to put that and more detail *into your question*. A solution in bash or even POSIX shell would be super easy, but I can't replicate your environment unless I know what it is. — ghoti, Oct 03 '18 at 16:05

score 1 · Answer 1 · answered Oct 03 '18 at 11:49

1

If the file is not too big, you can do it with GNU grep and the -z flag. From grep.info:

‘-z’
‘--null-data’
     Treat input and output data as sequences of lines, each terminated
     by a zero byte (the ASCII NUL character) instead of a newline.
     Like the ‘-Z’ or ‘--null’ option, this option can be used with
     commands like ‘sort -z’ to process arbitrary file names.

For example:

grep -zo '2018.*2018' infile

Output:

2018fffffffffffffffff
dddddd2018ddddddddddd
wwww2018wwwwwwwwwwwww
aaaaaaaaaaaa2018

answered Oct 03 '18 at 11:49

Thor

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does it not print aaaaaaaaaaaa2018aaaaa ?? – stack0114106 Oct 03 '18 at 13:16
@stack0114106: not with **GNU** grep – Thor Oct 04 '18 at 18:02

score 1 · Answer 2 · answered Oct 03 '18 at 13:23

1

Check this Perl solution

> cat 2018_match
sssssssssssssssssssss
ddddddddddddddddddddd
2018fffffffffffffffff
dddddd2018ddddddddddd
wwww2018wwwwwwwwwwwww
aaaaaaaaaaaa2018aaaaa
> perl -e ' BEGIN { $x=qx(cat 2018_match);$x=~s/(.+?)(2018.+2018)(.+)/$2/osmg; print "$x\n" ; exit } '
2018fffffffffffffffff
dddddd2018ddddddddddd
wwww2018wwwwwwwwwwwww
aaaaaaaaaaaa2018
>

answered Oct 03 '18 at 13:23

stack0114106

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thanks @stack0114106 and @Thor. Can you please help me with findstr solution. I wrote below code, but it retrieves only the lines matching 2018, but i want all lines between first and last occurrence of 2018. `findstr /s "2018" suFile.log > results.log` – Shawn Oct 03 '18 at 14:29

How to trim a log file using grep command

2 Answers2