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Darkmage
Asked: 2012-01-31 03:27:37 +0800 CST

get text between keywords in log file

  • 772

If I have a log file and want to dump only the text between 1234 and 9876 in to another file, how can i do this easily?

If I have a text file like this:

idsfsvcvs sdf sdf e e  sd vs d s g sg  s vc  d

slkdfnls 1234 keep me text 9876 das a g w eg dsf sd fsdf
sdfs fs dfsdf
sdfsdf sdf
sdf s fs
dfsf ds

I want to do somthing like this

$ getinfo "1234" "9876" log

$ cat log

keep me text

log-files text

4 Answers

  1. Best Answer

    One line of sed can do this for you:

    sed -n 's/.*1234 \(.*\)9876.*/\1/p' textfile.txt > log

    • 10

  2. normally you can do this with grep and the -o param. 

    -o, --only-matching
Print only the matched (non-empty) parts of a matching line, with each such part on a separate output line.

    so it would be something like: 

     grep -Po '1234.*9876' >> log

    Not 100% sure about the regex btw, I did not test it

    • 2

  3. This depends on your file content. You can start by something simple like:

    $ grep 1234 logfile | grep 9876 | cut -d ' ' -f 3,4,5

    This works for the provided example. You can work on it to cover other data formats if needed. You can also redirect the output to a file by appending > /path/to/output

    • 1

  4. For what it's worth, I would do the following:

    1. grep 1234.*9876 > myfile
    2. vim myfile
    3. :%s/^.*1234// (delete everything up to 1234)
    4. :%s/9876.*$// (delete everything after and including 9876)

    Not perfect, or the best way, but easy to remember if you often use vim.

    • 0