CODEWITHSUNDEEP
sedgrepmatching
Kiran Savadatti
Kiran Savadatti

Reputation: 23

grep after matching pattern but exclude period

I have the following sample strings:

Program exiting with code: 0.
Program exiting with code: 1.
Program exiting with code: 10.

I want grep to return values after the matching pattern "Program exiting with code:". However I do not need the period at the end. I have tried:

grep "Program exiting with exit code:" dataload.log | sed 's/^.*.: //'

The above command returns:

0.
1.
10.

I want to ignore the period at end. I picked up the above command from somewhere.

Can someone describe what each keyword represents and provide me with a regex that will only provide me with the value without period?

sed, awk, perl or any other way is fine with me.

Upvotes: 2

Views: 1106

Answers (4)

Jahid
Jahid

Reputation: 22428

This is another way:

grep -oE 'Program exiting with code:\s*[0-9]+' dataload.log |grep -oE '[0-9]+$'

Output of the first grep command is:

Program exiting with code: 0
Program exiting with code: 1
Program exiting with code: 10

Then you just grep the last digits.

Upvotes: 0

Adam Taylor
Adam Taylor

Reputation: 7783

sed 's/^.*.: //'

First of all this is a substition regular expression, as denoted by the s at the start. The part in the first / / is what to match and the second / / is what to replace it with.

The characters in the first (match) part of the expression are all special regular expression characters.

A full list of what they mean can be found here: http://www.rexegg.com/regex-quickstart.html

The match means:

^ - At the start of the line

. - Match any character

* - Any number of times

. - Match any character

: - Match a colon

- Match a space

And then that is all replaced with nothing. That is why the period at the end is kept because it removes everything up to Program exiting with code:, which leaves 1.

It's worth playing around with an interactive tool to test different regular expressions on your string, e.g. http://www.regexr.com/

You can probably just substitute/remove everything that is not a number in your case: 's/[^0-9]//g'.

Upvotes: 0

Hemang
Hemang

Reputation: 410

Your solution is just fine if you extend it with a cut command:

grep "Program exiting with code:" dataload.log | sed 's/^.*.: //' | cut -d"." -f1

Upvotes: -1

fedorqui
fedorqui

Reputation: 289795

Just use grep with a look-behind and catch only digits:

$ grep -Po '(?<=Program exiting with code: )\d*' file
0
1
10

Upvotes: 2

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