Grep specific words between quotation marks

Question

Lets say I have a line like this:

<group description="" name="voice_guidance" comment="Missing description!" status="0">

I wish to grep with a bash script only the comment saying:

Missing description!

Could you clarify your question? At least, add an example – Andrii Abramov Aug 01 '17 at 11:54 — Andrii Abramov, Aug 01 '17 at 11:54
Edited. Pressed enter a bit too soon. :) – Bert Aug 01 '17 at 11:57 — Bert, Aug 01 '17 at 11:57

score 3 · Accepted Answer · answered Aug 01 '17 at 12:06

3

If you only want the comment, you can use this: grep -Po 'comment="\K[^"]*'

\K means to only output the part following from there and then all the non-quote characters following are output.

answered Aug 01 '17 at 12:06

rubystallion

Excelent! Thank you very much! – Bert Aug 01 '17 at 12:10
@Bert Please read [What should I do when someone answers my question?](https://stackoverflow.com/help/someone-answers) – SLePort Aug 01 '17 at 12:42
Yeah, sorry. Completely forgot to accept it. – Bert Aug 02 '17 at 09:41

Andrii Abramov · Answer 2 · 2017-08-01T12:05:18.350

1

You can use straightforward solution:

grep '"Missing description!"' filename.html

Where filename.html is a target file.

Here is an example output:

P.S. In such when I see HTML and RegEx both present in the same question, I recommend to read this answer: https://stackoverflow.com/a/1732454/5091346

edited Aug 01 '17 at 12:05

answered Aug 01 '17 at 12:03

Andrii Abramov

Good idea, but the text is not always the same. – Bert Aug 01 '17 at 12:04
Sorry, @Bert. Seems that your question is not clear enough for me. – Andrii Abramov Aug 01 '17 at 12:05

James Brown · Answer 3 · 2017-08-01T12:20:50.637

1

If you can't use the perl-regexp (-P) in @rubystallion's fine answer, use:

$ grep -o "comment=\"[^\"]\+" file | grep -o "[^\"]\+$"
Missing description!

edited Aug 01 '17 at 12:20

answered Aug 01 '17 at 12:09

James Brown

3 Answers3