How do you search and print multiple matches per line with grep -o?

Question

I have a mkdocs instance and am writing a script to print internal links in a page. I cannot get grep to print only the matches if there are multiple per line.

This is what I currently have:

$ grep -Eon '\[([[:alpha:]]|[[:digit:]]|[[:space:]])*\]\((\/|\.).*\)' /path/to/file.md
10:[foo](../../relative_path/foobar.md) is the path to another file, also see [bar](/absolute/path/foobar.md)

I would like the output to look like this:

10:[foo](../../relative_path/foobar.md)
10:[bar](/absolute/path/foobar.md)

Is there a way to do this in grep or even another command like awk or sed?

Stalin Vignesh Kumar · Answer 1 · 2020-06-08T19:04:32.090

6

grep -Pno "[[[:alnum:]]*]\(.*?\)" /path/to/file.md

OR even better( this would match even ["foo anotherword"])

grep -Pno "\[([[:alnum:]]*[[:space:]]*)*?\]\(.*?\)"

-P => Perl Regex which is used to match non-greedy using ?

OR if don't want only alpha numeric and space but any character means

 grep -Pno "\[.*?\]\(.*?\)"

edited Jun 08 '20 at 19:04

answered Jun 08 '20 at 17:45

Stalin Vignesh Kumar

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Stalin, this seems to be what user binarysta has answered, but you dropped some constraints that OP used, such as `[[:space:]]` and `\/|\.` – Quasímodo Jun 08 '20 at 17:47
@Quasímodo :I answered it long back with -P option before binarysta, I just edited for some other thing.or may be in draft i don't know but was typing even before your answer ...It seems OP is looking for absolute or relative path , i think this would work... – Stalin Vignesh Kumar Jun 08 '20 at 17:51
what is the reason for omitting`[[:space:]]` from the OP's pattern? – binarysta Jun 08 '20 at 18:44
@binarysta bcz OP is using | inside ( and ) that matches either only spaces or digits or alphas which makes no sense ... – Stalin Vignesh Kumar Jun 08 '20 at 18:47
Ando so you down voted ? :( – Stalin Vignesh Kumar Jun 08 '20 at 18:49
@StalinVigneshKumar what do you mean makes no sense? we do not know the input. maybe there are `[space foo]` patterns. – binarysta Jun 08 '20 at 18:51
@binarysta : that would also print in the answer i provided after OR. – Stalin Vignesh Kumar Jun 08 '20 at 18:58
oh yes, you've edited. please make a small edit because my vote is blocked, i will upvote it. – binarysta Jun 08 '20 at 19:02
@binarysta : Thanks much – Stalin Vignesh Kumar Jun 08 '20 at 19:04

score 4 · Answer 2 · answered Jun 08 '20 at 17:41

\[([[:alpha:]]|[[:digit:]]|[[:space:]])*\]

would match [foo], that is OK. The mistake is that after it comes:

\((\/|\.).*\)

You need to be careful when you include .* in your regexes, because it is very, very greedy! That will match (../../relative_path/foobar.md) is the path to another file, also see [bar](/absolute/path/foobar.md). Concatenating, the whole line has been matched.

You should go for

grep -Eon '\[([[:alnum:]]|[[:space:]])*\]\((\.|\/)[^)]*\)'

The key was to replace .* by [^)]*, requiring the latter regex to stop short when if a closing parenthesis comes in its way. Also, I've applied this change:

[[:alpha:]]|[[:digit:]] can be collapsed into [[:alnum:]]

Output:

1:[foo](../../relative_path/foobar.md)
1:[bar](/absolute/path/foobar.md)

(I have 1: instead of 10: because it is the first line in my file.)

Stéphane Chazelas · Answer 3 · 2020-06-09T06:49:42.847

grep -on '\[[^]]*\]([^)]*)'

May just be enough in your case. Do you really need to restrict what characters may occur within [...] and (...)?

If you want to require the part inside [...] to only be made of alnums or whitespace and the part inside (...) to start with either a / or a ., that would simply be:

grep -on '\[[[:alnum:][:space:]]*\]([./][^)]*)'

In any case, note the [^)]* instead of .*) as .* would swallow the closing ) and everything up to the right-most ) on the line.

No need for -E's | alternation operator here. To match a single character, you can use the [set] bracket expression, where the set can include several characters or character classes (here [:alnum:], short for [:alpha:][:digit:] and [:space:]).

binarysta · Answer 4 · 2020-06-22T01:39:23.233

2

Need to use non-greedy grep

Added ? after .* in ((\/|\.).*?\)

grep -Pon '\[([[:alpha:]]|[[:digit:]]|[[:space:]])*\]\((\/|\.).*?\)' /path/to/file.md

10:[foo](../../relative_path/foobar.md)
10:[bar](/absolute/path/foobar.md)

-P for non greedy support. The regex should be in perl syntax

edited Jun 22 '20 at 01:39

answered Jun 08 '20 at 17:41

binarysta

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How do you search and print multiple matches per line with grep -o?

4 Answers4