sed regex to non-greedy replace?

Question

I am aware of another question that is quite similar, but for some reason I'm still having problems.

I have a GC log that I'm trying to trim out the Tenured section enclosed in [].

63.544: [GC 63.544: [DefNew: 575K->63K(576K), 0.0017902 secs]63.546: [Tenured: 1416K->1065K(1536K), 0.0492621 secs] 1922K->1065K(2112K), 0.0513331 secs]

I apply s/\[Tenured:.*\]//

And quite expectantly, the result is trimmed greedily through the remainder of the line:

63.544: [GC 63.544: [DefNew: 575K->63K(576K), 0.0017902 secs]63.546:

So let's try and be non-greedy not match a closing right bracket with s/\[Tenured:[^\]]*\]// but alas no match is made and sed skips over the line, producing the same original output:

63.544: [GC 63.544: [DefNew: 575K->63K(576K), 0.0017902 secs]63.546: [Tenured: 1416K->1065K(1536K), 0.0492621 secs] 1922K->1065K(2112K), 0.0513331 secs]

How do I non-greedily match and replace that section? Thanks,

score 5 · Accepted Answer · answered Nov 06 '09 at 05:34

5

Almost: s/\[Tenured:[^]]*\]//

The manual says:

To include a literal ']' in the list, make it the first character (following a possible '^').

i.e. No backslash is required in this context.

Raz

answered Nov 06 '09 at 05:34

Roland Turner

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score 3 · Answer 2 · answered Nov 06 '09 at 05:34

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sed -e 's/\[Tenured:[^]]*\]//'

Apparently you shouldn't escape the close square bracket. Wacky!

From man re_format:

A bracket expression is a list of characters enclosed in '[]' ... To include a literal ']' in the list, make it the first character (following a possible `^').

answered Nov 06 '09 at 05:34

Laurence Gonsalves

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Wow, that is wacky. I knew about the `^` situation, but not the `]`. – Jé Queue Nov 06 '09 at 05:42

score 0 · Answer 3 · edited May 23 '17 at 10:32

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Try .*? for the non-greedy variant of .*. (Not sure if it's supported in sed's regex engine or not, but it's worth a try.)

Edit: This previous SO question may be relevant - Non greedy regex matching in sed?

edited May 23 '17 at 10:32

Community

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answered Nov 06 '09 at 05:22

Amber

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I don't believe sed supports that, but regardless I just tried it again and no joy. – Jé Queue Nov 06 '09 at 05:24
See mentioned question http://stackoverflow.com/questions/1103149/non-greedy-regex-matching-in-sed – Krzysztof Wolny Jan 05 '12 at 14:38
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sed supports it with the -r (GNU) or -E (BSD) flags. – Will Tice Jul 22 '13 at 05:33

score 0 · Answer 4 · answered Nov 06 '09 at 05:34

0

This worked:

echo "63.544: [GC 63.544: [DefNew: 575K->63K(576K), 0.0017902 secs]63.546: [Tenured: 1416K->1065K(1536K), 0.0492621 secs] 1922K->1065K(2112K), 0.0513331 secs]" | sed -e s/\\[Tenured:[^\]]*\\]//

answered Nov 06 '09 at 05:34

I think that works because the shell was doing the escaping prior to getting to sed. The same would not be true unfortunately for a sed script file. – Jé Queue Nov 06 '09 at 05:44

sed regex to non-greedy replace?

4 Answers4

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