sed regex to non-greedy replace?

regex unix replace sed garbage-collection

10,009

Solution 1

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.

Solution 2

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 `^').

10,009

Author by

Jé Queue

Updated on June 04, 2022

Comments

Jé Queue almost 2 years
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,
Jé Queue over 14 years

I don't believe sed supports that, but regardless I just tried it again and no joy.
Jé Queue over 14 years

Wow, that is wacky. I knew about the ^ situation, but not the ].
Krzysztof Wolny over 12 years

See mentioned question stackoverflow.com/questions/1103149/…
Will Tice almost 11 years

sed supports it with the -r (GNU) or -E (BSD) flags.