问题
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,
回答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.
- Raz
回答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 `^').
回答3:
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?
回答4:
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:[^\]]*\\]//
来源：https://stackoverflow.com/questions/1685575/sed-regex-to-non-greedy-replace