问题
I have two strings, say foo_bar and foo_abc_bar. I would like to match both of them, and if the first one is matched I would like to emphasize it with = sign. So, my guess was:
echo 'foo_abc_bar' | sed -r 's/(foo).*(abc)?.*(bar)/\1=\2=\3/g'
> foo==bar
or
echo 'foo_abc_bar' | sed -r 's/(foo).*((abc)?).*(bar)/\1=\2=\3/g'
> foo==
But as output above shows none of them work.
How can I specify an optional group that will match if the string contains it or just skip if not?
回答1:
The solution:
echo 'foo_abc_bar' | sed -r 's/(foo)_((abc)_)?(bar)/\1=\3=\4/g'
Why your previous attempts didn't work:
.* is greedy, so for the regex (foo).*(abc)?.*(bar) attempting to match 'foo_abc_bar' the (foo) will match 'foo', and then the .* will initially match the rest of the string ('_abc_bar'). The regex will continue until it reaches the required (bar) group and this will fail, at which point the regex will backtrack by giving up characters that had been matched by the .*. This will happen until the first .* is only matching '_abc_', at which point the final group can match 'bar'. So instead of the 'abc' in your string being matched in the capture group it is matched in the non-capturing .*.
Explanation of my solution:
The first and most important thing is to replace the .* with _, there is no need to match any arbitrary string if you know what the separator will be. The next thing we need to do is figure out exactly which portion of the string is optional. If the strings 'foo_abc_bar' and 'foo_bar' are both valid, then the 'abc_' in the middle is optional. We can put this in an optional group using (abc_)?. The last step is to make sure that we still have the string 'abc' in a capturing group, which we can do by wrapping that portion in an additional group, so we end up with ((abc)_)?. We then need to adjust the replacement because there is an extra group, so instead of \1=\2=\3 we use \1=\3=\4, \2 would be the string 'abc_' (if it matched). Note that in most regex implementations you could also have used a non-capturing group and continued to use \1=\2=\3, but sed does not support non-capturing groups.
An alternative:
I think the regex above is your best bet because it is most explicit (it will only match the exact strings you are interested in). However you could also avoid the issue described above by using lazy repetition (matches as few characters as possible) instead of greedy repetition (matches as many characters as possible). You can do this by changing the .* to .*?, so your expression would look something like this:
echo 'foo_abc_bar' | sed -r 's/(foo).*?(abc).*?(bar)/\1=\2=\3/g'
回答2:
Maybe you could simply use:
echo 'foo_abc_bar' | sed -r 's/(foo|bar|abc)_?/\1=/g'
echo 'foo_bar' | sed -r 's/(foo|bar|abc)_?/\1=/g'
> foo=abc=bar=
> foo=bar=
This avoids the foo==bar you get with foo_bar and I found it a bit weird to show emphasis by putting = sometimes before the match, sometimes after the match.
来源：https://stackoverflow.com/questions/16719138/why-sed-doesnt-print-an-optional-group