Bash Pattern Mathcing string.*char

Question

I'm trying to match this pattern in grep that takes the form:

string.*char

But it doesn't match the char to the first occurrence.

Example:

command | grep -o '{id.*}'

The output of command is formatted something like:

*a lot of stuff I don't care about* {id: *comma sep. data*}, {*more data*}

I want it to output {id: *comma sep. data*} not {id: *comma sep. data*}, {*more data*}

Please add the output of `command`, the actual output of `grep` and your expected output. Add this by [edit]ing your question. — Socowi, Jul 07 '21 at 21:56
Are `{id` and `}` on the same line? `grep` matches a line at a time. — Barmar, Jul 07 '21 at 21:59
@Barmar yes they are on the same line and I'm aware grep only goes line by line. I did forget add -o flag though — David Wright, Jul 07 '21 at 22:08
Then your code looks like it should work. Please edit the question with the details requested by Socowi. — Barmar, Jul 07 '21 at 22:09
`.*` is greedy, it will match the longest possible string, not the shortest. — Barmar, Jul 07 '21 at 22:12
See https://stackoverflow.com/questions/3027518/how-to-do-a-non-greedy-match-in-grep — Barmar, Jul 07 '21 at 22:12

score 0 · Answer 1 · answered Jul 07 '21 at 22:18

0

As @Barmar replied, .* is greedy

grep -o '{id[}^]*}'

or

grep -o -P '{id.*?}'

will work to make it lazy

answered Jul 07 '21 at 22:18

David Wright

1 Answers1