How to do a non-greedy match in grep?

问题:

I want to grep the shortest match and the pattern should be something like:

<car ... model=BMW ...>
...
...
...
</car>

... means any character and the input is multiple lines.

回答1:

You're looking for a non-greedy (or lazy) match. To get a non-greedy match in regular expressions you need to use the modifier ? after the quantifier. For example you can change .* to .*?.

By default grep doesn't support non-greedy modifiers, but you can use grep -P to use the Perl syntax.

回答2:

Actualy the .*? only works in perl. I am not sure what the equivalent grep extended regexp syntax would be. Fortunately you can use perl syntax with grep so grep -P would work but grep -E which is same as egrep would not work (it would be greedy).

See also: http://blog.vinceliu.com/2008/02/non-greedy-regular-expression-matching.html

回答3:

My grep that works after trying out stuff in this thread:

echo "hi how are you " | grep -shoP ".*? "

Just make sure you append a space to each one of your lines

(Mine was a line by line search to spit out words)

回答4:

grep

For non-greedy match in grep you could use a negated character class. In other words, try to avoid wildcards.

For example, to fetch all links to jpeg files from the page content, you'd use:

grep -o '"[^" ]\+.jpg"'

To deal with multiple line, pipe the input through xargs first. For performance, use ripgrep.

回答5:

The short answer is using the next regular expression:

(?s)<car .*? model=BMW .*?>.*?</car>

• (?s) - this makes a match across multiline
• .*? - matches any character, a number of times in a lazy way (minimal match)

A (little) more complicated answer is:

(?s)<([a-z\-_0-9]+?) .*? model=BMW .*?>.*?</\1>

This will makes possible to match car1 and car2 in the following text

<car1 ... model=BMW ...>
...
...
...
</car1>
<car2 ... model=BMW ...>
...
...
...
</car2>

• (..) represents a capturing group
• \1 in this context matches the sametext as most recently matched by capturing group number 1