Question

I am trying to determine the correct RegEx syntax to perform the following. I have line in a file in which I want to match every character before the first occurrence of white space.

so for example in the line:

123abc xyz foo bar

it is unclear to me why the following:

^.*\s

is matching up to the b in the word bar:

123abc xyz foo

It appears to me that the \s is greedy, however I am not certain how I can make it not greedy and just match 123abc I have tried various forms of this regex in an attempt to make it non-greedy ^.*\s? or something like this, however I have been unsuccessful. Thank you in advance

Answer 1

That is because . can be any character, including space. You can try

^[^ ]*\s

or

^\S*\s

instead.

That is a greedy re. But you can make non-greedy re also:

^.*?\s

You mistake is that you have placed ? on a wrong place.

Examples:

$ echo aaaa bbb cccc dddd > re.txt
$ cat re.txt
aaaa bbb cccc dddd
$ egrep -o '^.*\s' re.txt
aaaa bbb cccc 
$ egrep -o '^\S*\s' re.txt
aaaa 
$ egrep -o '^[^ ]*\s' re.txt
aaaa 

And non-greedy search with perl:

$ perl -ne 'print "$1\n" if /^(.*?)\s/' re.txt
aaaa