CODEWITHSUNDEEP
bash
Vladimír Fencák
Vladimír Fencák

Reputation: 186

grep find letters between two spaces

I have to find words.

In my assignment a word is defined as letters between two spaces (" bla "). I have to find a decimalIntegerConstant like this but it has to be a word.

I use

 grep -E -o " (0|[1-9]+[0-9]*)([Ll]?) "

but it doesn't work on, for example:

bla 0l labl 2 3 abla0La 0L sfdgpočítačsd

Output is

0l
2 
0L

but 3 is missing.

Upvotes: 1

Views: 382

Answers (1)

n. m. could be an AI
n. m. could be an AI

Reputation: 119877

Matches don't overlap. Your regex have matched 2. The blank after 2 is gone. It won't be considered for further matches.

POSIX grep cannot do what you want in one step, but you can do something like this in two stages (simplified from your regex, doesn't support [lL])

grep -o ' [0-9 ]* ' | grep -E -o '[0-9]+'

That is, match a sequence of space-separated numbers with leading and trailing spaces, and from that, match individual numbers regardless of spaces. De-simplify the definition of number to suit your needs.

Perl-compatible regular expressions have a way to match stuff without consuming it, for example, as mentioned in the comments:

grep -oP " (0|[1-9]+[0-9]*)[Ll]?(?= )"

(?= ) is a lookahead assertion, which means grep will look ahead in the input stream and make sure the match is followed by a space. The space will not be considered a part of the match and will not be consumed. When no space is found, the match fails.

PCRE are not guaranteed to work in all implementations of grep.

Edit: -o is not specified by Posix either.

Upvotes: 1

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