expr string matching

Question

I want to match one or more occurance for ex. of blah using expr.

This works.

$ expr "blahblahblah" : 'blahblah'
8

What is wrong with this regular expression?

$ expr "blahblahblah" : '\(blah\)\+'
blah

I want the number of characters matched.

Answer 1

Since your question is tagged bash, there are better facilities than expr in modern versions of the shell, which do exactly what you want:

$ re='(blah)+'
$ [[ foo_blahblah_bar =~ $re ]] && echo "${#BASH_REMATCH[0]}"
8

Answer 2

Using sed:

echo "blahblahblah" | sed -n 's!\(\(blah\)*\).*!\1!p' | wc -c

Answer 3

First of all, you need \(\) instead of () and \+ instead of +. But that is not all.

You can't use groups () and get the length of matched string simultaneously.

Pattern matches return the string matched between ( and ) or null; if ( and ) are not used, they return the number of characters matched or 0.

You must use wc to get the length of the string:

$ expr "blahblahblahblah" : '\(\(blah\)\+\)' | wc -c
17

Or using parameters expansion:

$ m=$(expr "blahblahblahblah" : '\(\(blah\)\+\)') 
$ echo ${#m}
16

(wc -c counts the end of line also, hence the difference).

But if you can write the regular expression without groups, you get the length:

$ expr "blahhhhhbl" : "blah\+"
8