How can I match spaces with a regexp in Bash?

Question

Replace:

regexp="templateUrl:[\s]*'"

With:

regexp="templateUrl:[[:space:]]*'"

According to man bash, the =~ operator supports "extended regular expressions" as defined in man 3 regex. man 3 regex says it supports the POSIX standard and refers the reader to man 7 regex. The POSIX standard supports [:space:] as the character class for whitespace.

The GNU bash manual documents the supported character classes as follows:

Within ‘[’ and ‘]’, character classes can be specified using the syntax [:class:], where class is one of the following classes defined in the POSIX standard:

alnum alpha ascii blank cntrl digit graph lower print
punct space upper word xdigit

The only mention of \s that I found in the GNU bash documentation was for an unrelated use in prompts, such as PS1, not in regular expressions.

The Meaning of *

[[:space:]] will match exactly one white space character. [[:space:]]* will match zero or more white space characters.

The Difference Between space and blank

POSIX regular expressions offer two classes of whitespace: [[:space:]] and [[:blank:]]:

• [[:blank:]] means space and tab. This makes it similar to: [ \t].

• [[:space:]], in addition to space and tab, includes newline, linefeed, formfeed, and vertical tab. This makes it similar to: [ \t\n\r\f\v].

A key advantage of using character classes is that they are safe for unicode fonts.

Get rid of the square brackets in the regular expression:

regexp="templateUrl:\s*'"

With the square brackets present, the \s inside gets interpreted literally as matching either the \ or s characters, but your intent is clearly to match against the white space character class for which \s is shorthand (and therefore no square brackets needed).

$ uname -a
Linux noname 3.13.0-24-generic #47-Ubuntu SMP Fri May 2 23:30:00 UTC 2014 x86_64 x86_64 x86_64 GNU/Linux
$ bash --version
GNU bash, version 4.3.11(1)-release (x86_64-pc-linux-gnu)
Copyright (C) 2013 Free Software Foundation, Inc.
License GPLv3+: GNU GPL version 3 or later <http://gnu.org/licenses/gpl.html>

This is free software; you are free to change and redistribute it. 
There is NO WARRANTY, to the extent permitted by law.
$ cat test.sh
str="{templateUrl: '}" 
regexp="templateUrl:\s*'"

if [[ $str =~ $regexp ]]; then
  echo "yes"
else
  echo "no"
$ bash test.sh
yes 

This should work:

#!/bin/bash
str="{templateUrl: '}"
regexp="templateUrl:[[:space:]]*'"

if [[ $str =~ $regexp ]]; then
  echo "yes"
else
  echo "no"
fi

If you want to match zero or more whitespaces the * needs to added after [[:space:]].