How can I match spaces with a regexp in Bash?

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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.

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