How can I match spaces with a regexp in Bash?
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:]]
.