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← Revision 16 as of 2022-11-23 19:29:49 ⇥
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Comment: extglobs are only on by default inside [[ ]] in bash 4.1+
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# Bourne | |
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And here's one that uses `=~` (which requires bash 3.0 or higher): | In Bash and ksh, [[glob|Extended globs]] can also do this within a `[[` command: |
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regex='^(foo|bar|more)$' if [[ $var =~ $regex ]]; then |
# bash/ksh if [[ $var == @(foo|bar|more) ]]; then |
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This one '''only works in bash 3.1''' and '''some 3.2 revisions''' (it is untested in 3.0): | Extended globs are turned on by default inside the `[[` command in bash 4.1 and newer. If you need to target an older version of bash, you will need to turn them on in your script (`shopt -s extglob` outside of all functions or compound commands). Alternatively, you may loop over a list of patterns, checking each individually. |
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if [[ $var =~ '^(foo|bar|more)$' ]]; then ... fi |
#!/usr/bin/env bash # bash/ksh93 [[ -v BASH_VERSION ]] && shopt -s extglob # usage: pmatch string pattern [ pattern ... ] function any { [[ -n $1 ]] || return typeset pat match=$1 shift for pat; do [[ $match == $pat ]] && return done return 1 } var='foo bar' if any "$var" '@(bar|baz)' foo\* blarg; then echo 'var matched at least one of the patterns!' fi |
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The `=~` operator behavior changes drastically between 3.1 and 3.2, so be careful with it. The above expression is tested to work in bash ''3.1'' and ''3.2.{13,15,17}''; and it doesn't work in ''3.2.0''. Please also note that the regexp does not need to be quoted in the 3.2 revisions where it works. --[[redondos]] |
For logical conjunction (return true if `$var` matches all patterns), ksh93 can use the `&` pattern delimiter. |
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Normally I would never advocate [[BashFAQ/050|sticking code into a variable]] and attempting to use it -- lots of people have ''enormous'' trouble because they try to do that. In the case of `=~`, though, it seems to be required. Personally, I'd just stick with the `case`. --GreyCat | {{{ # ksh93 only [[ $var == @(foo&bar&more) ]] && ... }}} |
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The regexp works as long as it is quoted with bash version ''3.00.15(1)-release'' on CentOS4/RHEL4. I didn't bother checking whether bash is a patched version or not in CentOS4/RHEL4. --ellingsw | For shells that support only the ksh88 subset (extglob patterns), you may [[http://en.wikipedia.org/wiki/DeMorgan%27s_Law | DeMorganify]] the logic using the negation sub-pattern operator. {{{ # bash/ksh88/etc... [[ $var == !(!(foo)|!(bar)|!(more)) ]] && ... }}} But this is quite unclear and not much shorter than just writing out separate expressions for each pattern. ---- CategoryShell |
I want to check if [[ $var == foo || $var == bar || $var == more ]] without repeating $var n times.
The portable solution uses case:
# Bourne case $var in foo|bar|more) ... ;; esac
In Bash and ksh, Extended globs can also do this within a [[ command:
# bash/ksh if [[ $var == @(foo|bar|more) ]]; then ... fi
Extended globs are turned on by default inside the [[ command in bash 4.1 and newer. If you need to target an older version of bash, you will need to turn them on in your script (shopt -s extglob outside of all functions or compound commands).
Alternatively, you may loop over a list of patterns, checking each individually.
# bash/ksh93 [[ -v BASH_VERSION ]] && shopt -s extglob # usage: pmatch string pattern [ pattern ... ] function any { [[ -n $1 ]] || return typeset pat match=$1 shift for pat; do [[ $match == $pat ]] && return done return 1 } var='foo bar' if any "$var" '@(bar|baz)' foo\* blarg; then echo 'var matched at least one of the patterns!' fi
For logical conjunction (return true if $var matches all patterns), ksh93 can use the & pattern delimiter.
# ksh93 only [[ $var == @(foo&bar&more) ]] && ...
For shells that support only the ksh88 subset (extglob patterns), you may DeMorganify the logic using the negation sub-pattern operator.
# bash/ksh88/etc... [[ $var == !(!(foo)|!(bar)|!(more)) ]] && ...
But this is quite unclear and not much shorter than just writing out separate expressions for each pattern.