Bash Regular Expressions
May 26th, 2008 by Mitch Frazier in
When working with regular expressions in a shell script the norm is to use grep or sed or some other external command/program. Since version 3 of bash (released in 2004) there is another option: bash's built-in regular expression comparison operator "=~".
Bash's regular expression comparison operator takes a string on the left and an extended regular expression on the right. It returns 0 (success) if the regular expression matches the string, otherwise it returns 1 (failure).
In addition to doing simple matching, bash regular expressions support sub-patterns surrounded by parenthesis for capturing parts of the match. The matches are assigned to an array variable BASH_REMATCH. The entire match is assigned to BASH_REMATCH[0], the first sub-pattern is assigned to BASH_REMATCH[1], etc..
The following example script takes a regular expression as its first argument and one or more strings to match against. It then cycles through the strings and outputs the results of the match process:
#!/bin.bash
if [[ $# -lt 2 ]]; then
echo "Usage: $0 PATTERN STRINGS..."
exit 1
fi
regex=$1
shift
echo "regex: $regex"
echo
while [[ $1 ]]
do
if [[ $1 =~ $regex ]]; then
echo "$1 matches"
i=1
n=${#BASH_REMATCH[*]}
while [[ $i -lt $n ]]
do
echo " capture[$i]: ${BASH_REMATCH[$i]}"
let i++
done
else
echo "$1 does not match"
fi
shift
done
Assuming the script is saved in "bashre.sh", the following sample shows its output:
# sh bashre.sh 'aa(b{2,3}[xyz])cc' aabbxcc aabbcc
regex: aa(b{2,3}[xyz])cc
aabbxcc matches
capture[1]: bbx
aabbcc does not match
__________________________
Mitch Frazier is an Associate Editor for Linux Journal and the Web Editor for linuxjournal.com.
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bash-regular-expressions
On January 30th, 2009 kenb (not verified) says:
I wrote the suggested bash script and got the demonstrated result. However when I invoked:
sh bashre.sh 'aa(b{2,3}[xyz])cc' aabbxcc aabbcc aabbyccaabbbzcc
I expected to get two matches with the last parameter but I only got one. I'm surprised
that I'm the only one so what did I do wrong?
regex: aa(b{2,3}[xyz])cc
aabbxcc matches
capture[1]: bbx
aabbcc does not match
aabbyccaabbbzcc matches
capture[1]: bby
add color and better indentation to the output
On July 7th, 2008 Albert Bicchi (not verified) says:
#!/bin/sh if [[ $# -lt 2 ]]; then echo "Usage: regex PATTERN STRINGS..." exit 1 fi regex=$1 shift echo "regex: $regex" echo while [[ $1 ]] do if [[ $1 =~ $regex ]]; then echo -e "\t\E[42;37m${1} - matches\E[33;0m" i=1 n=${#BASH_REMATCH[*]} while [[ $i -lt $n ]] do echo -e "\t\t\E[43;37mcapture[$i]: ${BASH_REMATCH[$i]}\E[33;0m" let i++ done else echo -e "\t\E[41;37m${1} - does not match\E[33;0m" fi shift doneIs "(( $# < 2 ))" an
On June 25th, 2008 Anonymous (not verified) says:
Is "(( $# < 2 ))" an alternative conditional expression for the line "[[ $# -lt 2 ]]"?
Could you discuss BASH expressions with [[]] (()) and their valid operators. It seems the -lt, -gt, -a,
etc, can be replaced with <, >, &&, etc, if used with (()) --- replacing [] with (()) (numeric) and [[]] (strings).
Thank you.
PS: The captcha is really hard to read. It would be nice it there was an option to generate a new one that could possible be read by a mere human.
Is "(( $# < 2 ))" an
On June 25th, 2008 Anonymous (not verified) says:
Is "(( $# < 2 ))" an alternative conditional expression for the line "[[ $# -lt 2 ]]"?
Could you discuss BASH expressions with [[]] (()) and their valid operators. It seems the -lt, -gt, -a,
etc, can be replaced with <, >, &&, etc, if used with (()) --- replacing [] with (()) (numeric) and [[]] (strings).
Thank you.
That simple?!
On May 28th, 2008 Robert de Bock (not verified) says:
My god, Bash sure is a great tool! Thanks for the information.
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