Extract substring using regexp in plain bash
I'm trying to extract the time from a string using bash, and I'm having a hard time figuring it out.
My string is like this:
US/Central - 10:26 PM (CST)
And I want to extract the 10:26
part.
Anybody knows of a way of doing this only with bash - without using sed, awk, etc?
Like, in PHP I would use - not the best way, but it works - something like:
preg_match( ""(\d{2}\:\d{2}) PM \(CST\)"", "US/Central - 10:26 PM (CST)", $matches );
Thanks for any help, even if the answer uses sed or awk
Solution 1:
Using pure bash :
$ cat file.txt
US/Central - 10:26 PM (CST)
$ while read a b time x; do [[ $b == - ]] && echo $time; done < file.txt
another solution with bash regex :
$ [[ "US/Central - 10:26 PM (CST)" =~ -[[:space:]]*([0-9]{2}:[0-9]{2}) ]] &&
echo ${BASH_REMATCH[1]}
another solution using grep
and look-around advanced regex :
$ echo "US/Central - 10:26 PM (CST)" | grep -oP "\-\s+\K\d{2}:\d{2}"
another solution using sed :
$ echo "US/Central - 10:26 PM (CST)" |
sed 's/.*\- *\([0-9]\{2\}:[0-9]\{2\}\).*/\1/'
another solution using perl :
$ echo "US/Central - 10:26 PM (CST)" |
perl -lne 'print $& if /\-\s+\K\d{2}:\d{2}/'
and last one using awk :
$ echo "US/Central - 10:26 PM (CST)" |
awk '{for (i=0; i<=NF; i++){if ($i == "-"){print $(i+1);exit}}}'
Solution 2:
echo "US/Central - 10:26 PM (CST)" | sed -n "s/^.*-\s*\(\S*\).*$/\1/p"
-n suppress printing
s substitute
^.* anything at the beginning
- up until the dash
\s* any space characters (any whitespace character)
\( start capture group
\S* any non-space characters
\) end capture group
.*$ anything at the end
\1 substitute 1st capture group for everything on line
p print it
Solution 3:
Quick 'n dirty, regex-free, low-robustness chop-chop technique
string="US/Central - 10:26 PM (CST)"
etime="${string% [AP]M*}"
etime="${etime#* - }"
Solution 4:
If your string is
foo="US/Central - 10:26 PM (CST)"
then
echo "${foo}" | cut -d ' ' -f3
will do the job.