I'm trying to match some lines against regex that contains digits.
Bash version 3.2.25:
#!/bin/bash
s="AAA (bbb 123) CCC"
regex="AAA \(bbb \d+\) CCC"
if [[ $s =~ $regex ]]; then
echo $s matches $regex
else
echo $s doesnt match $regex
fi
Result:
AAA (bbb 123) CCC doesnt match AAA \(bbb \d+\) CCC
If I put regex="AAA \(bbb .+\) CCC" it works but it doesn't meet my requirement to match digits only.
Why doesn't \d+ match 123?
Either use standard character set or POSIX-compliant notation:
[0-9]
[[:digit:]]
As read in Finding only numbers at the beginning of a filename with regex:
\dand\wdon't work in POSIX regular expressions, you could use[:digit:]though
so your expression should be one of these:
regex="AAA \(bbb [0-9]+\) CCC"
# ^^^^^^
regex="AAA \(bbb [[:digit:]]+\) CCC"
# ^^^^^^^^^^^^
All together, your script can be like this:
#!/bin/bash
s="AAA (bbb 123) CCC"
regex="AAA \(bbb [[:digit:]]+\) CCC"
if [[ $s =~ $regex ]]; then
echo "$s matches $regex"
else
echo "$s doesn't match $regex"
fi
Let's run it:
$ ./digits.sh
AAA (bbb 123) CCC matches AAA \(bbb [[:digit:]]+\) CCC
$regex is replaced by the string containing its content? I.e. if [[ $s =~ "AAA \(bbb [[:digit:]]+\) CCC" ]]; then.Digit notation \d doesn't work with your bash version. Use [0-9] instead:
regex="AAA \(bbb [0-9]+\) CCC"
[0-9]+ doesn't work either, but [0-9]* does. Maybe + is unsupported?
[0-9]+ worked for me on older BASH 3.2 as well so not sure why Ubuntu BASH is not liking it.grep that doesn't handle + (at least not without additional options). I got confused because my script both uses bash regex matching and grep.grep by default uses BRE so + needs to be escaped. Otherwise you can use grep -E to support extended regex like above.
[0-9]{1,3} for 1 to three digit numbers work?