Regex in a bash scipt - regex

Question

I've got the following text file which contains:
12.3-456, test
test test test
If the line contains xx.x-xxx, then I want to print the line out. (X's are numbers)
I think I have the correct regex and have tested it here:
http://regexr.com/3clu3
I have then used this in a bash script but the line containing the text is not printed out.
What have I messed up?
#!/bin/bash
while IFS='' read -r line || [[ -n "$line" ]]; do
if [[ $line =~ /\d\d.\d-\d\d\d,/g ]]; then
echo $line
fi
done < input.txt

Answer 1

You need to use [0-9] instead of a \d in Bash regex. No regex delimiters are necessary, and the global flag is not necessary either. Also, you can contract it a bit using limiting quantifiers (like {3} that will match 3 occurrences of the pattern next to it). Besides, a dot matches any character in regex, so you need to escape it if you want to match a literal dot symbol.
Use
regex="[0-9]{2}\.[0-9]-[0-9]{3},"
if [[ $line =~ $regex ]]
...

Answer 2

This works:
#!/bin/bash
#regex="/\d\d.\d-\d\d\d,/g"
regex="[0-9\.\-]+\, [A-Za-z]+"
while IFS='' read -r line || [[ -n "$line" ]]; do
echo $line
if [[ $line =~ $regex ]]; then
echo "match"
fi
done
regex is [any of 0-9, '.', '-'] followed by ',' followed by alphachars. This could be refined in a number of ways - e.g. explicit places before/ after '-'.
Testing indicates:
$ ./sqltrace2.sh < input.txt
12.3-456, test
match
123.3-456, test
match
12.3-456,
test test test
test test test