Find strings in files with regex in bash

Question

I have a bunch of files, where template tags are used. I want to list all the different template tags used in all of the files.

This is my used bash script, which gives no match...

for f in EHS/*.html;  do
  value=`cat ${f}`;
  [[ $value =~ "({%.*?%})" ]]
  if [[ ${BASH_REMATCH[0]} ]]; then
    echo "Found: ${BASH_REMATCH[0]}";
  fi
done;

This is a snippet from one of the html files:

<p>
  The ordernumber is: {%OrderNumber%}<br>
  The partnumer is: {%PartNumber%}
</p>

So my goal is to just return all of the different tags used...

I think grep -o "{%[^%]*%}" would be more natural to get all the matches like that from the file. In Bash, it is combersome to extract multiple matches. — Wiktor Stribiżew
– Wiktor Stribiżew, Commented Jul 12, 2018 at 12:06
Seems to work :) Can you explain the difference with my regular expression? — Milkmannetje
– Milkmannetje, Commented Jul 12, 2018 at 12:11
Bash regex does not support lazy quantifiers, and when you quote the pattern after =~ inside [[...]], literal string matching is triggered, it is not regarded as a regex. You could use if [[ $s =~ \{%[^%]*%} ]]; but you will find a single match only. — Wiktor Stribiżew
– Wiktor Stribiżew, Commented Jul 12, 2018 at 12:15

choroba · Accepted Answer · 2018-07-12 12:11:00Z

4

Two problems:

The regex should be unquoted, but {} needs escaping, as their meaning is special.
bash doesn't support frugal quantifiers like *?.

It's easier to use grep:

grep -o '{%[^}]*%}'

The -o option only returns the matching parts, one per line.

Note that strings like {%ab%cd}ef%} aren't matched as there's no easy way how to prevent parts of multicharacter delimiters in standard grep. With pgrep, you can use

grep -o -P '{%.*?%}'

as you originally intended.

answered Jul 12, 2018 at 12:11

choroba

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