Bash Regular Expression Condition

Question

I have a regular expression that I need to verify. The regular expression has double quotes in it, but I can't seem to figure out how to properly escape them.

First attempt, doesn't work as the quotes are not escaped.

while read line
do
  if [[ $line =~ "<a href="(.+)">HTTP</a>" ]]; then
    SOURCE=${BASH_REMATCH[1]}
    break
  fi
done < tmp/source.html

echo "{$SOURCE}" #output = {"link.html"} (with double quotes)

How can I properly run this so the output is link.html without double quotes.

I have tried...

while read line
do
  if [[ $line =~ "<a href=/"(.+)/">HTTP</a>" ]]; then
    SOURCE=${BASH_REMATCH[1]}
    break
  fi
done < tmp/source.html

echo "{$SOURCE}" #output = {}

Without luck. Can someone please help me so I can stop beating my head on my desk? I am not great with Bash. Thank you!

Answer 1

It's always best to put your regex in a variable.

pattern='<a href="(.+)">HTTP</a>'
while read line
do
  if [[ $line =~ $pattern ]]; then
    SOURCE=${BASH_REMATCH[1]}
    break
  fi
done < tmp/source.html

echo "{$SOURCE}" #output = {link.html} (without double quotes)

If you quote the right hand side (the pattern), it changes the match from regex to a simple string equal (=~ effectively becomes ==).

As a side note, escaping is done with backslashes (\) rather than slashes (/), but that would not help your situation because of the outer quotes as mentioned in my previous paragraph.