How to get the URL from a file using a shell script

I have a file which consists of a URL. I'm trying to get the URL from that file using a shell script.

In the file, the URL is like this:

('URL', 'http://url.com');

I tried to use the following:

cat file.php | grep 'URL' | awk '{ print $2 }'

It gives the output as:

'http://url.com');

But I need to get only url.com in a variable inside the shell script. How can I accomplish this?

You can do everything with a simple grep:

grep -oP "http://\K[^']+" file.php

From man grep:

   -P, --perl-regexp
          Interpret  PATTERN  as  a  Perl  regular  expression  (PCRE, see
          below).  This is highly experimental and grep  -P  may  warn  of
          unimplemented features.
   -o, --only-matching
          Print  only  the  matched  (non-empty) parts of a matching line,
          with each such part on a separate output line.

The trick is to use \K which, in Perl regex, means discard everything matched to the left of the \K. So, the regular expression looks for strings starting with http:// (which is then discarded because of the \K) followed by as many non-' characters as possible. Combined with -o, this means that only the URL will be printed.

You could also do it in Perl directly:

perl -ne "print if s/.*http:\/\/(.+)\'.*/\$1/" file.php\

Something like this?

grep 'URL' file.php | rev | cut -d "'" -f 2 | rev

grep 'URL' file.php | cut -d "'" -f 4 | sed s/'http:\/\/'/''/g

To strip out http://.

Try this,

awk -F// '{print $2}' file.php | cut -d "'" -f 1

How to get the URL from a file using a shell script

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