In bash, how can I get a URL from a markup file without using regex lookarounds?

Question

I want to grep the URL out of a .asx file. The file would normally look like this.

<ASX VERSION="3.0">
<ENTRY>
<TITLE>Blah Blah</TITLE>
<AUTHOR>Someone</AUTHOR>
<COPYRIGHT>（C）2014 Someone Else</COPYRIGHT>
<REF HREF="mms://www.example.com/video/FilmName/FilmName.wmv"/>
</ENTRY>
</ASX>

I want to get the URL without the quotes, and stripping off the mms://

I came up with a regex that uses lookarounds that does this successfully:

((?<=\/\/).*?).(?=\")

but of course I can't use this with grep. So what is another approach that would be flexible to capture whatever comes between the mms:// and the " that I could put into a grep -o command?

Of course, [don't use regex to parse XML](http://stackoverflow.com/questions/1732348/regex-match-open-tags-except-xhtml-self-contained-tags/1732454#1732454). — tripleee, Jan 12 '14 at 17:08

score 1 · Accepted Answer · edited Jan 12 '14 at 18:09

1

but of course I can't use this with grep.

Why not? Modern grep versions supports -P switch for PCRE regex support.

Try this:

grep -oP '((?<=//).*?).(?=")' file
www.example.com/video/FilmName/FilmName.wmv

edited Jan 12 '14 at 18:09

Lightness Races in Orbit

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answered Jan 12 '14 at 16:15

anubhava

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Wouldn't `'(?<=://)[^"]+'` be sufficient for this? – Jonny 5 Jan 12 '14 at 16:32
1

Oh yes `grep -oP '(?<=://).*?(?=")'` would be fine too, I just used OP's regex. – anubhava Jan 12 '14 at 16:33
@LightnessRacesinOrbit: Thanks for correcting it, point well taken, cheers :) – anubhava Jan 12 '14 at 18:15

Jotne · Answer 2 · 2014-01-13T06:38:33.467

0

Like this:

awk -F '[:"]' '/REF HREF/ {print substr($3,3)}' file
www.example.com/video/FilmName/FilmName.wmv

edited Jan 13 '14 at 06:38

answered Jan 12 '14 at 16:11

Jotne

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Nice; perhaps `awk -F '[:"]' '/REF HREF/ {print substr($3,3)}' file` is simpler. – mklement0 Jan 12 '14 at 21:18

score 0 · Answer 3 · answered Jan 12 '14 at 18:22

With BASH, you can use the left/right pattern matching:

url='<REF HREF="mms://www.example.com/video/FilmName/FilmName.wmv"/>'
url=${url#<REF HREF=\"}
url=${url%\"/>}
echo "URL is '$url'"   # Prints URL is 'mms://www.example.com/video/FilmName/FilmName.wmv'

${VAR#pattern} strips off of $VAR the shortest left hand side glob that matches pattern. ${VAR##pattern}strips off of $var the largest left hand side glob that matches pattern. And, ${VAR%pattern} and ${VAR%pattern} do the same for the right hand side of $VAR.

An easy way to remember is that # is to the left of % on the keyboard. David Korn taught me that.

score 0 · Answer 4 · answered Jan 12 '14 at 21:14

0

Solution for OSX users, where grep (as of OSX 10.9) doesn't support -P and look-arounds are therefore not an option:

egrep -o '"[a-z]+://[^"]+' file | cut -d '/' -f 3-

answered Jan 12 '14 at 21:14

mklement0

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In bash, how can I get a URL from a markup file without using regex lookarounds?

4 Answers4