CODEWITHSUNDEEP
htmlxpath
alireza
alireza

Reputation: 557

Select URL from an attribute using XPath

I would like to retrieve the file url from a HTML meta-tag, given in the content attribute.

Here is the example HTML code: 

 <meta content="https://www.domain.com/player/player-viral.swf?config=
https://www.domain.com/configxml?id=133291&logo.
link=http://www.domain.org/Amin+Rostami/-/Havam+Toei&
image=https://www.domain.com/img/3lv68bc5w-1396897306.jpeg&provider=audio&
file=http://s10.domain.me/music/A/[one]/test-msusic.mp3" property="og:video"/>

I would like to get the file url, in this case http://s10.domain.me/music/A/[one]/test-msusic.mp3

Upvotes: 1

Views: 251

Answers (1)

alecxe
alecxe

Reputation: 473863

You can use substring-after() to extract the link after the file= from the content attribute of a meta tag:

substring-after(//meta/@content, "file=")

Demo (using xmllint):

$ cat input.xml
<meta content="https://www.domain.com/player/player-viral.swf?config=
https://www.domain.com/configxml?id=133291&amp;logo.
link=http://www.domain.org/Amin+Rostami/-/Havam+Toei&amp;
image=https://www.domain.com/img/3lv68bc5w-1396897306.jpeg&amp;provider=audio&amp;
file=http://s10.domain.me/music/A/[one]/test-msusic.mp3" property="og:video"/>

$ $ xmllint input.xml --xpath 'substring-after(//meta/@content, "file=")'
http://s10.domain.me/music/A/[one]/test-msusic.mp3

Upvotes: 2

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