Is there any easy way to extract this URL in bash/or PHP?
http://shop.image-site.com/images/2/format2013/fullies/kju_product.png
From this HTML code?
<a href="javascript: open_window_zoom('http://shop.image-site.com/image.php?image=http://shop.image-site.com/images/2/format2013/fullies/kju_product.png&pID=31777&download=kju.png&name=13011 KELLYS Kju: 490mm (19.5")',550,366);">
With perl you could do a match and a capture
perl -n -e 'print "$1\n" if (m/image=(.*?)\&/);'
This captures everything between image= and the next & and prints it $1.
For more on regular expressions, see perlre or http://www.regular-expressions.info/
In bash, you can try the following:
sed 's/.*image=\(http:\/\/[^&]*\).*/\1/g'
Update:
The solution above performs substitution rather than extraction. The line containing the pattern (required url) is replaced by the pattern itself. However, the substitution isn't in-place.
^ or $ in my solution.^ and $. It all comes down to greedy matching, like sputnick said.Whichever way you decide to dress it up, you could simply split with the delimiter equal to ?image= and then split the second token you receive (i.e. result[1]) with a simple & delimiter. The first result from that split is your answer.
However, a pure regex match would look something like: m#image=(a-z0-9\:/\.\-)&#i. You can take that regex and put it wherever you want to get your result stored in $1. Despite what a lot of people think, you do not have to match the beginning of a line and the end of a line to match a result.
Try doing this :
xmllint --html --xpath '//a/@href' file://file.html |
grep -oP 'image=\Khttp://.*?\.png'
You can use an URL instead of a local file :
http://domain.tld/path
Or if you had already extracted the line to parse in the $string variable :
grep -oP 'image=\Khttp://.*?\.png' <<< "$string"
