preg match for this pattern

Question

I am trying to get a link from a preg match but it is not giving any output

Here is pattern:

    <a href="http://media1.xyz.com/full5/Animals/Rabbits/rabbits-16a.jpg" download="rabbits-16a"><img onmouseover="showFullScreen('inline')" onmouseout="showFullScreen('none')" src="http://media1.xyz.com/full5/Animals/Rabbits/rabbits-16a.jpg" id="wall" border="0" align="middle"  width="1920" height="1080"  alt="Rabbits 1920x1080 Wallpaper # 17" title="Rabbits HD Wallpaper #17" /></a>

I have tried this:

     preg_match("/\<a href=/\"(.*)\">/",$str,$title);

I want to get the link of href only.

You should at least activate error reporting! http://3v4l.org/Y5kiL — deceze, Oct 14 '13 at 12:45

score -1 · Answer 1 · answered Oct 14 '13 at 12:47

-1

Using this match pattern should work

'/href=\"(.*?)\"/s'

I use it for matching the source in image tags I'm parsing from IMAP emails (in its original form '/src=\"(.*?)\"/s' But just changing src to href will match for you

answered Oct 14 '13 at 12:47

Dave

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score -1 · Accepted Answer · answered Oct 14 '13 at 12:52

$input = '<a href="http://media1.xyz.com/full5/Animals/Rabbits/rabbits-16a.jpg" download="rabbits-16a"><img onmouseover="showFullScreen(\'inline\')" onmouseout="showFullScreen(\'none\')" src="http://media1.xyz.com/full5/Animals/Rabbits/rabbits-16a.jpg" id="wall" border="0" align="middle"  width="1920" height="1080"  alt="Rabbits 1920x1080 Wallpaper # 17" title="Rabbits HD Wallpaper #17" /></a>';

preg_match('/href=\"(.+?)\"/',$input,$title);

echo $title[1];

Which will output:

http://media1.xyz.com/full5/Animals/Rabbits/rabbits-16a.jpg

http://3v4l.org/QnJ9e

preg match for this pattern

2 Answers2