PHP: links for URL preg_replace after %

Question

I have this expression:

$content = preg_replace('@(https?://([-\w\.]+)+(:\d+)?(/([\w/_\.]*(\?\S+)?)?)?)@', '<a href="$1" target="_blank">$1</a>', $content);

But a link like

http://photoartkalmar.com/Photoart%20Kalmar%20high%20res/Gigapixel/Vienna%2050%20Gigapixel/Donauturm.html

or that ends with something.jpg

The anchor link stops at % or . on the urls, so it doesnt make it all a link only half of it. How can i fix this?

PleaseStand · Accepted Answer · 2010-11-11 21:50:40Z

2

Of course, put a % symbol in the correct place in the regex: [\w/_\.] → [\w/_\.%]

Look at this whole line to see exactly where:

$content = preg_replace('@(https?://([-\w\.]+)+(:\d+)?(/([\w/_\.%]*(\?\S+)?)?)?)@', '<a href="$1" target="_blank">$1</a>', $content);

answered Nov 11, 2010 at 21:50

PleaseStand

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