preg_replace to replace a div for a minimum width

Question

I have a user input form where the user is allowed to use HTML, such as div, span, etc.

I would like to clean the input only if a user happens to put in a div with a style="width:" that is over the site width (which in this case is 800px.)

So I wanted to do a preg_replace searching for div style="width:" only when width is greater than 800px.

I already have all other fixes in place for spans, div width="", etc. but the preg_replace on the div style="width:" where the width is only changed if it is greater than 800 (such as 1500px, 2000px, 801px) eludes me - I cannot get it to work.

Is there any way to achieve this?

`(\d{4,}|9\d{2,}|8[^0]\d|80[^0])` – JNF Oct 16 '12 at 21:23 — JNF, Oct 16 '12 at 21:23

score 1 · Accepted Answer · edited May 23 '17 at 11:48

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The pony he comes...

How would you go about handling something like width:140%? Or width: 600px; margin: 200px;? Or just an image with no defined width but an intrinsic width of over 800px?

It's far easier to just put the user-supplied content in a wrapper:

<div style="width: 800px; overflow: hidden;">USER CONTENT HERE</div>

edited May 23 '17 at 11:48

Community

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answered Oct 16 '12 at 21:52

Niet the Dark Absol

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Yes, I think I was overthinking it - the overflow:hidden didn't even cross my mind. It is the best fix. Thank you. – user1751361 Oct 16 '12 at 22:01

preg_replace to replace a div for a minimum width

1 Answers1