Preg_replace help

[LemonInflux]

I'm currently doing a project where I have a page, and on it are images and links.

 

The links look like this:

 

mysite.com/path/file.extension

 

I want them to look like this:

 

myothersite.com/mysite.com/path/file.extension

 

I thought, 'yeah, let's just use preg_replace and just alter any links to that'.

 

However, there's a problem. With any images on the page, I still want to keep it as

 

mysite.com/path/image.extension

 

This would mean I'd need to parse the <a> tags, and not the <img> tags. I thought of doing <a href="(.*?)"> to <a href="myothersite.com/\\1">, but then you can have classes and stuff in an <a> tag, so if the person did, say, <a class="foo" href="">, that wouldn't parse it. I don't want to parse out anything like href="mysite.com/path/file.extension", because if by chance that was written on the page, it wouldn't work. Does anyone have any idea how to get around this problem?

[effigy]

If you use /<a[^>]+href=.../ it will find the "href" attribute no matter where it appears in the tag.

[LemonInflux]

So how would that look as a whole example of a link parser?

[effigy]

I thought of doing <a href="(.*?)"> to <a href="myothersite.com/\\1">, but then you can have classes and stuff in an <a> tag, so if the person did, say, <a class="foo" href="">, that wouldn't parse it.

 

/<a[^>]+href="(.*?)"/. Of course, this assumes that double quotes are always used. Are you expecting single quotes and/or no quotes at all?

[LemonInflux]

It could be any. Would I just use 3 different preg_replaces?

[effigy]

/<a[^>]+href=([\'"])?((?(1).+?|[^\s>]+))(?(1)\1)/