RegExp PHP Blocks

[RedBassett]

Greetings,

I have not written any RegExp in a long time, so forgive my slowly getting back into it.

I am trying to remove PHP blocks from an HTML/XML file. The best RegExp I can come up with tis this:

<\?php(.*?)\?>

However, I believe this is having trouble because of the closing "?". If I remove the closing tag, ("<\?php(.*?)") it just matches the opening tag. How can I match the tags and all of the code in between?

P.S. I understand that not matching anything that contains a PHP tag in a string (for example: echo "?>" is a lot harder, but this is not an issue in this application.

- RedBassett

[Chris]

Just a hunch, but I feel like linebreaks might be an issue. Try this instead?

<\?php([.\n\r]*?)\?>

[RedBassett]

Jan 13, 2012 11:31:02 GMT -8 Chris said:

Just a hunch, but I feel like linebreaks might be an issue. Try this instead?

<\?php([.\n\r]*?)\?>

Still didn't work I considered linebreaks before, but couldn;t find much on them. Here is a test line I am using:

echo preg_replace('/<\?php([.\n\r]*?)\?>/', '', $source);

$soruce is the output of file_get_contents. The file is a mix of PHP, HTML, and some custom code (plaintext).

[Charles Stover]

No need for the parenthesis, since you aren't using the contents in the replace.

Try this:

echo preg_replace('/\<\?php.*?\?\>/sU', '', $source);

[RedBassett]

Jan 23, 2012 16:22:00 GMT -8 Charles Stover said:

No need for the parenthesis, since you aren't using the contents in the replace.

Try this:

echo preg_replace('/\<\?php.*?\?\>/sU', '', $source);

Thanks, I think this is what I needed!