[RESOLVED] replace email

[zzz]

I need to parse a block of text (paragraph or two) and if email is present replace it with placeholder. I realize that there are many potential formats for email but the basic would be just fine.

I've found this regex that covers it [A-Z0-9._%-]+@[A-Z0-9.-]+.[A-Z]{2,4} not sure how to include it in my function...

$str = preg_replace("[A-Z0-9._%-]+@[A-Z0-9.-]+\.[A-Z]{2,4}","email removed",$str); 

[bpat1434]

Well, supposedly any email will have spaces on either side of it ... so just search for " (.@...*) " and replace it with " --email removed-- ".

<?php

$string = 'This is a string with an [e-mail] with some more text.';

$pattern = '~[\s]+([^\s]+@[^\s]+)[\s]+~iUs';

$string = preg_replace($pattern, ' --email removed-- ', $string);

echo $string;

Generates this output:

This is a string with an --email removed-- with some more text.

[nrg_alpha]

You can also go by:

$str = 'This is a string with an [e-mail] with some more text.';
echo $str . '<br />'; // to simply output the current string before any replacement...
$str = preg_replace('#[^\x20]+@[^\x20]+#', '*', $str);
echo $str; // ouput new version without email listed...

output:

This is a string with an [e-mail] with some more text.
This is a string with an * with some more text.

Come to think of it, best to use the \s instead of \x20. This way, ANY spacing will not be allowed.

preg_replace('#[^\s]+@[^\s]+#', '*', $str);

[zzz]

You guys rock!