\n (return carriages) in preg_replace

This adds a new line for every whitespace character:

$after=preg_replace('/\s+/','\n\n',$before);

I need it to replace any new line that contains extra whitespace characters with just a new line. I THOUGHT it should look something like this:

$after=preg_replace('/\n/\s+//\n','\n\n',$before);

But it doesn't work! Any ideas? Thanks!

 

Here the the correct format for preg_replace

$patterns = array ('/(19|20)(\d{2})-(\d{1,2})-(\d{1,2})/',
                   '/^\s*{(\w+)}\s*=/');
$replace = array ('\3/\4/\1\2', '$\1 =');
echo preg_replace($patterns, $replace, '{startDate} = 1999-5-27');

Code (markup):

$str = 'foo   o';
$str = preg_replace('/\s\s+/', ' ', $str);

Code (markup):

http://us3.php.net/preg_replace

 

Try this:
$after = preg_replace('/\A\s+\Z/','',$before);

Or this:
$after = preg_replace('/\n\s+\n/',"\n\n",$before);

 

hmm lets try this code.

 

Umm... yet another lapse on concentration.

The first one should be:

$after = preg_replace('/^\s+$/m', '', $before);

Code (markup):

The second one should still be fine if you prefer it.

 

Thanks for all your replies!

$after = preg_replace('/\n\s+\n/',"\n\n",$before);

Worked perfectly!