Parse out word only if it doesn't follow a specific word

[shadkeene]

I'm trying to parse out individual key words in a string ($dot_afd). However, in the case of "pass", there's a town named "GRANTS PASS" that I don't want to be included. I only want "PASS" highlighted if it doesn't come after the word "GRANTS".

I tried some lookbehind code, but unsuccessful. Here's my code that still highlights "PASS" even if it comes after "GRANTS".

PHP Code:

$patterns[2] = 'PASS';
$replacements = array();
$replacements[2] = '<b><i>PASS </i></b>';
$dot_afd = preg_replace($patterns, $replacements, $dot_afd); 


If you have any advice, I would appreciate it. Thanks for your time,

S

[kbluhm]

Quick and dirty... this will work, but there mayyy be a better way:

PHP Code:

// highlight ALL instances of `pass`
$dot_afd = preg_replace( '/(pass)/i', '<b><i>$1</i></b>', $dot_afd );

// now remove highlighting from instances that follow `grants`
$dot_afd = preg_replace( '/(grants)(\s+)\<b\>\<i\>(pass)\<\/i\>\<\/b\>/si', '$1$2$3', $dot_afd ); 


[gvre]

You should use negative lookbehind in your pattern. Try this one

PHP Code:

$dot_afd = "GRANTS PASS";
$pattern = '#(?<!GRANTS\s)PASS#';
$replacement = '<b><i>PASS </i></b>';
$dot_afd = preg_replace($pattern, $replacement, $dot_afd);
echo $dot_afd; 


[shadkeene]

Thanks very much. That works perfectly!