preg_replace pattern help

can anyone help me how i should write the pattern

i want to replace "xyz" with "yyy" but only if xyz stands alone. Also it should be case-insensitive.

"xyz what go www.xyz.com XyZ test"
->
"yyy what go www.xyz.com yyy test"

 

$str = "xyz what go www.xyz.com XyZ test";
$str = eregi_replace('xyz ', 'yyy ', $str);
$str = eregi_replace(' xyz', ' yyy', $str);
echo $str;

 

that was kinda smart but does that work if theres a line with "xyz" alone?

 

The eregi_* functions are old, ugly, slow, and deprecated (they will be removed in PHP 6). Get used to the preg_* functions. And besides that, they're meant for regular expressions, which you're not using. For simple replacements like yours, use str_replace() and str_ireplace().

Or do it this way:

$text = preg_replace('~(^|\s)xyz($|\s)~si', '$1yyy$2', $text);

PHP:

 

great solution, but failed on this:

$str = "xyz xyz xyz www.xyz.com xyz XyZ xyz";
$str = preg_replace('~(^|\s)xyz($|\s)~si', '$1yyy$2', $str);
echo $str;
//output -> yyy xyz yyy www.xyz.com yyy XyZ yyy

Code (markup):

$str = "xyz xyz xyz www.xyz.com xyz XyZ xyz";
$str = preg_replace('~(^|\s)xyz($|\s)~si', '$1yyy$2', $str);
$str = preg_replace('~(^|\s)xyz($|\s)~si', '$1yyy$2', $str);
echo $str;
//output -> yyy yyy yyy www.xyz.com yyy yyy yyy

Code (markup):

 

thx guys but writing the same line doesn't look like the perfect solution? must be better way

 

<?php
$str = "xyz xyz xyz www.xyz.com xyz XyZ xyz";
$str = preg_replace('/(^|\s)xyz($|\s)/i', '$1yyy$2', $str);
echo $str;
?>

Code (php):

I do think should do what you want. If not, could you expand? :-/

 

$str = "xyz xyz xyz www.xyz.com xyz XyZ xyz";
$str = preg_replace(array('/(^|\s)xyz($|\s)/i', '/(^|\s)xyz($|\s)/i'), '$1yyy$2', $str);
echo $str;

Code (markup):