preg_replace parse error please?

[zmola]

hi all. i'm using preg_replace in a function. this function takes in $word. $word is numeric and below is the pattern.

"/foo[(\'([^{)]{4,})\')]bar/",} <--- works when hard coded

"/foo[(\'([^{)]{$word,})\')]bar/",} <--- gives the following:

Parse error: parse error, expecting `'}'' in blah blah...

any help would be greatly appreciated. thanks!

[Bunkermaster]

try "/foo[(\'([^{)]{".$word.",})\')]bar/"}

[zmola]

great! worked. thanks!!!

just curious... could someone please explain why the above ".$word." is needed in preg_replace but in the below code $word works correctly without the ". ." while with the ". ." the if statement is ignored.

is it just that preg_replace needs that format whereas in the if statement ".$word." changes the numeric $word to string $word?

if(strlen($result[$cth])>=4) <--- works hard coded

if(strlen($result[$cth])>=$word) <--- works with var

if(strlen($result[$cth])>=".$word.") <--- does not work

thanks again. i was all over reading and searching for my previous mistake but just couldn't put my finger on my error.

[Bunkermaster]

Originally posted by zmola
if(strlen($result[$cth])>=".$word.") <--- does not work

doesn't work? :rolleyes:

is the code surounding the ".$word." a string?
what is tests now is if the length of your variable is equal to (if $word is 4 for example) '.4.' (between the single quotes)
see why it doesn't work? 😉

[zmola]

yep. got it. thanks!

[Bunkermaster]

May the Source be with you young padawan 😉