Smarty

[pointbweb]

Hi - I am a smarty newbie and am trying to us an if statement that checks to see if a string contains a specific substring. Smarty doesn't like LIKE and I don't think it allows wildcards either.

I tried this flawed syntax: {if $stringvariable Like "%substring%"}

Obviously flawed to you smartyer folks, but I'm sure you see what I need to do. Any help?

[messju]

you can use php-functions inside if:
{if substr_count($haystack, $needle)} ... {/if}

[pointbweb]

So it would look like this: {if ($stringvar like "%substring%")} ??
Somehow this doesn't look like it would work either, but I'll try it. Remember - the problem I have is using LIKE and WILDCARDS. Your example didn't contain that, so if you could clarify, that would be great!

Thanks.

[messju]

it uses http://www.php.net/manual/en/function.substr-count.php as i said.

it counts the number of occurences of $needle inside $haystack. if the number of occurences is greater than 0, than the if-condition is true, else it's false. so it should obviously read:

{if substr_count($stringvar, "substring")} ... {/if}

to match your example.

[pointbweb]

Ok - I missed that. I'm new to PHP as well

Thanks a bunch.