what is the simplest method to do the following
Look at a block of text and anywhere it finds %sometxt% replace that with something else,
So basically any text that is enclosed within a set of percentage signs similar to how Smarty do this?
The simplest method you asked for:
$blockOfText = str_replace('%sometxt%', 'something else', $blockOfText);
You just replace it with str_replace.
$newText = preg_replace('/%.+?%/', 'replacedText', $text);
Should do it. replacedText needs to be whatever you want to replace %sometxt% with.
The regexp %.+?% means
Fine a
%sign, followed by any character until I find another%sign.
The ? makes the expression 'lazy', so it will match up to the first occurrence of %.