May someone please explain the difference between using $1 and \1 in this scenario?
<?php
$text = "You know I mean *seriously*!";
$newText = preg_replace('/\*([^\*]+)\*(!)/','$2<em>$1</em>', $text);
echo $newText;
Explanations on the net are a bit mind-boggling, I’m getting similar results using either $ or \ in the code above anyway.
Thanks in advance.
They refer to the capture groups in your regex expression, which are denoted by (). The first captures [^*]+ and the second captures !
No I think you have misunderstood what I’m asking about here 
I’ve used $2<em>$1</em> in the above code but I’m asking how would it be different if I used \2<em>\1</em> instead? I see no obvious difference between them and they both “apparently” seem to be the same…
They are “back references”
AFAIK, for PHP at least
replacement may contain references of the form \\n or $n, with the latter form being the preferred one.
I’m guessing the backslash way is like a “go get it” and the dollar sign way is more like a “use what’s already in memory”. In other words, though the end result is the same the latter is more efficient.
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