help with a preg_replace regular-expression

Question

Hello all, I'm really confused as to using preg_replace but slowly learning.

I need help with the function:

$str= preg_replace('#\W\d+#', '\W \d+', $str);

The idea is that Im looking for numbers that have letters directly before them with no spaces,

ie abc123. (NOT 'abc 123' and NOT '123abc')

and how do I simply include a space or character in between so abc123 becomes abc 123 or abc@@123

thanks everyone!

Answer 1

You want

$str= preg_replace('#([A-Za-z]+)(\d+)#', '$1 $2', $str);

\W doesn't do what you think it does; \w means "word character" and \W means "not word character". Since you specifically want letters, though, neither does what you want, and you need [A-Za-z], which is the character class matching all letters. You also need a + after it in order to match "one or more".

Parentheses are how you "capture" part of the expression for later reference; $1 and $2 refer to your parenthesis-captured text.

Answer 2

<?php

$str = 'abc()1234';
$str= preg_replace('#([a-zA-Z()])(\d+)#', '$1 $2', $str);

echo $str;

Output:

abc() 1234

$1 and $2 are backreferences and refer to the first and second captured groups, respectively. More info @ http://www.regular-expressions.info/brackets.html.

Answer 3

You can see the difference between /([A-Za-z])(\d+)/ and /([A-Za-z]+)(\d+)/ in the online PHP Regegular expressor tester. Both work with preg_replace in your case. But the latter also works with preg_match_all, so it's more robust.