Using regex to insert a space in a string in php

Question

I have a string that is not properly formed and am attempting to correct it. An example of the string is: -

A Someone(US)B Nobody(US)

I am attempting to correct it to: -

A Someone(US) B Nobody(US)

I am using the below code to match ")" followed by a capital letter and using php's preg_replace function to do the match and add the space. However I'm completely rubbish at regex and cannot get the space added in the correct place.

$regex = "([\)][A-Z])";
$replacement = ") $0";
    $str = preg_replace($regex, $replacement, $output);

Can anyone suggest a better method? I realise the space is not adding correclty because $0 contains the data I am matching, is there a way to manipulate $0?

$0 is capture group zero which is ([\)][A-Z]), just move the paranthesis like this [\)]([A-Z]) and use $1 I think. — Yet Another Geek
– Yet Another Geek, Commented Sep 24, 2011 at 20:15
I don't think there is a capture group $1, I've tried this and it chops off the capital letter, so produces an output of A Someone(US) Nobody(US). — KryptoniteDove
– KryptoniteDove, Commented Sep 24, 2011 at 20:28

Tim Pietzcker · Accepted Answer · 2011-09-24 20:31:16Z

2

$str = preg_replace('/(?<=\))(?=\p{Lu})/u', ' ', $output);

inserts a space between a closing parenthesis (\)) and an uppercase letter (\p{Lu}). You don't need $0 (or $1 etc.) at all since you're just inserting something at a position between two characters, and this regex matches exactly this (zero-width) position. Check out lookaround assertions.

answered Sep 24, 2011 at 20:31

Tim Pietzcker

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2 Comments

KryptoniteDove Over a year ago

How would you adapt this to lowercase letter preceding a capital letter again no space? I tried /(?<=))(?=\p{Lu})/u but no joy..

Tim Pietzcker Over a year ago

Do you mean "insert a space between a lowercase and an uppercase letter"? $str = preg_replace('/(?<=\p{Ll})(?=\p{Lu})/u', ' ', $output);

Kent · Accepted Answer · 2012-08-02 08:36:15Z

0

how about regex="(?<=\))[A-Z]"
and replacement=" $0"

edited Aug 2, 2012 at 8:36

user212218

answered Sep 24, 2011 at 20:14

Kent

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1 Comment

KryptoniteDove Over a year ago

This regex causes an Unknown modifier '[' error when used with preg_replace.

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