Question

I would like to create a regular expression that will replace a normal space with hard-breaking space according to my country typographical rules. It shouldn’t touch HTML tags.

All spaces in these examples should be hard-breaking:

Numbers

1 000
10 000
100 000
1 000 000
etc..

Dates

17. 6.
17. 6. 2012

I came up with:

$pattern = '/((\d\.?)\s(\d))(?=[^>]*(<|$))/';
$text = preg_replace($pattern, '$2&nbsp;$3', $text);

It can do all of these but:
17. 6. 2012
It won’t replace the second space, so result is ->

17. 6. 2012

Thank you for any help!

Answer 1

The (\d) is eating the first trailing digit, and if the trailing number only has a single digit, there won’t be enough of it left to match on the next pass.

For instance, the following work fine: 12 34 56, 12. 34. 56. But these do not: 1 2 3, 1. 2. 3

The (\d) can be moved into the lookahead, to avoid eating it:

$pattern = '/((\d\.?)\s)(?=\d[^>]*(<|$))/';
$text = preg_replace($pattern, '$2&nbsp;$3', $text);

Combining this with @flec’s use of a lookbehind yields:

$pattern = '/(?<=\d)(\.?)\s(?=\d[^>]*(<|$))/';
$text = preg_replace($pattern, '$1&nbsp;', $text);