PHP / Regex：获取星星之间的内容，但如果有反斜杠，则不会
I would like to get everything between two stars - except of they have a leading backslash.
So for example:
*hello* world
should return "hello", but
*hello \* world*
should return "hello * world"
I tried the following regex:
/(?<!\\)\*(.+?)(?<!\\)\*/s
which works perfect on http://regex101.com/ but php returns:
Warning: preg_replace(): Compilation failed: missing ) at offset 21
What am I doing wrong?
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EDIT 1: Here's my PHP-Code for that:
var_dump(preg_replace('/(?<!\\)\*(.+?)(?<!\\)\*/s', '<strong>$1</strong>', '*hello world*'));
You are not escaping the backslashes correctly which results in escaping the ) character.
To match a \ in PHP you need 4 backslashes
/(?<!\\\\)\*(.+?)(?<!\\\\)\*/s
It must be done like this because every backslash in a C-like string must be escaped by a backslash. That would give us a regular expression with 2 backslashes, as you might have assumed at first. However, each backslash in a regular expression must be escaped by a backslash, too. This is the reason that we end up with 4 backslashes.
Or use a character class with 2 backslashes
/(?<![\\])\*(.+?)(?<![\\])\*/s
A literal backslash can also be matched using preg_match() by using a character class instead. Backslashes are not escaped when they appear within character classes in regular expressions. Therefore (“[\]“) would match a literal backslash. The backslash must still be escaped once by another backslash because it is still a C-like string.
Edit Found this article which explains why this is necessary. Also, added explanations.
