如何使用PHP检查RegEx搜索模式的前一个字符？

Question

I would like to check if the preceding character of a search pattern is an alphanumeric character.

If true, do nothing.

If fasle, remove the preceding space in the search pattern .

For example:

$string1 = "This is a test XYZ something else";

$string2 = "This is a test? XYZ something else";

$pattern = " XYZ";

In $string1 scenario, the preceding character of the search pattern is t and considered a match, nothing will be perform.

In $string2 scenario, the preceding character of the search pattern is ? and considered a non-match, and I'm removing the extra space in searhc pattern .

Making it:

$string2 = "This is a test?XYZ something else";

How can this be accomplished in PHP?

Answer 1

You may use a \\B XYZ pattern and use a preg_replace_callback to trim the match value and insert it back:

$string1 = "This is a test XYZ something else";
$string2 = "This is a test? XYZ something else";
$pattern = " XYZ";
echo preg_replace_callback('~\B'.$pattern.'~', function($m) { return trim($m[0]); }, $string1) . PHP_EOL;
// => This is a test XYZ something else
echo preg_replace_callback('~\B'.$pattern.'~', function($m) { return trim($m[0]); }, $string2);
// => This is a test?XYZ something else

See the PHP demo

Since \\B matches at the locations other than those matched with a word boundary (a non-word boundary), the pattern \\B XYZ will only match after a non-word char.

More details : your pattern starts with a space. This is a non word char. By adding \\B before it we require that the character before the space should also be a non word char. Else, we'll get no match. The word char is a char from [a-zA-Z0-9_] range. If you need to customize the boundary, use a lookbehind like (?<![a-zA-Z0-9]) to exclude the underscore from the boundary characters.

For more information on non-word boundary see this What are non-word boundary in regex ( \\B ), compared to word-boundary? SO thread .