preg_quote忽略preg_match_all中的尾随点

Question

Regex

preg_match_all('@(\b' . preg_quote($needle,'@') . '\b)@is', $haystack, $matches);

Haystack

(Job: NAS-Inkrementell) Operation succeeded.

Needle

Operation succeeded --> works

Operation succeeded. --> does not work

I did some tests:

Alternative Haystack

(Job: NAS-Inkrementell) Op.eration succeeded.

Alternative Needle

Op.eration succeeded --> works

So, the needle does get escaped correctly, as I can see by dump(preg_quote($needle,'@')); --> Operation succeeded\.

What would be the correct way to include a trailing dot?

For quick reference: http://www.phpliveregex.com/p/hzl

Thanks

EDIT

I also need to distinguish between My-NAS and My-NAS-2 in the haystack when the needle is My-NAS, so I need the trailing \b as well.

EDIT2

ideone.com/Suzz2E I want it to be 1 1 0 1 instead of 1 1 1 1, so MyNAS should not be found if there is only MyNAS-2 in the haystack.

Answer 1

The . is not a word character, and \.\b pattern requires a word char after .. You need to implement custom word boundaries.

Since you need to match strings that are not preceded with a word char or a hyphen and that are not followed with a word char or hyphen, you may use

preg_match_all('@(?<![\w-])' . preg_quote($needle1,'@') . '(?![\w-])@is', $haystack, $matches)

See the regex demo.

See the PHP demo:

$haystack1 = "(Job: NAS-Inkrementell) Operation succeeded. MyNAS-2 reports duty.";

$needle1 = "Operation succeeded";
echo preg_match_all('@(?<![\w-])' . preg_quote($needle1,'@') . '(?![\w-])@is', $haystack1, $matches1) ."
";

$needle2 = "Operation succeeded.";
echo preg_match_all('@(?<![\w-])' . preg_quote($needle2,'@') . '(?![\w-])@is', $haystack1, $matches2) ."
";

$needle3 = "MyNAS";
echo preg_match_all('@(?<![\w-])' . preg_quote($needle3,'@') . '(?![\w-])@is', $haystack1, $matches3) ."
";

$needle4 = "MyNAS-2";
echo preg_match_all('@(?<![\w-])' . preg_quote($needle4,'@') . '(?![\w-])@is', $haystack1, $matches4) ."
";