将 2 个正则表达式模式与前面或后面的 1 个模式匹配

Question

我有一个正则表达式匹配前面或后面有 1 个模式的 2 个模式，但第一个数组 return 有 2 个空索引。它为什么这样做，我如何阻止它这样做？

$text = "i did";
preg_match("~(?:(did) (.+)|(.+) (did))~", $text, $match);
print_r($match);

echo "<br>";

$text = "did i";
preg_match("~(?:(did) (.+)|(.+) (did))~", $text, $match);
print_r($match);

结果：

Array ( [0] => i did [1] => [2] => [3] => i [4] => did ) 
Array ( [0] => did i [1] => did [2] => i )

想要的结果：

Array ( [0] => i did [1] => i [2] => did ) 
Array ( [0] => did i [1] => did [2] => i )

Answer 1

您可以使用 branch reset (?|...):

Alternatives inside a branch reset group share the same capturing groups. The syntax is (?|regex) where (?| opens the group and regex is any regular expression.

您的 preg_match 看起来像：

preg_match("~(?|(did) (.+)|(.+) (did))~", $text, $match);

见IDEONE demo

结果：

Array
(
    [0] => i did
    [1] => i
    [2] => did
)

我猜你的正则表达式是一个示例。如果您需要匹配 did 之后或之前的单词，请使用 \w shorthand class:

preg_match("~(?|(did) (\w+)|(\w+) (did))~", $text, $match);

见another demo

Answer 2

这是一个修改后的版本，可以按照您的意愿运行：

$text1 = "i did";
preg_match("~(did|\w+(?= did)) (did|(?<=did )\w+)~", $text1, $match1);
print_r($match1);

$text2 = "did i";
preg_match("~(did|\w+(?= did)) (did|(?<=did )\w+)~", $text2, $match2);
print_r($match2);

$text3 = "did x, x did";
preg_match_all("~(did|\w+(?= did)) (did|(?<=did )\w+)~", $text3, $match3);
print_r($match3);

$text4 = "a a";
preg_match("~(did|\w+(?= did)) (did|(?<=did )\w+)~", $text4, $match4);
print_r($match4);

在线版本here

注意：正则表达式利用了正则表达式中OR的行为，第一个匹配的结果停止正则表达式引擎走得更远。

将 2 个正则表达式模式与前面或后面的 1 个模式匹配

Matching 2 regex patterns with 1 pattern in the front or back

php

regex

preg-match