匹配2个正则表达式图案，正面或背面有1个图案

I have a regex that matches 2 patterns with 1 pattern in the front or back, but the first array return 2 empty indexes. Why is it doing that, and how would I stop it from doing that?

$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);

Result:

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

Want Result:

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

You can use a 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.

Your preg_match will look like:

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

See IDEONE demo

Results:

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

I guess your regex is a sample one. If you need to match a word after or before did, use the \w shorthand class:

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

See another demo

This is a modified version that behaves as you wish:

$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);

An online version here

Note: the regex takes advantage of the behaviour of OR in regex, the first results matched stop the regex engine to go further.