我知道您說過您剛剛刪除了添加空格的功能，但我仍然想發布解決方案。需要明確的是，我不一定認為您應該使用此代碼，因為它可能更容易使事情變得簡單，但我認為它仍然應該有效。
您的主要問題是，幾乎每次提及都會引起兩次查找，因為@bob johnson went to the store可能是bob或bob johnson，并且如果不訪問數據庫就無法確定這一點。幸運的是，緩存將大大減少這個問題。
下面是一些通?？梢酝瓿赡趯ふ业牟僮鞯拇a。為了清晰和可重復性，我僅使用數組制作了一個假數據庫。內聯代碼注釋應該是有意義的。
function mentionUser($matches)
{
// This is our "database" of users
$users = [
'bob johnson',
'edward',
];
// First, grab the full match which might be 'name' or 'name name'
$fullMatch = $matches['username'];
// Create a search array where the key is the search term and the value is whether or not
// the search term is a subset of the value found in the regex
$names = [$fullMatch => false];
// Next split on the space. If there isn't one, we'll have an array with just a single item
$maybeTwoParts = explode(' ', $fullMatch);
// Basically, if the string contained a space, also search only for the first item before the space,
// and flag that we're using a subset
if (count($maybeTwoParts) > 1) {
$names[array_shift($maybeTwoParts)] = true;
}
foreach ($names as $name => $isSubset) {
// Search our "database"
if (in_array($name, $users, true)) {
// If it was found, wrap in HTML
$ret = '<span>@' . $name . '</span>';
// If we're in a subset, we need to append back on the remaining string, joined with a space
if ($isSubset) {
$ret .= ' ' . array_shift($maybeTwoParts);
}
return $ret;
}
}
// Nothing was found, return what was passed in
return '@' . $fullMatch;
}
// Our search pattern with an explicitly named capture
$pattern = '#@(?<username>\w+(?:\s\w+)?)#';
// Three tests
assert('hello <span>@bob johnson</span> test' === preg_replace_callback($pattern, 'mentionUser', 'hello @bob johnson test'));
assert('hello <span>@edward</span> test' === preg_replace_callback($pattern, 'mentionUser', 'hello @edward test'));
assert('hello @sally smith test' === preg_replace_callback($pattern, 'mentionUser', 'hello @sally smith test'));
