正则表达式：如何删除除另一个正则表达式之外的所有内容？

Question

i have some regexp (like ^\w+[\w-.] \@\w+((-\w+)|(\w )).[az]{2,3}$, to match correct emails), but i cant figure out how to remove everything that doesn't match the regexp in my string.

Keeping the email example, i need a way to, given a sting like

$myString = "This is some text, the email is here example@example.com, and other things over here";

i need to return just 'example@example.com', or boolean false, if there is no email in the strings.

Of course, the email is just an example, some others times I'll need to remove everything except integer/floating numbers, etc...

I've googled around so much but didn't find anything.

Answer 1

If you surround your regex in parentheses and use preg_match or preg_match_all it will only return the part of the string that was matched by the regular expression:

$myString = "This is some text, the email is here example@example.com, and other things over here";
preg_match("/(\w+[\w-.]\@\w+((-\w+)|(\w)).[a-z]{2,3})/", $myString, $matches);
print_r($matches);

Note I also took off the beginning and end of string delimeters as in this case they are not needed.

Answer 2

Use the preg_match_all function to match all occurrences. The preg_match_all function itself returns the number of matches or false if an error occured.

So:

$myString = "This is some text, the email is here example@example.com, and other things over here";
if (($num = preg_match_all('/\w+[\w-.]\@\w+((-\w+)|(\w)).[a-z]{2,3}/', $myString, $matches)) !== false) {
    echo $num.' matches have been found.';
    var_dump($matches);
} else {
    // error
}

Answer 3

Your regex needs to have $ and ^ removed in order to work, those symbols cannot be found in the middle of the string. If your regex matches it cannot possibly contain anything else but email.