正则表达式PHP，在特定位置查找字符

Question

I explain my problem : I'm working on different kind of address

" 25 Down Street 15000 London "

" 25 B Down Street 15000 London "

" Building A 25 Down Street 15000 London "

I found a way to determine which is the number of the street on all case with this regex :

 `^([1-9][0-9]{0,2}(?:\s*[A-Z])?)\b`

But now i got a problem that i can't solve, i need when the case is real to determine characters which are before the street's number .

Example : " Building 2 25 Down Street 15000 London " i need here to find only "Building 2"

I understand that i have to find characters before the first number of this string.

Keep searching on my own but will be great if someone got a solution for me .

Thank you .

Edit my code now is :

preg_match('/^(.*?)\d+\s+\D+/', $cleanAdressNode, $result, PREG_OFFSET_CAPTURE,0);
        print $result[0][0];

        return $result[0][0];

and the result now is : Résidence Les Thermes 1 15 boulevard Jean Jaurès instead of only : Résidence Les Thermes 1

Answer 1

How about:

preg_match('/^(\D*)/', $str, $match);

You will find in $match[1] everything that is not a digit at the begining of the string.

According to your example:

preg_match('/^(.*?)\d+\s+\D+/', $str, $match);

Answer 2

If you only want to match the first non-numeric characters, ^([^0-9]*) should do the trick. It uses class negation to grab every non-numeric characters at the start of the string.