Warning: preg_match() expects parameter 2 to be string, array given in e:\\test.php on

system · November 7, 2008, 8:59am

Hi,

Normally we use preg_match() to search the given string, and it returns 1 or 0.

while using it to search for array values,
for eg,
$arr = array();
$tt = preg_match(“[0-9]”, $arr);

it shows warning as,

Warning: preg_match() expects parameter 2 to be string, array given in e:\ est.php on line 27.

How to check for array values???

oddz · November 7, 2008, 9:11am

What do you mean by array values the array in your example is empty?

If you head over to the manual you should see that preg_match accepts only strings as the first two arguments.

ro88o · November 7, 2008, 9:55am

If I understand correctly what you’re trying to do, you could loop through the array, check each string value in the array with preg_match and break out of it if you find a match.

e.g.


$arr = array();
//Add data into the array here

for ($i = 0; $i < count($arr); $i++) {
    $tt = preg_match("[0-9]", $arr[$i]);
    if ($tt == 1) {
          break;
    }
}

tdolsen · November 7, 2008, 10:25am

You could take a look at preg_grep.

decowski · November 7, 2008, 11:18am

ro88o:

If I understand correctly what you’re trying to do, you could loop through the array, check each string value in the array with preg_match and break out of it if you find a match.

e.g.
for ($i = 0; $i < $arr.length; $i++)
{
     $tt = preg_match("[0-9]", $arr);
     if ($tt == 1)
          break;
} 

You, of course, meant:


for ($i = 0; $i < count($arr); $i++) {
	$tt = preg_match("[0-9]", $arr[$i]);
	if ($tt == 1) {
          break;
	}
}

ro88o · November 7, 2008, 12:55pm

Ha, yep I did Been spending too much time developing JS recently and in too much of a rush to check my code!