image - How to continue a php script after using exit() / die()

Question

For example i have the following script.

in this case i want to get the value of X1 and Y1 but the the exit() is not letting me to do so

please help thanks in advance and special thanks to Mark Setchell :P (if he sees this)

image link

 $im = imagecreatefromjpeg("omr.jpg");
for($x=0;$x<100;$x++)   {

    for($y=0;$y<100;$y++)   
                            {

$rgb = imagecolorat($im,$x,$y);
      $r = ($rgb >> 16) & 0xFF;
      $g = ($rgb >> 8) & 0xFF;
      $b = $rgb & 0xFF;
      if($r<128 && $g<128 && $b<128){
         printf("%d,%d: %d,%d,%d
",$x,$y,$r,$g,$b);
         exit;
                                    }
                            }
                        }

 for($x1=1170;$x1<1270;$x1++){

 for($y1=0;$y1<100;$y1++){

          $rgb = imagecolorat($im,$x1,$y1);

          $r1 = ($rgb >> 16) & 0xFF;
          $g1 = ($rgb >> 8) & 0xFF;
          $b1 = $rgb & 0xFF;
          if($r1<128 && $g1<128 && $b1<128){
          printf("%d,%d: %d,%d,%d
",$x1,$y1,$r1,$g1,$b1);
          exit;
        }
       }
    }

current output : 30,53: 123,119,118

Desired output : 30,53: 123,119,118

1217,55: 115,114,112

Answer 1

You can use break 2; to break from two nested for loops and your next for loop for x1 and y1 will also execute.