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Home/ Questions/Q 3492570
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Editorial Team
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Editorial Team
Asked: May 18, 20262026-05-18T11:46:28+00:00 2026-05-18T11:46:28+00:00

I have little cycle. This take some image to the screen. Every image have

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I have little cycle. This take some image to the screen. Every image have an id which i stored in $id variable.

When user click on an image they get a popup window. Now im use query string. When user click, get a popup with id from query string.

But this is not a good way, ‘coz if user reload the page, with the query string..they get the popup every time.

I need the $id when show the popup. How can i do this without querystring? How can i check if click on image and which image clicked on?

for j=1 .....{
...
..

 for i=1....... {
  $id=array[j,i];

   echo "<a href=test.php><img style='z-index:$z; position:absolute; left: $lf; top: $tf;' src='images/$src' width='$width' height='$heigth' title='$title' /></a>";


 }
}
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  1. Editorial Team
    Editorial Team
    2026-05-18T11:46:29+00:00Added an answer on May 18, 2026 at 11:46 am

    You can use JavaScript to trigger the popup when the user clicks on the image. This way, nothing is sent back to the server, and there is no query string.

    I don’t see where you print the id for each picture to the page, but because it is psuedocode, I’ll assume it works. Using javascript would look something like this:

    for j=1 .....{
      ...
      ..
    
     for i=1....... {
      $id=array[j,i];
    
       echo "<a href='javascript:alert(\"$id)\"'><img style='z-index:$z; position:absolute; left: $lf; top: $tf;' src='images/$src' width='$width' height='$heigth' title='$title' /></a>";
    
     }
    }
    

    Clicking on the image would then generate a popup with the id of the image, and no information would be sent to the server.

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