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Home/ Questions/Q 7766101
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Editorial Team
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Editorial Team
Asked: June 1, 20262026-06-01T15:19:38+00:00 2026-06-01T15:19:38+00:00

I followed a tutorial that went like so: ajaxRequest.onreadystatechange = function(){ if(ajaxRequest.readyState == 4){

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I followed a tutorial that went like so:

        ajaxRequest.onreadystatechange = function(){
            if(ajaxRequest.readyState == 4){
                //use the value  passed in to return response to correct button
                e = document.getElementById("RB" + number);
                e.value = ajaxRequest.responseText + number;
            }

        }

        ajaxRequest.open("GET", "Reject.php" + number, true);
        ajaxRequest.send(null); 

I need to pass number for a mysql function in Reject.php. However, despite the example working for the tutor, my file wants to append the number onto the reject.php giving me a 404 error. What am I doing wrong? In my Reject.php:

   $ID = $_GET['number'];
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  1. Editorial Team
    Editorial Team
    2026-06-01T15:19:39+00:00Added an answer on June 1, 2026 at 3:19 pm

    You need to pass a querystring, just like any other URL:

    "Reject.php?number=" + number
    
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