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Editorial Team
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Asked: 2026-05-27T19:40:37+00:00

Okay, what have I done wrong here? I’m trying to .POST data from a

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Okay, what have I done wrong here?
I’m trying to .POST data from a Form with JQuery $.ajax function and return json data. But every time the post data is null.

Here is the HTML/JQuery Part

<!DOCTYPE HTML>
<html>
<head>
<link rel="stylesheet" type="text/css" href="http://ajax.googleapis.com/ajax/libs/jqueryui/1/themes/redmond/jquery-ui.css">
<script src="http://ajax.googleapis.com/ajax/libs/jquery/1/jquery.min.js"></script>
<script src="http://ajax.googleapis.com/ajax/libs/jqueryui/1/jquery-ui.min.js"></script>
</head>

<body>
<form action="testform" method="post">
OLD ID:
<input id="OldID" name="OldID" type="text"><br>
NewID:
<input id="NewID" name="NewID" type="text"><br>

<input id="do" name="do" type="button" value="Get New ID">

</form>

<script type="text/javascript">

            var CID = $('#OldID').val();

        $(document).ready(function(){
                $('#do').click(function(){
                    sendValue();   
                }); 

        function sendValue(CID){

               $.ajax({
                  type: 'POST',
                  url: 'getNewID.php',
                  data: {OldID: CID},
                  dataType: 'json',
                  cache: false,
                  success:  
                    function(data){
                    $('#NewID').val(data.NewID);
                    }
              }) }
        });

</script>


</body>
</html>

Here’s the PHP page

<?php 

if (isset($_POST['OldID'])){

    $value = $_POST['OldID'];   

}else{
    $value = "ELSE VALUE";
}
echo json_encode(array("NewID"=>$value));  

?>

I’m sure this is a simple mistake on my part but I’m just going in circles at this point.

Thanks in advance!

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1 Answer

  1. Editorial Team

    Your mistake is that you are getting the value of the OldID field in your script, and not on some event handler method. This causes when the script is loaded (when the page is being viewed), the value of the text input is kept on variable CID. and when the user clicks on the #do button, then the value of CID is still the same as the one when the page was loaded. So no value is stored in CID variable, so no value is sent via AJAX.
    Define another function (Say sendOldID) that reads the value of the text input in the CID variable, and then calls the sendValue() method with CID paramter. The use that function as the event handler of the #do button clicked.

    <script type="text/javascript">
    $(document).ready(function(){
            $('#do').click(function(){
                sendOldID();   
            });

    function sendOldID() {
        var CID = $('#OldID').val();
        sendValue(CID);
    }

    function sendValue(CID) {
           $.ajax({
              type: 'POST',
              url: 'getNewID.php',
              data: {OldID: CID},
              dataType: 'json',
              cache: false,
              success:  
                function(data){
                $('#NewID').val(data.NewID);
                }
          }) }
    });
</script>
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