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Editorial Team
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Asked: 2026-05-25T17:20:40+00:00

index.html <html> <head> </head> <body> <input type=text id=inputtext><input type=button value=submit id=submit> <br> <div id=response>

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index.html

    <html>
<head>
</head>
<body>


        <input type="text" id="inputtext"><input type="button" value="submit "id="submit">
                <br>                
        <div id="response">
        </div>


    <script src="http://ajax.googleapis.com/ajax/libs/jquery/1.3.2/jquery.min.js" type="text/javascript"></script>
    <script type="text/javascript" src="thescript.js"></script>
</body>
</html>

thescript.js

$(document).ready(function(){

    $("#submit").click(function(){

        var dataString = $("#inputtext").val();



           $.ajax({
             type: "POST",
             url: "greet.php",
             data: dataString,
             success: function(){
                 $("#response").load("greet.php");
              }
           });   


    });




});

greet.php

<?PHP

print "hello " . $_POST['dataString'];

?>

The value that was entered in the textfield should show when the greet.php is loaded. not sure why it won’t work

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1 Answer

  1. Editorial Team

    I think it can easier:

    $("#submit").click(function()
{
    var dataString = $("#inputtext").val();

           $.ajax({
             type: "POST",
             url: "greet.php",
             data: "dataString=" + dataString,
             success: function(result){
                 $("#response").html(result);
              }
           });   
    });

    And for php:

    <?PHP
if(isset($_POST["dataString"]))
{
    echo "hello " . $_POST['dataString'];
}
?>
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