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Home/ Questions/Q 6615577
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Editorial Team
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Editorial Team
Asked: May 25, 20262026-05-25T20:28:46+00:00 2026-05-25T20:28:46+00:00

I have this code in jQuery: <script type=text/javascript src=http://ajax.googleapis.com/ajax/libs/jquery/1.3.0/jquery.min.js></script> <script type=text/javascript > $(function() {

  • 0

I have this code in jQuery:

<script type="text/javascript" src="http://ajax.googleapis.com/ajax/libs/jquery/1.3.0/jquery.min.js"></script>

<script type="text/javascript" >
$(function() {
  $("input[type=submit]").click(function() {
    var name = $("#problem_name").val();
    var problem_blurb = $("#problem_blurb").val();

    var dataString = 'problem_name='+ name + '&problem_blurb=' + problem_blurb;

    if(name=='' || problem_blurb == '') {
      $('.success').fadeOut(200).hide();
      $('.error').fadeOut(200).show();
    }
    else {
      $.ajax({
        type: "POST",
        url: "/problems/add_problem.php",
        data: dataString,
        success: function() {
          $('.success').fadeIn(200).show();
          $('.error').fadeOut(200).hide();

          // Here can update the right side of the screen with the newly entered information
          alert (dataString);
        }
      });
    }

    return false;
  });
});
</script>

And I have an AJAX call that I want to make JSON so I can unpack the JSON in the success case of the call to the AJAX.

What I don’t understand is how to create the JSON and how to make the front end detect it, and how to unpack it and display it. Any advice with that?

My backend return will be simple. Just a name and description for every item returned.

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1 Answer

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  1. Editorial Team
    Editorial Team
    2026-05-25T20:28:46+00:00Added an answer on May 25, 2026 at 8:28 pm

    Make an array of what you want to return in php, preferably an associative array but it could be integer indexed and it will work just as well.

    $arr = array();
    // Fill array with return data with PHP.
    $arr['name'] = 'John';
    
    echo json_encode($arr);
    

    That will render your array as output in JSON format, which your jQuery success function will pick up in a parameter. Below, I name the parameter json but you can name it whatever you like.

    ...
    success: function(json) {
       var ob = $.parseJSON(json);
       // At this point, ob is a JavaScript object that should look pretty much the same 
       // as the PHP object you created.
       alert(ob.name); // In this example, will alert 'John'
    }
    ...
    

    Also note that $.parseJSON was not introduced in jQuery until version 1.4.1.

    Please see: PHP Manual: json_encode()

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