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jQuery Ajax passing value on php same page

Posted by: admin July 12, 2020 Leave a comment

Questions:

I am kinda confused on it, when trying to send value on the same page.

 <script>
      $("select[name='sweets']").change(function () {
      var str = "";
      $("select[name='sweets'] option:selected").each(function () {
            str += $(this).text() + " ";

          });

            jQuery.ajax({
            type: "POST",
            data:  $("form#a").serialize(),

            success: function(data){
                jQuery(".res").html(data);

                $('#test').text($(data).html());


            }
            });  
            var str = $("form").serialize();
            $(".res").text(str);
    });
    </script>
 <div id="test">
 <?php
  echo $_POST['sweets'];
  ?>
  </div>
<form id="a" action="" method="post">
  <select name="sweets" >
   <option>Chocolate</option>
   <option selected="selected">Candy</option>
   <option>Taffy</option>
   <option>Caramel</option>
   <option>Fudge</option>
  <option>Cookie</option>
</select>
</form>

Well it will display if its in the top of html tag but if its inside the body it will display null.

How to&Answers:

Here is the working code for you. To send ajax request to the same page you can keep url parameter empty, which you are already doing. If you are trying to make the script behave differently when $_POST has value then use isset as I have used below.

 <?php
  if(isset($_POST['sweets'])) 
  {
    echo $_POST['sweets'];
    exit;
  }
  ?>

    <script>
        $(function(){
          $("select[name='sweets']").change(function () {
          var str = "";
          $("select[name='sweets'] option:selected").each(function () {
                str += $(this).text() + " ";

              });

                jQuery.ajax({
                type: "POST",
                data:  $("form#a").serialize(),

                success: function(data){
                    jQuery(".res").html(data);

                    $('#test').html(data);


                }
                });  
                var str = $("form").serialize();
                $(".res").text(str);
        });
        });
        </script>


 <div id="test">

  </div>

<form id="a" action="" method="post">
  <select name="sweets" >
   <option>Chocolate</option>
   <option selected="selected">Candy</option>
   <option>Taffy</option>
   <option>Caramel</option>
   <option>Fudge</option>
  <option>Cookie</option>
</select>
</form>

Answer:

You should wrap your code with

$(document).ready(function(){
   // your code here
});

This way, it will only run when the browser finishes processing the structure of your HTML.

UPDATE

There was a lot of debug stuff on your code, try this (requires Firebug to see the output of the ajax request):

<script>
$(document).ready(function(){
    $("select[name='sweets']").change(function () {
        jQuery.ajax({
            type: "POST",
            data:  $("form#a").serialize(),
            success: function(data) {
                // Check the output of ajax call on firebug console
                console.log(data);
            }
        });
    });
});
</script>