jQuery Ajax passing value on php same page

Question

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.

Noenoel

Please could you post the full PHP code? – Thermo

im on a testing phase that is the full php code. if you have a further question just add a comment. thanks – Noenoel

From the discussion on my answer, it became clear we need more info to help you. What are you trying to accomplish, exactly? Why do you need ajax if you are posting to the same page? – Roundworm

Answer 1

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>

Amicable

Answer 2

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>

Roundworm

i seem that wont work, i tried adding it. but the question is how do send value from jquery ajax to php without reloading it. – Noenoel

You code already does that. I suggested using document.ready because you seem to be trying to add the change handler to the <select> before the browser knows what that element is. – Roundworm

oh yeah, but it appears that it display the whole form >_< can you make an example of it? – Noenoel

why it does the output did not print out? i added echo on it? – Noenoel

echo $_POST['sweets']; should print the value of the selected option, not the option text. If it's empty, it's printing an empty string. – Roundworm

yeah i see that it print in the firebug but it didnt print in the browser can u do a further sample? – Noenoel

Answer 3

How to retrieve:

<div id="test">
  <?php
    if (isset($_POST['sweets'])) 
    {
      ob_clean();
      echo $_POST['sweets'];
      exit;
    }
  ?>
</div>

I'm looking for this:

$string = $_POST['sweets'];
<h2><?= $string;</h2> 

<form id="a" action="" method="post">
  <select name="sweets" onchange="change()" id="select1">
    <option >Chocolate</option>
    <option selected="selected">Candy</option>
    <option >Taffy</option>
    <option >Caramel</option>
    <option >Fudge</option>
    <option >Cookie</option>
  </select>
</form>

Korte

If you have a new question, please ask it by clicking the Ask Question button. Include a link to this question if it helps provide context. - From Review – Replace

Answer 4

I believe there is another way to get an ajax response on the same page. But some changes apply.

Like this easy and short

<script>
  function change() {
    var sweets = $("#select1").val();
    $.ajax({
      type: "POST",
      data: { sweets: sweets },
      success: function(data) {
        $("#test").html(data);
      }
    });
  }
</script>
<div id="test">
  <?php
    if (isset($_POST['sweets'])) 
    {
      ob_clean();
      echo $_POST['sweets'];
      exit;
    }
  ?>
</div>
<form id="a" action="" method="post">
  <select name="sweets" onchange="change()" id="select1">
    <option >Chocolate</option>
    <option selected="selected">Candy</option>
    <option >Taffy</option>
    <option >Caramel</option>
    <option >Fudge</option>
    <option >Cookie</option>
  </select>
</form>

I checked and it's working well

Wrung