Jscript code for php page

Question

I have webpage. I need to create 2 types of view for it. Depending of dropdown form I have to choose with content to show.

So, I have created form:

        <form action="" method="post" class="form-horizontal">

        <div class="form1">
            <label class="label1" for="viewid">
                View option
            </label>

            <div class="controls">
                <select name="choose_viewid" id="viewid">

                        <option value="var1" >Full view</option>
                        <option value="var2" >Short view</option>


                </select>
            </div>
        </div>
    </form>

After that I have created php if:

if ($_POST[id]= "var1"): {
  include "var1.html";
}


else:
{
  include "var2.html";
}

endif;

I have coded js script at the bottom:

  <script language="JavaScript" type="text/javascript"> 

function chooseopt() {
var id= document.getElementById('viewid');

$.ajax({
type: 'POST',
url: '',
data:{id:id.value},
});
return false;
};
  </script>

But something is wrong. Who can advise needed actions? Thank you.

UPD: All this code I have at the same page.

Answer 1

change

if ($_POST[id]= "var1"):

to

if ($_POST['id']== "var1"):

Answer 2

You need to call chooseopt js method when you change option. You can use jquery handler ou old onchange action.

Answer 3

Change your javascript function like this.

function chooseopt() 
{
    var objSelected = document.getElementById("viewid");
    var id = objSelected.options[objSelected.selectedIndex].value;

    $.ajax({
    type: 'POST',
    url: '',
    data:{id:id},
    });
    return false;
};