AJAX/Javascript Select element change on page load in PHP

Question

I am using the following Javascript to populate another form based on a dynamic select using PHP/Mysql.

This works great but I also want this to trigger when I send a URL string use _GET. I've configured the select element to use this as default:

<option value="<?php echo $_GET['EmpNumber'] ?>"><?php echo $_GET['EmpFirstName'] ?> <?php echo $_GET['EmpLastName'] ?></option>

But when I call the page with values for EmpNumber,EmpFirstName and EmpLastName in the URL, the below script does not trigger. How do I achieve this as well?

<script>
      $(document).ready(function() {    
$('select').on('change', function (e) {
    var optionSelected = $("option:selected", this);
    var valueSelected = this.value;
   if(valueSelected){
           $.ajax({
            type: "POST",
            cache: false,
            url: "getinfo.php",
            data: "data="+valueSelected,
            dataType: "html",
            success: function(res) 
            {

               //alert(res);
          $('#Name').val(res + '\'s');
            },
            error: function(xhr, ajaxOptions, thrownError)
            {

            },
            async: false
        });
   }

   else {
        $('#Name').val('Value');
   }




   //alert(valueSelected);
});
});
</script>

Anthony Grist · Accepted Answer · 2014-03-05 14:54:09Z

2

If I'm understanding the question correctly you want to trigger the same functionality for the change event when the page first loads. That's really simple:

$('select').on('change', function() {
    // some code
}).trigger('change');

That last part, .trigger('change') (or just .change() for short), will execute the function you just bound to the change event for those elements, as if the user had selected a new value.

answered Mar 5, 2014 at 14:54

Anthony Grist

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2 Comments

Rocco The Taco Over a year ago

Where should I put the trigger in the context of the code? This did not work for me: async: false }).trigger('change');

Anthony Grist Over a year ago

@RoccoTheTaco It goes immediately after the closing bracket for the call to .on(), as I demonstrated in my answer. In your code, that would be the second to last line as far as I can make out, the one just below the commented out alert.

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