Help with adding a jquery alert box after form is correctly submitted

[eluser]LuckyFella73[/eluser]
would be nicer if you would start and post what you allready tried
and where to optimise ..

But I give you a starting point:

Your form could look like:

Code:

<p>
    &lt;input type="text" name="item" id="item" value="" /&gt;&lt;br />
    &lt;input type="submit" name="submit_item" id="submit_item" value="submit item" /&gt;
</p>

Place that js in your header (don't forget the js wrapping tags)
and include the jquery.js before this !

Code:

$(document).ready(function()
{
    $("#submit_item").click(function ()
    {
        var item = $('#item').val(); // for each form item you will need an entry here

        $.post("form/process", { "item" : item }, // extend the brackets {} with your items seperated with ","
        
        function(data) { // when your callback func. answers here we go
            // in this example your callback function returned variable "result"
            // in json encoded

            alert(data.result);

            // return a flag (from callback funct.) when the formdata is saved and everything is fine
            // print message here -> depending on the flag ("success" or "try again")
        }, 'json');

    });
});

Your submit item (submit button) has to have the id "submit_item"

In your controller (called "form" in this example) you need the callback
function. Could look like:

Code:

function process()
    {
        $item = $this->input->post('item');
        $item = 'From The Controller '.$item;
        $array = '{ "result" : "'.$item.'" }';
        echo $array;
    }

For I had no php 5 running and this I couldn't use "json_encode()"
I manually encoded the data ...

That should give you a good starting point - I hope.

The js function is more or less a example from the jquery website.