Codeigniter ajax request fail on output

Question

I'm trying to make an ajax request with jquery/codeigniter. Well, there seems to be a problem. i do get it to work, but it seems like the POST is not sent, for some reason...

Javascript

    $('#profileUpdateButton').click(function() {

        $.ajax({ // Starter Ajax Call

            method: "POST", 
            url: baseurl + 'profile/statusUpdate', 
            data: $('#statusUpdateForm').serialize(),
            success: function(data) { 
                alert(data);
            }

        });

        return false;

    });

From the codeigniter controller

    if($this->input->is_ajax_request()) {
        echo $this->input->post('profileUpdate');
    }

If i replace the "echo $this->" etc.. with just echo "Hello" i do get an alertbox with "Hello", why don't i get the text from the box?

html

    <div id="statusUpdate">
        <?php echo form_open(base_url() . 'profile/statusUpdate', array('name' => 'statusUpdateForm')); ?>
            <input type="text" value="Hva tenker du på?" name="profileUpdate" id="profileUpdate" onfocus="if(this.value == 'Hva tenker du på?')this.value=''" onblur="if(this.value == '')this.value='Hva tenker du på?'" />
            <input type="submit" value="" name="profileUpdateButton" id="profileUpdateButton" />
        <?php echo form_close(); ?>
    </div>

have you tried to alert $('#statusUpdateForm').serialize() into the click event handler? — Dalen
– Dalen, Commented Jul 21, 2011 at 20:10
What happened when you ran var_dump($_POST)? Should be one of your your default debugging tools: php.net/manual/en/function.var-dump.php — No Results Found
– No Results Found, Commented Jul 21, 2011 at 21:06
if i var_dump i get array(0){} and alert $('#statusUpdateForm').serialize() give me prntscr.com/2ecwh — Dexty
– Dexty, Commented Jul 23, 2011 at 14:49

jondavidjohn · Accepted Answer · 2011-07-21 21:31:14Z

2

Change your html markup to this...

<div id="statusUpdate">                   <!-- Use ID not NAME here |v| -->
    <?php echo form_open(base_url() . 'profile/statusUpdate', array('id' => 'statusUpdateForm')); ?>
        <input type="text" value="Hva tenker du på?" name="profileUpdate" id="profileUpdate" onfocus="if(this.value == 'Hva tenker du på?')this.value=''" onblur="if(this.value == '')this.value='Hva tenker du på?'" />
        <input type="submit" value="" name="profileUpdateButton" id="profileUpdateButton" />
    <?php echo form_close(); ?>
</div>

you were setting the form's name and you were selecting $('#statusUpdateForm') which means the id of statusUpdateForm, not the name attribute...

edited Jul 21, 2011 at 21:31

answered Jul 21, 2011 at 21:15

jondavidjohn

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