Form / ajax / PHP issue

Question

I am trying to make a small form that lets the user pick one element from 3 different radiobutton lists to set one element as the users active element (that will be stored to MySQL). Somehow along the way it does not work and I can not seem to figure out why, perhaps someone of you can see what I did wrong?

HTML:

<form name="activeForm1" method="post">
<fieldset data-role="controlgroup">
    <div class="ui-radio">
        <input type="radio" name="active" value="1" id="1">
        <label for="1"></label></input>
    </div>
    <div class="ui-radio">
        <input type="radio" name="active" value="2" id="2">
        <label for="2"></label></input>
    </div>
    <div class="ui-radio">
        <input type="radio" name="active" value="3" id="3">
        <label for="3"></label></input>
    </div>
</fieldset>
<div data-role="footer">
    <input type="submit" href="#" onclick="setActive(1)"/>
</div>
</form>

JavaScript / Ajax call

function setActive(formid) 
    {
            $.ajax(
            {
                type:'POST',
                url:'active.php',
                data:$('#activeForm'+formid).serialize(),
                success:function(response)
                {
                }
            }
        );
    }

PHP code:

session_start();
include('connectToDb.php');
$id = $_SESSION['id'];

if (isset($_POST['active']))
{
    $formValue = $_POST['active'];
    mail('[email protected]','Test',$formValue,'From: [email protected]');
    mysql_query(/* UPDATE MySQL */);
    header("Location: main.php");
}
else
{
    mail('[email protected]','Test','No data recieved!','From: [email protected]');
}

So it works up until the if (isset($_POST['active'])) but then mails me that no data was recieved. I already have 2 similar forms on the same page and they are way bigger and has no problems running. Can't figure out what I did wrong here.

Are you making sure that at least one of the radio buttons is checked before clicking the submit button? — asprin
– asprin, Commented Jul 27, 2012 at 6:26

Sumesh TG · Accepted Answer · 2012-07-27 06:31:19Z

2

Wrong code :

  data:$('#activeForm'+formid).serialize(),

#activeForm is not an id, it is the name of the form tag,

Correct the form tag to,

<form name="activeForm1" id="activeForm1" method="post">

answered Jul 27, 2012 at 6:31

Sumesh TG

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2 Comments

asprin Over a year ago

Damn..How did i miss that? Nice catch

Vlad Over a year ago

Aaaaah! That makes perfect sense! Cheers! Been bugging me for hours now, can't believe I missed that!

scessor · Accepted Answer · 2012-07-27 06:31:51Z

2

Replace following line

data:$('#activeForm'+formid).serialize(),

with

data: $('form[name="activeForm'+formid+'"]').serialize(),

answered Jul 27, 2012 at 6:31

scessor

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Comments

Dr. Dan · Accepted Answer · 2012-07-27 06:26:58Z

0

change

<input type="submit" href="#" onclick="setActive(1)"/>

to

<input type="button" href="#" onclick="setActive(1)"/>

and then it should work

answered Jul 27, 2012 at 6:26

Dr. Dan

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