Form posts data twice

I have a feedback form in a pop-up div that otherwise works fine but processes SQL twice when the form results in error at first instance.

This is the html form:

<div id="stylized" class="myform">
    <form id="form" method="post" name="form">
        <p>Report:
            <select id="fbtype" name="fbtype"> 
            <option>Bug</option>
            <option>Suggestion</option>
            <option>Discontentment</option>
            <option>Appreciation</option>
            </select>
        </p>
        <p>Brief description:
            <textarea name="comments" id="comments" cols="45" rows="10"></textarea>
        </p>
        <span class="error" style="display:none">Please describe your feedback.</span>
        <span class="success" style="display:none">We would like to thank you for your valuable input.</span>
   <input type="button" value="Submit" class="submit" onclick="feedback_form_submit()"/>
    </form>
</div>

The feedback_form_submit() function is:

function feedback_form_submit() {
$(function() {
$(".submit").click(function() {
var fbtype = $("#fbtype").val();
var comments = $("#comments").val();
var dataString = 'fbtype='+ fbtype + '&comments=' + comments;

if(fbtype=='' || comments=='' )
    {
    $('.success').fadeOut(200).hide();
    $('.error').fadeOut(200).show();
    }
    else
    {
    $.ajax({
    type: "POST",
    url: "processfeedback.php",
    data: dataString,
    success: function(){
    $('.success').fadeIn(200).show();
    $('.error').fadeOut(200).hide();
    }
    });
    }
return false;
});
});

}

And the processfeedback.php has:

include "./include/session_check.php";
include_once "./include/connect_to_mysql.php";

if (isset($_POST['fbtype'])){

$userid =$_SESSION['id'];
$fbtype=$_POST['fbtype'];
$comments=$_POST['comments'];
$sql = mysql_query("INSERT INTO feedback (userid, type, comments) 
    VALUES('$userid','$fbtype','$comments')") or die (mysql_error());
}

Could anyone figure out why does the form submits twice? And any suggestion to control this behaviour?开发者_高级运维

If this is actually the code you're using, you seem to have wrapped your onclick function around the $.click event-adding function:

function feedback_form_submit() {
    $(function() {

        //  This adds a click handler each time you run feedback_form_submit():
        $(".submit").click(function() {

            // ... //

            return false;
        });
    });
}

When I tried this on jsFiddle.net, I clicked Submit the first time and nothing happened, then the second time it posted twice, the third click posted three times, etc.

You should just keep it simple: take out the onclick attribute:

<input type="button" value="Submit" class="submit" />

and remove the feedback_form_submit() wrapper:

$(function() {

    $(".submit").click(function() {

        // ... //

        return false;
    });
});

This way the $.click handler function will be applied just once, when the page loads, and will only run once when Submit is clicked.

EDIT:

If your form is loaded via AJAX in a popup DIV, you have two options:

Keep your onclick but remove the $.click wrapper instead:

function feedback_form_submit() {

    // ... //

}

and

<input type="button" value="Submit" class="submit" onclick="feedback_form_submit()" />

Note that you only need to return false if you're using <input type="submit" ... >; when using <input type="button" ... >, the browser does not watch the return value of onclick to determine whether to post the form or not. (The return value may affect event propagation of the click, however ...).

Alternatively, you can use jQuery's $.live function:

$(function() {
    $('.submit').live('click',function() {

        // ... //

    });
});

and

<input type="button" value="Submit" class="submit" />

This has the effect of watching for new DOM elements as they are added dynamically; in your case, new class="submit" elements.

Your feedback_form_submit function doesn't return false and on submit click you're also posting to the server. There is no need to have onClick in:

 <input type="button" value="Submit" class="submit" onclick="feedback_form_submit()"/>

Change that to:

 <input type="button" value="Submit" class="submit"/>

And change your code to:

// Update: Since you're loading via AJAX, bind it in the success function of the
// AJAX request.

// Let's make a function that handles what should happen when the popup div is rendered
function renderPopupDiv() {
    // Bind a handler to submit's click event
    $(".submit").click(function(event) {
        var fbtype = $("#fbtype").val();
        var comments = $("#comments").val();
        var dataString = 'fbtype='+ fbtype + '&comments=' + comments;

        if(fbtype=='' || comments=='' )
        {
            $('.success').fadeOut(200).hide();
            $('.error').fadeOut(200).show();
        }
        else
        {
            $.ajax({
                type: "POST",
                url: "processfeedback.php",
                data: dataString,
                success: function(){
                    $('.success').fadeIn(200).show();
                    $('.error').fadeOut(200).hide();
                }
            });
        }
        return false;
    });

// This is the success function for the AJAX request that loads the popup div
success: function() {
    ... 
    // Run our "onRender" function
    renderPopupDiv();
}