Submitting a jQuery ajax form with two submit butt

问题:

I have a form that looks like this:

<form action="/vote/" method="post" class="vote_form">
    <input type="hidden" name="question_id" value="10" />
    <input type="image" src="vote_down.png" class="vote_down" name="submit" value="down" />
    <input type="image" src="vote_up.png" class="vote_up" name="submit" value="up" />
</form>

When I bind to the form's submit ($("vote_form").submit()), I don't seem to have access to which image the user clicked on. So I'm trying to bind to clicking on the image itself ($(".vote_down, .vote_up").click()), which always submits the form, regardless of whether I try

• return false;
• event.stopPropogation(); or
• event.preventDefault();

because all of those are form events.

1. Should I attach my $.post() to the form.submit() event, and if so, how do I tell which input the user clicked on, or

2. Should I attach my $.post() to the image click, and if so, how do I prevent the form from submitting also.

Here is what my jQuery code looks like now:

$(".vote_up, .vote_down").click(function (event) {
    $form = $(this).parent("form");
    $.post($form.attr("action"), $form.find("input").serialize() + {
        'submit': $(this).attr("value")
    }, function (data) {
        // do something with data
    });
    return false; // <--- This doesn't prevent form from submitting; what does!?
});

回答1:

Based on Emmett's answer, my ideal fix for this was just to kill the form's submit with Javascript itself, like this:

$(".vote_form").submit(function() { return false; });

And that totally worked.

For completeness, some of my JS code in the original post need a little love. For example, the way I was adding to the serialize function didn't seem to work. This did:

    $form.serialize() + "&submit="+ $(this).attr("value")

Here's my entire jQuery code:

$(".vote_form").submit(function() { return false; });
$(".vote_up, .vote_down").click(function(event) {
    $form = $(this).parent("form");
    $.post($form.attr("action"), $form.serialize() + "&submit="+ $(this).attr("value"), function(data) {
        // do something with response (data)
    });
});

回答2:

Another solution is to use a hidden field, and have the onclick event update its value. This gives you access from javascript, as well as on the server where the hidden field will get posted.

回答3:

You can trigger the form submit on the click of the images. This will work with the preventDefault().

var vote;
$(".vote_up, .vote_down").click(function(event) {
    vote = $(this).attr("class");
    $(".vote_form").trigger("submit");
});

$(".vote_form").submit(function(event) { 
    $form = $(this);
    $.post($form.attr("action"), $form.serialize() + "&submit="+ vote, function(data) {
        // do something with response (data)
    });     
    event.preventDefault();
});

回答4:

I don't get how return false and preventDefault failed to do their job. Maybe try replacing the image buttons with linked images:

<a href="#" class="vote_down"><img src="vote_down.png"/></a>

$('#vote_form > a').click(function(e) {
    e.preventDefault();

    //one way to know which image was clicked
    alert($(this).attr('class'));

    $.post(...
});

You can always ensure that a form does not submit by binding to the submit event, e.g.:

$('#vote_form').submit(function() {
    return false;
});

回答5:

Try adding onsubmit="return false;" to your form, and then submitting your form with javascript:

<form action="/vote/" method="post" name="vote_form" class="vote_form" onsubmit="return false;">
    <input type="hidden" name="question_id" value="10" />
    <input type="image" src="vote_down.png" class="vote_down" name="submit" value="down" onclick="imageClicked(this)"/>
    <input type="image" src="vote_up.png" class="vote_up" name="submit" value="up" onclick="imageClicked(this)"/>
</form>

<script>
function imageClicked(img) {
    alert(img.className);
    document.forms["vote_form"].submit();
}
</script>

回答6:

You can also try out the jquery From Plugin which as all sorts of nice tools for submitting forms via ajax.

http://malsup.com/jquery/form/

回答7:

Without resorting to the Form plugin (which you should use) you should be handling the submit event instead. The code would stay pretty close to the original:

$("form").submit(function()) {
  $.post($(this).attr("action"), $(this).serialize(), function(data) {
    // work with the response
  });
  return false;
});

回答8:

var form = jQuery('#myform');

var data = form.serialize();

// add the button to the form data
var btn = jQuery('button[name=mybuttonname]').attr('value');
data += '&yourpostname=' + btn;

var ajax = jQuery.ajax({
    url: url,
    type: 'POST',
    data: data,
    statusCode: {
        404: function () {
            alert("page not found");
        }
    }
});
... rest of your code ...