Welcome to OGeek Q&A Community for programmer and developer-Open, Learning and Share
Welcome To Ask or Share your Answers For Others

Categories

0 votes
194 views
in Technique[技术] by (71.8m points)

python - Django Ajax Jquery Call

This may be basic, but I've spent two days, read countless tutorials and I still can not get this to work. For simplicitly I tried to accomplish a basic task just to see it work. I want to send make an ajax call to my donate view. I see that it successfully passes through but I was expecting that my template would also get updated to "TRUE" but it remains "FALSE". Any help or suggestions I appreciate.

my jquery...

$.ajax({
    type: "POST",
    url:"/donate/",
    data: {
    'test': 'success',
    },
    success: function(){
      alert('test')
     },
    error: function(){
        alert("Error");
});

this is my view...

def donate(request):
    if request.is_ajax():
        test = "TRUE"

    if request.method == 'POST':
        form = DonateForm(request.POST)
        if form.is_valid():
            form.save()
    else:
        form = DonateForm()
        test = "FALSE"

    return render_to_response('donate_form.html', {'form':form,'test':test}, context_instance=RequestContext(request))

my template include this...

<h1>{{ test }}</h1>

Update/Solution

Like mentioned in the comments on this question, I was not doing anything the the returned data. I updated my success call to the following and it worked

    $.ajax({
    type: "POST",
    url:"/donate/",
    data: {
    'zip': zip,
    },
    success: function(data){            

        results = $(data).find('#results').html()               
        $("#id_date").replaceWith("<span>" + results + "</span >");

    },
    error: function(){
        alert("Error");
    },
See Question&Answers more detail:os

与恶龙缠斗过久,自身亦成为恶龙;凝视深渊过久,深渊将回以凝视…
Welcome To Ask or Share your Answers For Others

1 Reply

0 votes
by (71.8m points)

I think the problem is at where you pass the data. Do you use Firebug? An excellent tool for checking if you pass anything in POST, it is an excellent tool for web development in general.

Here's a working example for sending Ajax call from a form

$("#form").submit(function(event) {
        var $form = $(this),
        $inputs = $form.find("input, select, button, textarea"),
        serializedData = $form.serialize();

        $.ajax({
            url: "/donate/",
            type: "post",
            data: serializeData,
            success: function(response) {
                alert(response)
            }
        })
        event.preventDefault();
    });

and then your view can look something like this

if request.is_ajax() or request.method == 'POST':
        form = DonateForm(data=request.POST)
        if form.is_valid():
            return HttpResponse("Success")
        else:
            return HttpResponse("Fail")

Btw, unless you really need all the extras from $.ajax() I would suggest you to use $.post() instead as it seems more straight forward to use.


与恶龙缠斗过久,自身亦成为恶龙;凝视深渊过久,深渊将回以凝视…
OGeek|极客中国-欢迎来到极客的世界,一个免费开放的程序员编程交流平台!开放,进步,分享!让技术改变生活,让极客改变未来! Welcome to OGeek Q&A Community for programmer and developer-Open, Learning and Share
Click Here to Ask a Question

...