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我在装上一种表格,使用舱载()功能,迄今为止,它正在做出色的工作。我要补充的是,如果提交表格时出现错误,就会把错误标记重新装入人口编组,而不是改用目前实际的纸页。

有些人建议我使用jquery-form。我认为,这将完全奏效。我不知道如何执行。

此处为有效载荷功能:

$(document).ready(function(){
        $(".create").on("click", function(){
            $("#popupContact").load("/cookbook/createrecipe #createform");
        });
});

The page thatloads the form:

<div id="popupContact" class="popup">
        <a id="popupContactClose" style="cursor:pointer;float:right;">x</a>
        <p id="contactArea">
        </p>
</div>
<div id="backgroundPopup">
</div>  
<div id="col2-footer">
{% paginate %}
</div>

我的形式模板如下:

<div id="createform">
<h1>Create New Recipe</h1>
    <form id="createrecipe" action="{% url createrecipe %}" method="POST">
        <table>
            {% csrf_token %}
            {{ form.as_table }}
        </table>
        <p><input type="submit" value="Submit"></p>
    </form>
</div>

here is my attempt to use the jquery-form:

<script> 
    // wait for the DOM to be loaded 
    $(document).ready(function() { 
        var options ={
            target:  .popup ,
    };
    $( #createrecipe ).submit(function() {
        $(this).ajaxSubmit(options); 
        return false;
    }); 
});
</script>

形成反感:

def createrecipe(request):
    if not request.user.is_authenticated():
        return HttpResponseRedirect( /index/ )
    else:
        if request.method ==  POST :
            print 1
            form = RecipeForm(request.POST)
            if form.is_valid():
                print 2
                recipe = form.save(commit=False)
                recipe.original_cookbook = request.user.cookbooks.all()[0]
                recipe.pub_date = datetime.datetime.now()
                recipe.save()
                user = request.user
                cookbooks = user.cookbooks
                cookbook = cookbooks.all()[0]
                cookbook.recipes.add(recipe)
                return HttpResponseRedirect( /account )
        else:
            form = RecipeForm()

        return render_to_response( cookbook/createrecipe.html ,
                                    { form :form},
                              context_instance=RequestContext(request))

thank you snackerfish

Answer 1

i) 我的javascript的辛醇全都是ok。

Answer 2

既然你使用 j,你就应该通过麻省提交你表格,否则你将改头:

 $( #createform form ).submit(function(e) {
 e.preventDefault();
 $.post("/your views url/", { 
            data: $( #createform form ).serialize(),
            },function(data){ //this is the successfunction 
                              //data is what your view returns}
 });

Your view should return the errors as json so that you can process them inside your javascript:

from django.utils import simplejson

def ajax_recipe(request):
    if request.method ==  POST :
        form = YourForm(request.POST)
        if form.is_valid():
            form.save()
            return HttpResponse("ok")
        else:
            errors = form.errors
            return HttpResponse(simplejson.dumps(errors))
    else:
        data = "error"
        return HttpResponse(data)

有了这一标志,你就能够把你想要的那类错误放在你身上,利用破碎后的成功功能。

if(data == "ok"){do something}
// else error handling
var errors = jQuery.parseJSON(data)
if(errors.somefield){//place errors.somefield anywhere}

Note the code is not tested. But thats the way I would go.

Edit

请注意,要完成这项工作,你必须确定一个习惯X-CSRF Token Header to the Value of the CSRF token on each XMLHttpRequest。 http://docs.djangoproject.com

您可以如何提出错误,以证明您的表态。

def clean_somefield(self):
        somefield = self.cleaned_data.get( some field )
        if not somefield:
            raise forms.ValidationError(u This field is required. )

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