Django表单is_valid()失败

Joe*_*Fox 9 forms django post

我是网络开发的真正初学者.以下代码在is_valid()检查时失败.但我不明白为什么:表单应该从POST数据中填充数据?

模型:

class Statement(models.Model):
    text = models.CharField(max_length=255)
    user = models.ForeignKey(User)
    time = models.DateField()
    views = models.IntegerField()
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的ModelForm:

class StatementForm(ModelForm):
    class Meta: 
        model = Statement
        widgets = {
                   'time':forms.HiddenInput(),
                   'user':forms.HiddenInput(), 
                   'views':forms.HiddenInput(), 
        }
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查看功能:

def new(request):  
    if request.method == 'POST': # If the form has been submitted...
        form = StatementForm(request.POST) # A form bound to the POST data
        if form.is_valid():
            stmt = form.save()
            path = 'stmt/' + stmt.id
            return render_to_response(path, {'stmt': stmt})
    else:  
        c = {}
        c.update(csrf(request))
        loggedin_user = request.user 
        d = datetime.now()
        form = StatementForm(request.POST, initial={'time': d.strftime("%Y-%m-%d %H:%M:%S"), 'user':loggedin_user, 'views':0})
        return render_to_response('new_stmt.html', {'form': form, },context_instance=RequestContext(request))
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我找到了类似的主题,并尝试了很多.这就是我认为应该如何运作的方式.我真的需要建议.

sta*_*alk 10

您的模型的所有字段都是必需的.因此,form.is_valid()如果所有字段都填充了正确的值且未被消隐,则为True.你已经宣布领域time,user,views为隐藏字段.你确定,你已经填写了模板表格吗?此外,您可能想要自动标记字段time = models.DateField().修改您的模型字段,如

time = models.DateField(auto_now=True)`. 
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在此之后,您不必自己以模板形式填写它.

您的视图必须在所有情况下都返回HttpResponse对象.如果您的表单无效,即if form.is_valid()将返回False,那么HttpResponse您的视图将不会返回任何对象.这可能是您失败的根源.添加else语句if form.is_valid():

from django.http import Http404
def new(request):  
    if request.method == 'POST': # If the form has been submitted...
        form = StatementForm(request.POST) # A form bound to the POST data
        if form.is_valid():
            stmt = form.save()
            path = 'stmt/' + stmt.id
            return render_to_response(path, {'stmt': stmt})
        else:
            # Do something in case if form is not valid
            raise Http404 
    else: 
        # Your code without changes
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