Django 验证并返回预览 URL？

发布于 2024-08-01 19:42:11 字数 947 浏览 5 评论 0原文

我再次询问:)，我不知道如何做到这一点。

我的英语不太好，但我会尝试问这个：

如何验证表单并返回预览网址（相同的视图表单）并显示验证错误？我问这个是因为我有 2 个表单，第一个表单的操作将转到第二个表单（POST），但是在第二个表单（视图？）中，我需要验证第一个表单，如果第一个表单有效，我想显示第二个表单，否则将显示第一个表单，其中有错误。

我不知道我是否清楚，抱歉。

我正在考虑这个：

def secondForm(request):
    if request.method =='POST':
       form = FirstForm(request.POST)
       if form.is_valid():
          fields = request.POST.copy()
          # showing the second form?? with x differents fields, i have 2.
          if fields['xvalue']=='1': # this is from radio buttons
             form2 = xSecondForm()
          elif fields['xvalue']=='2':
             form2 = ySecondForm()
       else:
         # here go back with erros msgs????
         #return render_to_response('firstFormTemplate.html',{'form': form}, context_instance=RequestContext(request))
    else:
      #return something or go back ???

谢谢大家 PD：我很高兴，我正在编码 Django + python :D

原文

im asking again :), i don't know how make this.

My English is not too good, but ill try to asking this:

how ill validate a form and go back to the preview url (the same view form) and show the validation errors?, im asking this because i have 2 form's, the first form's action is going to a second form (POST), but in this second form (view?) i need to validate the first form, if the first form is valid i want to show the second form else ill show the first form with errors.

i dont know if im clear , im sorry.

im thinking about this:

def secondForm(request):
    if request.method =='POST':
       form = FirstForm(request.POST)
       if form.is_valid():
          fields = request.POST.copy()
          # showing the second form?? with x differents fields, i have 2.
          if fields['xvalue']=='1': # this is from radio buttons
             form2 = xSecondForm()
          elif fields['xvalue']=='2':
             form2 = ySecondForm()
       else:
         # here go back with erros msgs????
         #return render_to_response('firstFormTemplate.html',{'form': form}, context_instance=RequestContext(request))
    else:
      #return something or go back ???

Thanks guys
PD: im happy, im coding Django + python :D

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以往的大感动 2024-08-08 19:42:11

是的，formwizard 可能是您的答案，但您可能可以在视图中使用一些逻辑来做到这一点。

就像是：

def your_view(request):
    context = {}
    data = request.method == 'POST' and request.POST or None
    form = FirstForm(data=data)
    # If the first form is valid, build the second.
    if form.is_valid():
        if form.cleaned_data['xvalue'] == 1:
            form2 = xSecondForm(data=data)
        else:
            form2 = ySecondForm(data=data)
        context['form2'] = form2
    # If both forms are valid, everything is done.
    if form.is_valid() and form2.is_valid():
        # TODO: put save/processing logic here
        # Now redirect.
        return http.HttpResponseRedirect(...)
    # If we get here, either there hasn't been a post yet, the second form hasn't
    # been entered, or there was an error in either form.
    context['form'] = form
    return render_to_response(...)

Yes, formwizard might be your answer, but you could probably do it with some logic in a view.

Something like:

def your_view(request):
    context = {}
    data = request.method == 'POST' and request.POST or None
    form = FirstForm(data=data)
    # If the first form is valid, build the second.
    if form.is_valid():
        if form.cleaned_data['xvalue'] == 1:
            form2 = xSecondForm(data=data)
        else:
            form2 = ySecondForm(data=data)
        context['form2'] = form2
    # If both forms are valid, everything is done.
    if form.is_valid() and form2.is_valid():
        # TODO: put save/processing logic here
        # Now redirect.
        return http.HttpResponseRedirect(...)
    # If we get here, either there hasn't been a post yet, the second form hasn't
    # been entered, or there was an error in either form.
    context['form'] = form
    return render_to_response(...)

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