Django - 提交的链接

Question

我有一个POST提交按钮的窗体。我想在提交表单和form.is_valid（）时链接到新页面。

在视图中还是在模板中创建链接会更好吗？怎么做？

view.py：

from django.shortcuts import render_to_response 
from ezmapping.models import * 
from django.forms.models import modelformset_factory 

def setName(request): 
    ezAppFormSet = modelformset_factory(ezApp, extra=1, fields=('name')) 
    formset = ezAppFormSet(queryset=ezApp.objects.none()) 
    if request.method == 'POST': 
     formset = ezAppFormSet(request.POST, request.FILES) 
     if formset.is_valid(): 
      formset.save() 
    return render_to_response("project/manage_new.html", {'formset': formset, 'title': "New"}, context_instance=RequestContext(request)) 

template.html

{% extends "basemap.html" %} 
{% block content %} 
<table border="1"> 
    <tr> 
    <td> 
    <h1>Define new App options</h1> 
    {% if formset.errors %} 
     <p style="color: red;"> 
      Please correct the error{{ formset.errors|pluralize }} below. 
     </p> 
    {% endif %} 
    <form method="post" action="." encrypt="multipart/form-data">{% csrf_token %} 
      {{ formset.as_p }} 
     <input type="submit" value="Submit"> 
    </form> 
</td> 
</tr> 
</table> 
{% endblock %} 

Answer 1

你可以在这样的视图中使用HttpResponseRedirect（）：

from django.shortcuts import render_to_response 
from ezmapping.models import * 
from django.forms.models import modelformset_factory 

def setName(request): 
    ezAppFormSet = modelformset_factory(ezApp, extra=1, fields=('name')) 
    formset = ezAppFormSet(queryset=ezApp.objects.none()) 
    if request.method == 'POST': 
     formset = ezAppFormSet(request.POST, request.FILES) 
     if formset.is_valid(): 
      formset.save() 
      next_link = u"/next/" 
      return HttpResponseRedirect(next_link) 

    return render_to_response("project/manage_new.html", {'formset': formset, 'title': "New"}, context_instance=RequestContext(request)) 

Answer 2

肯定。只要您的条件内返回一个选择render_to_response为request.method == 'POST'，或可选，刚刚成立的背景和模板变量，以手握成选择render_to_response这样：

意译：

def foo(request): 
    template = 'template1.html' 
    # form and formsets here set into `context` as a dictionary 

    if request.method == 'POST': 
     template = 'template2.html' 

    return render_to_response(template, context) 

[编辑]

当我重新阅读你的问题时，如果你想重定向到不同的视图，如果你的表单是有效的，然后返回一个HttpResponseRedirect。

Answer 3

如果我正确理解你的问题。

您的看法：

from django.shortcuts import render_to_response 
from ezmapping.models import * 
from django.forms.models import modelformset_factory 

def setName(request): 
    ezAppFormSet = modelformset_factory(ezApp, extra=1, fields=('name')) 
    formset = ezAppFormSet(queryset=ezApp.objects.none()) 
    if request.method == 'POST': 
     formset = ezAppFormSet(request.POST, request.FILES) 
     if formset.is_valid(): 
      formset.save() 
      submit_link = True # 2* 
     else: 
      submit_link = False 
    return render_to_response("project/manage_new.html", {'submit_link': submit_link, 'formset': formset, 'title': "New"}, context_instance=RequestContext(request)) 

您的模板：

{% extends "basemap.html" %} 
{% block content %} 
<table border="1"> 
    <tr> 
    <td> 
    <h1>Define new App options</h1> 
    {% if formset.errors %} 
     <p style="color: red;"> 
      Please correct the error{{ formset.errors|pluralize }} below. 
     </p> 
    {% endif %} 
    <form method="post" action="." encrypt="multipart/form-data">{% csrf_token %} 
      {{ formset.as_p }} 
     <input type="submit" value="Submit"> 
    </form> 

    {% if submit_link %} 
     <a href='/next/'>Data is saved, let's continue.</a> 
    {% endif %} 
</td> 
</tr> 
</table> 
{% endblock %} 

更新。

是的，如果你想重定向（不显示下一个链接）只需将：

from django.shortcuts import redirect 
... 
return redirect('you-url') 

而不是2 *。