小编典典

CSRF验证失败。请求中止

django

我尝试建立一个非常简单的网站,在该网站中可以将数据添加到sqlite3数据库中。我有两个文本输入的POST表单。

index.html:

{% if top_list %}
    <ul>
    <b><pre>Name    Total steps</pre></b>
    {% for t in top_list %}
        <pre>{{t.name}} {{t.total_steps}}</pre>
    {% endfor %}
    </ul>
    {% else %}
    <p>No data available.</p>
{% endif %}
<br>
<form action="/steps_count/" method="post">
    {% csrf_token %}
    Name: <input type="text" name="Name" /><br />
    Steps: <input type="text" name="Steps" /><br />
   <input type="submit" value="Add" />
 </form>

forms.py:

from django import forms
from steps_count.models import Top_List

class Top_List_Form(forms.ModelForm):
    class Meta:
        model=Top_List

views.py:

# Create your views here.
from django.template import Context, loader
from django.http import HttpResponse
from steps_count.models import Top_List
from steps_count.forms import Top_List_Form
from django.template import RequestContext
from django.shortcuts import get_object_or_404, render_to_response

def index(request):

if request.method == 'POST':
    #form = Top_List_Form(request.POST)
    print "Do something"
else:
    top_list = Top_List.objects.all().order_by('total_steps').reverse()
    t = loader.get_template('steps_count/index.html')
    c = Context({'top_list': top_list,})
    #output = ''.join([(t.name+'\t'+str(t.total_steps)+'\n') for t in top_list])
    return HttpResponse(t.render(c))

但是,当我单击“提交”按钮时,出现403错误:

CSRF verification failed. Request aborted.

我已经包含{% csrf_token %}在index.html中。但是,如果这是一个RequestContext问题,我真的不知道在哪里以及如何使用它。我希望所有事情都在同一页面上发生(index.html)。


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2020-03-31

共1个答案

小编典典

使用自动添加的render快捷方式RequestContext。

from django.http import HttpResponse
from django.shortcuts import get_object_or_404, render
from steps_count.models import Top_List
from steps_count.forms import Top_List_Form


def index(request):

    if request.method == 'POST':
        #form = Top_List_Form(request.POST)
        return HttpResponse("Do something") # methods must return HttpResponse
    else:
        top_list = Top_List.objects.all().order_by('total_steps').reverse()
        #output = ''.join([(t.name+'\t'+str(t.total_steps)+'\n') for t in top_list])
        return render(request,'steps_count/index.html',{'top_list': top_list})
2020-03-31