CSRF令牌错误，同时要求对POST请求一个JSON

Question

我试图呈现从Django中的POST请求的JSON响应，使得我的看法是

from django.shortcuts import render 
from django.http import HttpResponse 
from django.views.generic import View 

class TaskView(View): 
    def get(self,request,*args,**kwargs): 
     return HttpResponse("hello",content_type="application/json") 
    def post(self,request,*args,**kwargs): 
     return HttpResponse("hello",content_type="application/json") 

这必须返回两个你好JSON响应获取以及作为发布请求。

但我做了这个工作得到请求。但是，如果我做一个post请求，然后 我收到以下错误

Forbidden (403) 
CSRF verification failed. Request aborted. 
You are seeing this message because this HTTPS site requires a 'Referer header' to be sent by your Web browser, but none was sent. This header is required for security reasons, to ensure that your browser is not being hijacked by third parties. 
If you have configured your browser to disable 'Referer' headers, please re-enable them, at least for this site, or for HTTPS connections, or for 'same-origin' requests. 
Help 
Reason given for failure: 
    Referer checking failed - no Referer. 

In general, this can occur when there is a genuine Cross Site Request Forgery, or when Django's CSRF mechanism has not been used correctly. For POST forms, you need to ensure: 
Your browser is accepting cookies. 
The view function uses RequestContext for the template, instead of Context. 
In the template, there is a {% csrf_token %} template tag inside each POST form that targets an internal URL. 
If you are not using CsrfViewMiddleware, then you must use csrf_protect on any views that use the csrf_token template tag, as well as those that accept the POST data. 
You're seeing the help section of this page because you have DEBUG = True in your Django settings file. Change that to False, and only the initial error message will be displayed. 
You can customize this page using the CSRF_FAILURE_VIEW setting. 

我中间件类设置

MIDDLEWARE_CLASSES = (
    'django.contrib.sessions.middleware.SessionMiddleware', 
    'django.middleware.common.CommonMiddleware', 
    'django.middleware.csrf.CsrfViewMiddleware', 
    'django.contrib.auth.middleware.AuthenticationMiddleware', 
    'django.contrib.auth.middleware.SessionAuthenticationMiddleware', 
    'django.contrib.messages.middleware.MessageMiddleware', 
    'django.middleware.clickjacking.XFrameOptionsMiddleware', 
) 

我需要重构的角度来处理JSON POST请求任何部分。

Answer 1

您需要在您的要求实施明确的CSRF保护：

// using jQuery 
function getCookie(name) { 
    var cookieValue = null; 
    if (document.cookie && document.cookie != '') { 
     var cookies = document.cookie.split(';'); 
     for (var i = 0; i < cookies.length; i++) { 
      var cookie = jQuery.trim(cookies[i]); 
      // Does this cookie string begin with the name we want? 
      if (cookie.substring(0, name.length + 1) == (name + '=')) { 
       cookieValue = decodeURIComponent(cookie.substring(name.length + 1)); 
       break; 
      } 
     } 
    } 
    return cookieValue; 
} 
var csrftoken = getCookie('csrftoken'); 

function csrfSafeMethod(method) { 
    // these HTTP methods do not require CSRF protection 
    return (/^(GET|HEAD|OPTIONS|TRACE)$/.test(method)); 
} 

function sameOrigin(url) { 
    // test that a given url is a same-origin URL 
    // url could be relative or scheme relative or absolute 
    var host = document.location.host; // host + port 
    var protocol = document.location.protocol; 
    var sr_origin = '//' + host; 
    var origin = protocol + sr_origin; 
    // Allow absolute or scheme relative URLs to same origin 
    return (url == origin || url.slice(0, origin.length + 1) == origin + '/') || 
     (url == sr_origin || url.slice(0, sr_origin.length + 1) == sr_origin + '/') || 
    // or any other URL that isn't scheme relative or absolute i.e relative. 
    !(/^(\/\/|http:|https:).*/.test(url)); 
} 
$.ajaxSetup({ 
    beforeSend: function(xhr, settings) { 
     if (!csrfSafeMethod(settings.type) && !this.crossDomain) { 
      xhr.setRequestHeader("X-CSRFToken", csrftoken); 
     } 
    } 
}); 

参见：Cross Site Request Forgery protection

Answer 2

如果您使用表单提交或ajax进行POST调用。您需要将{{csrf_token}}与表单提交一起传递。

可以共享模板代码..