在 Django 框架中发出 ajax Post 请求时出现 403 Forbidden 错误 [英] 403 Forbidden error when making an ajax Post request in Django framework

问题描述

我正在尝试将 jquery 集成到我使用 Django 框架制作的 Web 应用程序中.然而，我很难尝试通过简单的 ajax 调用来工作.我的模板文件包含表单 html 和 javascript 来处理 ajax 调用，如下所示:

I am trying to integrate jquery into a web application I am making with Django framework. I am however having a hard time trying to make a simple ajax call to work. My template file that contains the form html and javascript to handle the ajax call looks like:

<script type="text/javascript">
$(document).ready(function() {
$( "#target" ).submit(function() {
console.log('Form was submitted');
$.ajax({
        type: "POST",
        url: "/hello/",  // or just url: "/my-url/path/"
        data: {
            query: $( "#query" ).val()   
        },
        success: function(data) {
            console.log(data);
        }
    });
return false;
  });   
  })
</script>
<form id="target" action="." method="post">{% csrf_token %}
 <input id= "query" type="text" value="Hello there">
 <input type="submit" value="Search Recent Tweets">
</form>

我应该处理 ajax 调用的 views.py 看起来像:

My views.py that is supposed to handle the ajax call looks like:

 from django.core.context_processors import csrf
 from django.shortcuts import render_to_response
 from django.template.loader import get_template
 from django.template import Context,RequestContext
 from django.views.decorators.csrf import ensure_csrf_cookie
 from django.http import HttpResponse

 # access resource
 def hello(request):
  c = {}
  c.update(csrf(request))
  if request.is_ajax():
        t = get_template('template.html')
        #html = t.render(Context({'result': 'hello world'}))
        con = RequestContext(request, {'result': 'hello world'})
        return render_to_response('template.html', c, con)
  else:
        return HttpResponse('Not working!') 

我已尝试遵循 上的官方文档跨站点请求伪造保护，并查看了解决类似问题的几个计算器溢出问题.我在我的 html 模板文件中包含了 {% csrf_token %} 但它似乎仍然不起作用.我在控制台中收到一条错误消息，提示 ajax 调用失败:

I have tried to follow the official documentation on Cross-Site Request Forgery Protection and also looked at several stackoverflow questions addressing a similar problem. I have included the {% csrf_token %} in my html template file but it still doesn't seem to be working. I get an error in the console suggesting that the ajax call failed:

POST http://127.0.0.1:8000/hello/ 403 (FORBIDDEN)   

如何将 result 变量与我的 http 响应一起传递并让 ajax 调用顺利运行?任何帮助深表感谢.

How do I pass the result variable along with my http response and get the ajax call to work smoothly? Any help is deeply appreciated.

编辑-1

我应该没有将 csrf 令牌与我的发布请求一起传递.所以根据文档，我将以下代码添加到我的模板 javascript 中:

I wasn't supposedly passing the csrf token along with my post request. SO as per the documentation I added the following code to my template javascript:

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');
console.log(csrftoken);

//Ajax call
function csrfSafeMethod(method) {
// these HTTP methods do not require CSRF protection
return (/^(GET|HEAD|OPTIONS|TRACE)$/.test(method));
}
$.ajaxSetup({
    crossDomain: false, // obviates need for sameOrigin test
    beforeSend: function(xhr, settings) {
        if (!csrfSafeMethod(settings.type)) {
            xhr.setRequestHeader("X-CSRFToken", csrftoken);
        }
    }
});

我在浏览器中刷新模板html页面时，在控制台得到null，提示cookie没有设置或者没有定义.我错过了什么?

When I refresh the template html page in the browser, I get null in the console, suggesting that the cookie is not set or not defined. What am I missing?

推荐答案

因为你没有发布csrfmiddlewaretoken，所以Django禁止你.本文档可以帮助您.

Because you did not post the csrfmiddlewaretoken, so Django forbid you. this document can help you.

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