Question

I use Tastypie for Django s API. but it returns error. my code is bellow.

$.ajax({
      type : "POST", 
      url : "http://192.168.1.130:8000/api/user/author/", 
      data :  {"first_name": "111","second_name": "222"} , 
      success: function(){
        alert( Submit Success )
      },
      dataType :  json , 
      contentType :  application/json ,
      processData: false
    });

我的pi.py像这样:

class AuthorResource(ModelResource):
    class Meta:
        queryset = Author.objects.all()
        resource_name = author 
        fields = [ first_name , last_name ]
        filtering = {
             first_name : ALL,
        }
        authentication = Authentication()
        authorization = Authorization()

返回200, 寄什么也没有。我怎么能重新爱它呢?

Answer 1

这是"https://stackoverflow.com/ questions/10138169/ returning-data-on-post-in-django-tattipie"的假称,“在django-tattipie中返回POST的数据 < /a > 。

在您的元资源中添加 always_return_data = true

Answer 2

如果对象成功创建, 对象 uri 将在响应信头中显示在位置字段中。

Answer 3

URL 应该是 < a href=" http://192168.1.130:800/api/traces/"rel="nofollow" > http://192168.1.130:800/api/traces/ http://192168.1.130:800/api/user/tracs/

友情链接