Question

我在404张贴JSON参数时收到https://api.hackerearth.com/codemonk/v1/topicdetail/错误。服务器使用POST HTTP方法获取topic的详细信息＆amp;成功时，预计会有JSON响应。 POST参数是id对象的topic。 POST参数应为JSON格式。我正在使用AFNetworking如下 -

AFHTTPRequestOperationManager *manager = [AFHTTPRequestOperationManager manager];
manager.requestSerializer = [AFJSONRequestSerializer serializer];

NSDictionary *params = [NSDictionary dictionaryWithObject:@1 forKey:@"id"];

NSString *str = @"https://api.hackerearth.com/codemonk/v1/topic-detail/";

NSString *encodedStr = [str stringByAddingPercentEscapesUsingEncoding:NSUTF8StringEncoding];

[manager POST:encodedStr
   parameters:params
      success:^(AFHTTPRequestOperation * _Nonnull operation, id  _Nonnull responseObject) {

    NSLog(@"responseObject : %@",responseObject);


} failure:^(AFHTTPRequestOperation * _Nullable operation, NSError * _Nonnull error) {

    NSLog(@"error : %@", error.localizedDescription);

}];

这是常规的东西，但不知道为什么我现在似乎无法纠正它。我只收到404 Page Not Found错误。这肯定不是服务器端问题。有帮助吗？

Answer 1

也许这会有所帮助。

AFJSONRequestSerializer *requestSerializer = [AFJSONRequestSerializer serializer];

[requestSerializer setValue:@"application/json" forHTTPHeaderField:@"Content-Type"];
[requestSerializer setValue:@"application/json" forHTTPHeaderField:@"Accept"];

operationManagerInstance.requestSerializer = requestSerializer;

============= UPDATE

复制网址时我有404。这是因为topic-detail之间的连字符符号实际上不是连字符。这是一些不起作用的特殊角色。

https://api.hackerearth.com/codemonk/v1/topic - 细节/

相反，我删除了它并手动输入连字符，它工作正常。

Answer 2

删除以下行

NSString *encodedStr = [str stringByAddingPercentEscapesUsingEncoding:NSUTF8StringEncoding];

尝试以下方法：

AFHTTPRequestOperationManager *manager = [AFHTTPRequestOperationManager manager];
manager.requestSerializer = [AFJSONRequestSerializer serializer];

NSDictionary *params = [NSDictionary dictionaryWithObject:@1 forKey:@"id"];

NSString *str = @"https://api.hackerearth.com/codemonk/v1/topic-detail/";

[manager POST:str
   parameters:params
      success:^(AFHTTPRequestOperation * _Nonnull operation, id  _Nonnull responseObject) {

    NSLog(@"responseObject : %@",responseObject);


} failure:^(AFHTTPRequestOperation * _Nullable operation, NSError * _Nonnull error) {

    NSLog(@"error : %@", error.localizedDescription);

}];

iOS使用JSON参数从AFNetworking HTTP POST方法获取JSON响应

2 个答案: