如何在ios中发布对象并获得响应答案

【问题标题】：How to post object and get response in ios如何在ios中发布对象并获得响应
【发布时间】：2016-08-09 05:45:04
【问题描述】：

在我的应用程序中，我需要将数据发布到服务器并需要接收响应。但是我在发布数据后得到空值。下面是我的完整代码。提前致谢。

{
        NSString *post =[[NSString alloc] initWithFormat:@"%@%@%@%@%@",[self.username_reg text],[self.emailid_reg text],[self.phone_reg text],[self.password_reg text],[self.confirmpassword_reg text]];
        NSLog(@"PostData: %@",post);
        NSMutableURLRequest *request = [[NSMutableURLRequest alloc] init];
        NSURL *url=[NSURL URLWithString:@"https://servlet/URL"];

        NSData *postData = [post dataUsingEncoding:NSASCIIStringEncoding allowLossyConversion:YES];

        NSString *postLength = [NSString stringWithFormat:@"%lu", (unsigned long)[postData length]];


        [request setURL:url];
        [request setHTTPMethod:@"POST"];
        [request setValue:postLength forHTTPHeaderField:@"Content-Length"];
        [request setValue:@"application/json" forHTTPHeaderField:@"Accept"];
        [request setValue:@"application/x-www-form-urlencoded" forHTTPHeaderField:@"Content-Type"];


    NSMutableDictionary *postDict = [[NSMutableDictionary alloc] init];
    [postDict setValue:_username_reg.text forKey:@"UserName"];
    [postDict setValue:_emailid_reg.text forKey:@"Email"];
    [postDict setValue:_phone_reg.text forKey:@"Phone"];
    [postDict setValue:_password_reg.text forKey:@"Pass"];
    [postDict setValue:_confirmpassword_reg.text forKey:@"ConPass"];


    NSData *jsonData = [NSJSONSerialization dataWithJSONObject:postDict options:0 error:nil];

    // Checking the format
    NSString *urlString =  [[NSString alloc] initWithData:jsonData encoding:NSUTF8StringEncoding];


    // Convert your data and set your request's HTTPBody property
    NSString *stringData = [[NSString alloc] initWithFormat:@"jsonRequest=%@", urlString];
    NSData *requestBodyData = [stringData dataUsingEncoding:NSUTF8StringEncoding];

    request.HTTPBody = requestBodyData;

        NSLog(@"bcbc:%@",requestBodyData);
        [NSURLConnection sendAsynchronousRequest:request queue:[NSOperationQueue mainQueue] completionHandler:^(NSURLResponse *response, NSData *data, NSError *connectionError)
         {
             NSString* newStr = [[NSString alloc] initWithData:jsonData encoding:NSUTF8StringEncoding];
             NSLog(@"new:%@",newStr);
             NSError *error;
            NSDictionary *json_Dict = [NSJSONSerialization JSONObjectWithData:data options:kNilOptions error:&error];
                    NSLog(@"%@",json_Dict);
         }];

    }

请求给出的格式是：

{"UserName":"sony","Email":"ronyv@example.in","Phone":"7358700457","Pass":"sony88","ConPass":"sony88"}

需要得到的响应：

{"responseHeader":{"responseCode":0,"responseMessage":"Success"}}

【问题讨论】：

使用AFNetworking怎么样？这可能不是您问题的答案。但是它很容易用于网络用途，最好的！
你能解释一下如何使用AFNetworking
AFNetworking ：查看示例。
试试这个answer它可能会有所帮助。

标签： ios json post http-post

【解决方案1】：

试试这个代码：

NSData* jsonData = [NSJSONSerialization dataWithJSONObject:@"your dictionary name" options:NSJSONWritingPrettyPrinted error:&error];

NSString *jsonString = [[NSString alloc] initWithData:jsonData encoding:NSUTF8StringEncoding];
NSLog(@"jsonString: %@", jsonString);

NSData *requestData = [jsonString dataUsingEncoding:NSUTF8StringEncoding];

NSMutableData *body = [NSMutableData data];
[body appendData:requestData];

NSString *postLength = [NSString stringWithFormat:@"%lu", (unsigned long)[body length]];

NSURL *url = [NSURL URLWithString:@"your url"];
NSMutableURLRequest *request = [[NSMutableURLRequest alloc] initWithURL:url];
[request setHTTPMethod:@"POST"];
[request setHTTPShouldHandleCookies:NO];

[request setValue:@"application/json" forHTTPHeaderField:@"Content-type"];

[request setValue:postLength forHTTPHeaderField:@"Content-Length"];
[request setHTTPBody:body];

NSURLSession *session = [NSURLSession sessionWithConfiguration:[NSURLSessionConfiguration defaultSessionConfiguration]];
[[session dataTaskWithRequest:request
            completionHandler:^(NSData *data,
                                NSURLResponse *response,
                                NSError *error)
  {
      if (error) {
          failure(error);
      } else {
          NSDictionary * jsonDic =[NSJSONSerialization JSONObjectWithData:data options:NSJSONReadingMutableContainers error:nil];
         NSLog(@"%@",jsonDic);
          if ([jsonDic objectForKey:@"error"]) {

          }
          else{

          }

      }
  }] resume];

【讨论】：