IOS用AFN发送字符串形式的Json数据给服务器实例

编辑: admin 分类: 苹果教程 发布时间: 2022-03-30 来源:互联网

每个人遇到后台的需求不一样,我只能写出自己目前遇到的需求和做法,我的后台是需要一个Json的key,对应一个字符串类型的字典,后台要求的是带双引号和冒号的字典,所以就拼接了一下

AFHTTPSessionManager *manager = [AFHTTPSessionManager manager];
  // 发送post请求
//  manager.responseSerializer = [AFJSONResponseSerializer serializer];
  manager.responseSerializer = [AFJSONResponseSerializer serializer];
  manager.responseSerializer.acceptableContentTypes = [NSSet setWithObject:@"text/html"];
  //
  manager.requestSerializer=[AFHTTPRequestSerializer serializer];
  NSDictionary *json = [NSDictionary dictionary];
   NSString *str = [NSString stringWithFormat:@"{\"city\":\"%@\",\"country\":\"%@\",\"headimgurl\":\"%@\",\"language\":\"%@\",\"openid\":\"%@\",\"province\":\"%@\",\"sex\":%@,\"unionid\":\"%@\",\"userName\":\"%@\"}",self.userInfoDict[@"city"],self.userInfoDict[@"country"],self.userInfoDict[@"headimgurl"],self.userInfoDict[@"language"],self.userInfoDict[@"openid"],self.userInfoDict[@"province"],self.userInfoDict[@"sex"],self.userInfoDict[@"unionid"],self.userInfoDict[@"nickname"]];
  json = @{
//      @"city":self.userInfoDict[@"city"],
//      @"country":self.userInfoDict[@"country"],
//      @"headimgurl":self.userInfoDict[@"headimgurl"],
//      @"language":self.userInfoDict[@"language"],
//      @"openid":self.userInfoDict[@"openid"],
//      @"province":self.userInfoDict[@"province"],
//      @"sex":self.userInfoDict[@"sex"],
//      @"unionid":self.userInfoDict[@"unionid"],
//      @"userName":self.userInfoDict[@"nickname"],
      @"json":str,
//      @"json":@"json"
      
          };
  NSLog(@"post调试%@",json);
  NSString *URLStr = @"http://10.62.16.247:8080/LoginServlet.bmh";
  
  [manager POST:URLStr parameters:json progress:^(NSProgress * _Nonnull uploadProgress) {
    
  } success:^(NSURLSessionDataTask * _Nonnull task, id _Nullable responseObject) {
    NSString *str = responseObject;
    NSLog(@"post回调====%@",str);
    NSLog(@"post成功");
  } failure:^(NSURLSessionDataTask * _Nullable task, NSError * _Nonnull error) {
    NSLog(@"post失败");
  }];

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