Question

你好朋友我用post方法传递JSON参数我很厌倦这个代码..

NSDictionary *parameters = @{@"cmscontent" : @{
                                         @"access_name"   : @"about u",
                                         }
                                 };


    NSData *data = [NSJSONSerialization dataWithJSONObject:parameters options:0 error:nil];

    NSMutableURLRequest *request = [NSMutableURLRequest requestWithURL:[NSURL URLWithString:@"http://.........."]];
    [request setHTTPMethod:@"POST"];
    [request setValue:@"application/json;charset=UTF-8" forHTTPHeaderField:@"content-type"];

    NSURLSession *session = [NSURLSession sessionWithConfiguration:[NSURLSessionConfiguration defaultSessionConfiguration]];
    NSURLSessionUploadTask *dataTask = [session uploadTaskWithRequest: request
                                                             fromData: data completionHandler:^(NSData *data, NSURLResponse *response, NSError *error)

                                        {
                                            NSDictionary *json = [NSJSONSerialization JSONObjectWithData:data options:0 error:nil];
                                            NSLog(@"%@", json);
                                        }];

    [dataTask resume];

和我的json字典参数是..

 {
  "cmscontent": 
    {
          "access_name": "about us"
    }
}

所以请告诉我如何传递这类参数??

Answer 1

<强>先决条件

确保您的API方法在后端POST。
确保您的API方法接受JSON数据。
确保所有参数名称与API方法完全相同。

使用以下方法将数据发布到API

NSMutableURLRequest *request = [[NSMutableURLRequest alloc] initWithURL:@"http://....."];

NSDictionary *params= @{@"cmscontent" :
                                    @{
                                        @"access_name"   : @"about u"
                                     }
                             };
 NSError *err = nil;
 NSData *requestData = [NSJSONSerialization dataWithJSONObject:params options:NSJSONWritingPrettyPrinted error:&err];

[request setHTTPMethod:@"POST"];
[request setValue:@"application/json" forHTTPHeaderField:@"Accept"];
[request setValue:@"application/x-www-form-urlencoded" forHTTPHeaderField:@"Content-Type"];
[request setValue:[NSString stringWithFormat:@"%lu", (unsigned long)[requestData length]] forHTTPHeaderField:@"Content-Length"];
[request setHTTPBody:requestData];

[NSURLConnection sendAsynchronousRequest:request
                                   queue:[NSOperationQueue mainQueue]
                       completionHandler:^(NSURLResponse *response, NSData *data, NSError *error) {
                           if([data length] > 0)
                           {
                               NSError *err = nil;
                               NSDictionary *dictResponse = [NSJSONSerialization JSONObjectWithData:data options:NSJSONReadingAllowFragments error:&err];
                               NSLog(@"[%@]", dictResponse);                                  
                           }
                           else{
                               NSLog(@"Failed To Get Response for : %@",params);
                           }
                       }];

Answer 2

我不确定将NSDictionary对象转换为NSString对象，然后将此字符串对象作为参数发布，就像其他字符串参数一样。我希望这会对你有帮助......

NSString *postDictionaryAsSting = [[NSString alloc]initWithData:[NSJSONSerialization dataWithJSONObject:dictObj options:NSJSONWritingPrettyPrinted error:nil] encoding:NSUTF8StringEncoding];

NSMutableURLRequest *request = [NSMutableURLRequest requestWithURL:url
             cachePolicy:NSURLRequestUseProtocolCachePolicy timeoutInterval:60.0];


NSData *requestData = [postDictionaryAsSting dataUsingEncoding:NSUTF8StringEncoding];

[request setHTTPMethod:@"POST"];
[request setValue:@"application/json" forHTTPHeaderField:@"Accept"];
[request setValue:@"application/json" forHTTPHeaderField:@"Content-Type"];
[request setValue:[NSString stringWithFormat:@"%d", [requestData length]] forHTTPHeaderField:@"Content-Length"];
[request setHTTPBody: requestData];

NSURLConnection *connection = [[NSURLConnection alloc]initWithRequest:request delegate:self];
if (connection) {
 receivedData = [[NSMutableData data] retain];
}

如何使用post方法将参数作为字典传递给url

2 个答案: