如何在http post body上发布json字符串？在对象 c答案

【问题标题】：How to post json string on http post body ? in obj c如何在http post body上发布json字符串？在对象 c
【发布时间】：2015-10-20 06:54:43
【问题描述】：

您好，我是 ios 开发新手，我想通过 http post 请求在我的 Web 服务上发送一些 json 数据，其中数据将在请求正文中而不是在参数中？

感谢任何帮助，谢谢。

【问题讨论】：

Sending a JSON via POST in NSURLRequest的可能重复

标签： ios http-post afnetworking nsurlconnection afnetworking-2

【解决方案1】：

试试这个代码

    NSURL *url = [NSURL URLWithString:@"Your Json URL"];

     NSMutableURLRequest *request = [NSMutableURLRequest requestWithURL:url
                                                                   cachePolicy:NSURLRequestUseProtocolCachePolicy
                                                               timeoutInterval:60.0];
            [request setValue:@"application/json" forHTTPHeaderField:@"Accept"];
            [request setValue:@"application/json" forHTTPHeaderField:@"Content-Type"];

            [request setHTTPMethod:@"POST"];

            NSURLResponse *response;
            NSError *err;
            NSData *responseData = [NSURLConnection sendSynchronousRequest:request returningResponse:&response error:&err];
            NSString *str = [[NSString alloc] initWithData:responseData encoding:NSUTF8StringEncoding];
            NSLog(@"responseData in string : %@",str);

          NSMutableDictionary *dictTemp = [NSJSONSerialization JSONObjectWithData:responseData options:0 error:nil];

NSLog(@"responseData in Dictionary : %@",dictTemp);

【讨论】：

【解决方案2】：

> Try this code

    -(void)answerCode
{
    NSError * error ;
    NSURLResponse * urlResponse;

    NSURL * postUrl =[NSURL URLWithString:YourUrl];//enter your url

//your body here 
NSString * body =[NSString stringWithFormat:@"email=dharasis"];



    NSMutableURLRequest * request =[[NSMutableURLRequest alloc]initWithURL:postUrl cachePolicy:NSURLRequestReloadIgnoringCacheData timeoutInterval:50];
    [request setHTTPMethod:@"POST"];
    [request setHTTPBody:[body dataUsingEncoding:NSUTF8StringEncoding allowLossyConversion:YES]];

    NSData * data =[NSURLConnection sendSynchronousRequest:request returningResponse:&urlResponse error:&error];
    if (!data) {

        return;
    }

    id json =[NSJSONSerialization JSONObjectWithData:data options:NSJSONReadingAllowFragments error:&error];

  NSLog(@"Json result is:%@",josn);

}

【讨论】：