NSInvalidArgumentException - "unrecognized selector sent to instance "传递 json 数据时出错 IOS7

Question

嗨，在我的应用程序中，我试图将json数据传递给我的tableview但它显示像这样发送给实例的无法识别的选择器

我有一个json数据，像这样。

"youtube_videos":[{"id":"1","name":"Little Flower Public School","youtube":"FFTi2Sl8hHw","link":"http://img.youtube.com/vi/FFTi2Sl8hHw/default.jpg"}]

这个json必须是youtube视频id和视频的缩略图现在我想在imageview中查看缩略图在tableview中查看但它给出了错误，如

Terminating app due to uncaught exception 'NSInvalidArgumentException', reason: '-[__NSCFDictionary objectAtIndex:]: unrecognized selector sent to instance 0xa72e290

这是用来获取json数据的代码。

    -(void) retrieveData
     {
      NSURL * url = [NSURL URLWithString:getDataURL];
      NSData * data = [NSData dataWithContentsOfURL:url];
       json = [NSJSONSerialization JSONObjectWithData:data options:kNilOptions error:nil];
       imgarray =[[NSMutableArray alloc]init];
      for (int i=0; i<json.count; i++) {
       NSString * dd = [[json objectAtIndex:i]objectForKey:@"youtube"];
       NSString * sspp = [[json objectAtIndex:i]objectForKey:@"link"];
       NSString * plae =[[json objectAtIndex:i]objectForKey:@"name"];
       NSLog(@"%@",dd);
       vv *myimg =[[vv alloc]initWithvideo:dd andtitle:plae andlink:sspp];
       [imgarray addObject:myimg];
    }
     [self.mytableview reloadData];

My Json NSlog.

json :{
"youtube_videos" =     (
            {
        id = 1;
        link = "http://img.youtube.com/vi/FFTi2Sl8hHw/default.jpg";
        name = "Little Flower Public School";
    },
            {
        id = 2;
        link = "http://img.youtube.com/vi/4oI7xC2wSpw/default.jpg";
        name = "Little Flower Public School";
    },
            {
        id = 3;
        link = "http://img.youtube.com/vi/XCPKMWyF1Lo/default.jpg";
        name = Function;
    }
  );
}

请告诉我如何解决这个问题，我已经卡在这里很长时间了。

谢谢。

Answer 1

我认为你应该像下面这样更新你的代码。因为你的JSON看起来像Dictionary。

-(void) retrieveData
{
    NSURL * url = [NSURL URLWithString:getDataURL];
    NSData * data = [NSData dataWithContentsOfURL:url];
    id json = [NSJSONSerialization JSONObjectWithData:data options:kNilOptions error:nil];
    imgarray = [[NSMutableArray alloc]init];
    NSArray * respArray = [json objectForKey:@"youtube_videos"];
    for (int i=0; i<respArray.count; i++)
    {
        NSString * dd = [[respArray objectAtIndex:i]objectForKey:@"youtube"];
        NSString * sspp = [[respArray objectAtIndex:i]objectForKey:@"link"];
        NSString * plae =[[respArray objectAtIndex:i]objectForKey:@"name"];
        NSLog(@"%@",dd);
        vv *myimg =[[vv alloc]initWithvideo:dd andtitle:plae andlink:sspp];
        [imgarray addObject:myimg];
    }
    [self.mytableview reloadData];
}

Answer 2

不知道你的反应是什么。在json响应中获得单个或多个记录。试试下面的代码:

-(void) retrieveData
 {
  NSURL * url = [NSURL URLWithString:getDataURL];
  NSData * data = [NSData dataWithContentsOfURL:url];
   json = [NSJSONSerialization JSONObjectWithData:data options:kNilOptions error:nil];
   imgarray =[[NSMutableArray alloc]init];
   NSString * dd = [json objectForKey:@"youtube"];
   NSString * sspp = [json objectForKey:@"link"];
   NSString * plae =[json objectForKey:@"name"];
   NSLog(@"%@",dd);
   vv *myimg =[[vv alloc]initWithvideo:dd andtitle:plae andlink:sspp];
   [imgarray addObject:myimg];

 [self.mytableview reloadData];
 }