Question

下午好，

我试图存储来自JSON输出的响应，因为我想在＆＃34; ProfileViewController＆＃34;中显示。用户的统计数据以及我尝试使用以下功能来存储信息。

目前输出正常，因为数据可以取决于用户，但现在我必须将每个统计数据存储在＆＃34;变量＆＃34;对于每个统计数据（我有3个），但是当我运行代码时，它没有显示我的星标，追随者和图片的NSLog ......

你可以帮帮我吗？响应很好，现在我只想将每个统计数据存储在一个变量中。我怎样才能做到这一点？我的代码有什么问题？

ProfileViewController - ＆gt;的 fetchJson ：

-(void)fetchJson {

    NSString *usersPassword = [SSKeychain passwordForService:@"login" account:@"account"];

    NSString *post =[[NSString alloc] initWithFormat:@"usersPassword=%@",usersPassword];
    NSLog(@"PostData: %@",post);

    NSURL *url=[NSURL URLWithString:@"http://website.com/profile.php"];

    NSData * data = [NSData dataWithContentsOfURL:url];

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

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

    NSMutableURLRequest *request = [[NSMutableURLRequest alloc] init];
    [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"];
    [request setHTTPBody:postData];

    [NSURLRequest setAllowsAnyHTTPSCertificate:YES forHost:[url host]];

    NSError *error = [[NSError alloc] init];
    NSHTTPURLResponse *response = nil;
    NSData *urlData=[NSURLConnection sendSynchronousRequest:request returningResponse:&response error:&error];

    NSLog(@"Response code: %ld", (long)[response statusCode]);
    if ([response statusCode] >=200 && [response statusCode] <300)
    {
        NSString *responseData = [[NSString alloc]initWithData:urlData encoding:NSUTF8StringEncoding];
        NSLog(@"Response ==> %@", responseData);

        SBJsonParser *jsonParser = [SBJsonParser new];
        NSDictionary *jsonData = (NSDictionary *) [jsonParser objectWithString:responseData error:nil];

        NSInteger success = [(NSNumber *) [jsonData objectForKey:@"success"] integerValue];
        NSLog(@"%ld",(long)success);

        if(success == 1)
        {
            NSLog(@"Login SUCCESS");

            [_jsonArray removeAllObjects];
            _jsonArray = [NSJSONSerialization
                          JSONObjectWithData:data
                          options:NSJSONReadingMutableContainers|NSJSONReadingMutableLeaves
                          error:&error];

            for(int i=0;i<_jsonArray.count;i++)
            {
                NSDictionary * jsonObject = [_jsonArray objectAtIndex:i];
                NSString* stars = [jsonObject objectForKey:@"stars"];
                NSLog(@"Stars ==> %@", stars);

                NSDictionary * jsonObject2 = [_jsonArray objectAtIndex:i];
                NSString* followers = [jsonObject2 objectForKey:@"followers"];
                NSLog(@"Followers ==> %@", followers);

                NSDictionary * jsonObject3 = [_jsonArray objectAtIndex:i];
                NSString* photos = [jsonObject3 objectForKey:@"photos"];
                NSLog(@"Pictures ==> %@", photos);
            }
        }
    }
}

JSON输出：

{"success":1,"stars":50,"photos":50,"followers":50}

提前致谢。

Answer 1

  _jsonArray = [NSJSONSerialization JSONObjectWithData:myNSData options:kNilOptions error:&error];
    NSDictionary * jsonObject  = (NSDictionary *)_jsonArray;

然后检查字典对象。

Answer 2

jsonData已包含已解析的响应，并且您已成功从中检索成功值，如下所示

NSInteger success = [(NSNumber *) [jsonData objectForKey:@"success"] integerValue];
NSLog(@"%ld",(long)success);

您可以按如下方式检索其他值

[[jsonData objectForKey:@"stars"] integerValue]

等等。

您没有必要使用以下方法

创建基础对象

NSJSONSerialization
                  JSONObjectWithData:data
                  options:NSJSONReadingMutableContainers|NSJSONReadingMutableLeaves
                  error:&error];

并且不需要以下代码块，

for(int i=0;i<_jsonArray.count;i++)
{
      NSDictionary * jsonObject = [_jsonArray objectAtIndex:i];
      NSString* stars = [jsonObject objectForKey:@"stars"]
      NSLog(@"Stars ==> %@", stars);

      NSDictionary * jsonObject2 = [_jsonArray objectAtIndex:i];
      NSString* followers = [jsonObject2 objectForKey:@"followers"];
      NSLog(@"Followers ==> %@", followers);

      NSDictionary * jsonObject3 = [_jsonArray objectAtIndex:i];
      NSString* photos = [jsonObject3 objectForKey:@"photos"];
      NSLog(@"Pictures ==> %@", photos);
}

使用JSON在变量中存储数据

2 个答案: