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将Json转换为Ios格式_Ios_Json_Parsing_Jsonp - Fatal编程技术网

将Json转换为Ios格式

将Json转换为Ios格式,ios,json,parsing,jsonp,Ios,Json,Parsing,Jsonp,我尝试在iOS中解析json,但出现了错误。 这是我从api获得的json data = ( { LogIn = { Emails = "abc@xyz`.com"; Latitude = "37.33233141"; Longitude = "-122.03121860"; Password = 123456; }; Status = Success; } ); 我想要的所有细节,如电子邮件,纬度,经度和密码,以及状态。我正在尝试下面的json代码 login

我尝试在iOS中解析json,但出现了错误。 这是我从api获得的json

data =     (
{
LogIn =             {
Emails = "abc@xyz`.com";
Latitude = "37.33233141";
Longitude = "-122.03121860";
Password = 123456;
};
Status = Success;
}
);
我想要的所有细节,如电子邮件,纬度,经度和密码,以及状态。我正在尝试下面的json代码

loginurl= [NSURL URLWithString:string1];
NSLog(@"url is %@",string1);
dispatch_async(kBgQueue, ^{
NSData *data = [NSData dataWithContentsOfURL:loginurl];
[self performSelectorOnMainThread:@selector(fetchedData:) withObject:data waitUntilDone:YES];
});
这是我的获取数据方法-

-(void)fetchedData:(NSData *)responseData
{   
    NSError * error;
    NSDictionary *json = [NSJSONSerialization JSONObjectWithData:responseData      options:kNilOptions error:&error];

    NSLog(@"JSON: %@", json);
    NSDictionary *data = [json objectForKey:@"data"];
    NSArray *currentConditions = [data objectForKey:@"LogIn"];

    NSDictionary *conditions = [currentConditions objectAtIndex:0];
    NSLog(@"%@",conditions);
 }

首先,您的JSON格式无效。应该是这样的:

{
    "data": {
        "LogIn": {
            "Emails": "abc@xyz`.com",
            "Latitude": "37.33233141",
            "Longitude": "-122.03121860",
            "Password": 123456
        },
        "Status": "Success"
    }
}
您可以使用验证JSON数据

第二,

[data objectForKey:@"LogIn"];
是一个字典,而不是数组

尝试:

NSDictionary *currentConditions = [data objectForKey:@"LogIn"];
NSLog(@"Longitude: %@", [currentConditions objectForKey@"Longitude"]);