ios - 从NSJSONSerialization中获取数据中的值

Question

我有一些从URL提取的JSON数据。我编写的代码可以很好地下载JSON并进行解析，但是我似乎也无法按需访问它，尤其是在数据包含在另一个元素中的情况下。

这是JSON格式:

{
    address = "<null>";
    city = "<null>";
    country = UK;
    "country_code" = GB;
    daylight = 1;
    for = daily;
    items =     (
                {
            asr = "5:22 pm";
            "date_for" = "2013-7-1";
            dhuhr = "1:01 pm";
            fajr = "2:15 am";
            isha = "11:47 pm";
            maghrib = "9:24 pm";
            shurooq = "4:39 am";
        }
    );
    latitude = "50.9994081";
    link = "http://muslimsalat.com/UK";
    longitude = "0.5039011";
    "map_image" = "http://maps.google.com/maps/api/staticmap?center=50.9994081,0.5039011&sensor=false&zoom=13&size=300x300";
    "postal_code" = "<null>";
    "prayer_method_name" = "Muslim World League";
    "qibla_direction" = "119.26";
    query = "51.000000,0.500000";
    state = "<null>";
    timezone = 0;
    title = UK;
    "today_weather" =     {
        pressure = 1020;
        temperature = 14;
    };
}

(这些是伊斯兰祈祷时间。)

到目前为止，我的Objective-C是:

-(CLLocationCoordinate2D) getLocation{
    CLLocationManager *locationManager = [[[CLLocationManager alloc] init] autorelease];
    locationManager.delegate = self;
    locationManager.desiredAccuracy = kCLLocationAccuracyBest;
    locationManager.distanceFilter = kCLDistanceFilterNone;
    [locationManager startUpdatingLocation];
    CLLocation *location = [locationManager location];
    CLLocationCoordinate2D coordinate = [location coordinate];

    return coordinate;
}

//class to convert JSON to NSData
- (IBAction)getDataFromJson:(id)sender {
    //get the coords:
    CLLocationCoordinate2D coordinate = [self getLocation];
    NSString *latitude = [NSString stringWithFormat:@"%f", coordinate.latitude];
    NSString *longitude = [NSString stringWithFormat:@"%f", coordinate.longitude];

    NSLog(@"*dLatitude : %@", latitude);
    NSLog(@"*dLongitude : %@",longitude);

    //load in the times from the json
    NSString *myURLString = [NSString stringWithFormat:@"http://muslimsalat.com/%@,%@/daily/5.json", latitude, longitude];
    NSURL *url = [NSURL URLWithString:myURLString];
    NSData *jsonData = [NSData dataWithContentsOfURL:url];

    if(jsonData != nil)
    {
        NSError *error = nil;
        id result = [NSJSONSerialization JSONObjectWithData:jsonData options:NSJSONReadingMutableContainers error:&error];
        NSArray *jsonArray = (NSArray *)result; //convert to an array
        if (error == nil)
            NSLog(@"%@", result);
            NSLog(@"%@", jsonArray);
            for (id element in jsonArray) {
                NSLog(@"Element: %@", [element description]);

            }
    }
}

运行此代码时，我得到的唯一输出是元素名称的列表( address, city, country，依此类推)。给出了 items，但未给出其子元素。我了解这是我要使用的代码:

for (id element in jsonArray) {
    NSLog(@"Element: %@", [element description]);
}

但我不知道该如何进行下一步。

我唯一需要的数据值实际上是时间本身(例如 items>asr， items>dhuhr等)。

我如何才能自己获取这些值，然后将它们另存为可以使用的值？

谢谢!

Answer 1

NSArray *jsonArray = (NSArray *)result; //convert to an array
这不是“转换”，只是您向编译器保证result实际上是NSArray。在这种情况下，这是一个谎言。

您的代码当前仅在JSON中返回的字典中打印键列表。尝试此操作以获取项目列表(它是一个数组，因此您需要处理可能存在多个条目):

NSDictionary *result = [NSJSONSerialization ...

for (NSDictionary *itemDict in result[@"items"]) {
    NSLog(@"item: %@", itemDict);
}

然后，您可以提取时间。