iPhone - 从纬度和经度获取城市名称

Question

我想在我的 iPhone 应用程序中获取我当前的城市名称。

我目前正在使用 CLLocationManager 获取纬度和经度，然后将我的坐标传递给 CLGeocoder。

    CLGeocoder * geoCoder = [[CLGeocoder alloc] init];
    [geoCoder reverseGeocodeLocation:location completionHandler:^(NSArray *placemarks, NSError *error) {
        for (CLPlacemark * placemark in placemarks) {
            UIAlertView *alert = [[UIAlertView alloc] initWithTitle:@"Current City" message:[NSString stringWithFormat:@"Your Current City:%@",[placemark locality]] delegate:self cancelButtonTitle:@"OK" otherButtonTitles:@"Cancel", nil];
            [alert  show];
        }  
    }];

这在 iOS 5.0 中工作正常但在 iOS 4.3 中不工作。

作为替代方案，我开始使用 Google 网络服务

-(void)findLocationFor:(NSString *)latitudeStr 
          andLontitude:(NSString *)longtitudeStr{
    if ([self connectedToWiFi]){
        float latitude  = [latitudeStr floatValue];
        float longitude = [longtitudeStr floatValue];
        NSMutableDictionary *parameters = [NSMutableDictionary dictionaryWithObjectsAndKeys: 
                                           [NSString stringWithFormat:@"%f,%f", latitude, longitude], @"latlng", nil];
        NSMutableURLRequest *request = [NSMutableURLRequest requestWithURL:[NSURL URLWithString:@"http://maps.googleapis.com/maps/api/geocode/json"]];
        [parameters setValue:@"true" forKey:@"sensor"];
        [parameters setValue:[[NSLocale currentLocale] objectForKey:NSLocaleLanguageCode] forKey:@"language"];
        NSMutableArray *paramStringsArray = [NSMutableArray arrayWithCapacity:[[parameters allKeys] count]];

        for(NSString *key in [parameters allKeys]) {
            NSObject *paramValue = [parameters valueForKey:key];
            [paramStringsArray addObject:[NSString stringWithFormat:@"%@=%@", key, paramValue]];
        }

        NSString *paramsString = [paramStringsArray componentsJoinedByString:@"&"];
        NSString *baseAddress = request.URL.absoluteString;
        baseAddress = [baseAddress stringByAppendingFormat:@"?%@", paramsString];
        [request setURL:[NSURL URLWithString:baseAddress]];

        NSError        *error    = nil;
        NSURLResponse  *response = nil;
        NSData *returnData = [NSURLConnection sendSynchronousRequest:request returningResponse:&response error:&error];
        if (response == nil) {
            if (error != nil) {
            }
        }
        else {
            NSDictionary *responseDict = [returnData objectFromJSONData];
            NSArray *resultsArray = [responseDict valueForKey:@"results"]; 
            NSMutableArray *placemarksArray = [NSMutableArray arrayWithCapacity:[resultsArray count]];
            for(NSDictionary *placemarkDict in resultsArray){
                NSDictionary *coordinateDict = [[placemarkDict valueForKey:@"geometry"] valueForKey:@"location"];

                float lat = [[coordinateDict valueForKey:@"lat"] floatValue];
                float lng = [[coordinateDict valueForKey:@"lng"] floatValue];

                NSMutableDictionary *dict = [[NSMutableDictionary alloc] init];
                [dict setObject:[NSString stringWithFormat:@"%.f",lat] forKey:@"latitude"];
                [dict setObject:[NSString stringWithFormat:@"%.f",lng] forKey:@"longitude"];
                [dict setObject:[placemarkDict objectForKey:@"formatted_address"] forKey:@"address"];

                [placemarksArray addObject:dict];
                [dict release];
            }
            NSDictionary *placemark = [placemarksArray objectAtIndex:0];
        }
    }
}

但是我得到的响应太长，这意味着我仍然无法获得城市名称，因为在某些情况下，此 Web 服务会提供有关坐标的所有其他信息，但不包括城市名称。

谁能帮帮我？

Answer 1

根据文档 CLGeocoder在 iOS5 以下不起作用。你需要走另一条路来同时支持 iOS4 和 iOS5。

可以看看MKReverseGeocoder ，但是它在 iOS5 中已被弃用，但仍然可以达到目的。为了确认，您可以查看 so called question

+(NSString *)getAddressFromLatLon:(double)pdblLatitude withLongitude:(double)pdblLongitude
{
    NSString *urlString = [NSString stringWithFormat:@"http://maps.google.com/maps/geo?q=%f,%f&output=csv",pdblLatitude, pdblLongitude];
    NSError* error;
    NSString *locationString = [NSString stringWithContentsOfURL:[NSURL URLWithString:urlString] encoding:NSASCIIStringEncoding error:&error];
    locationString = [locationString stringByReplacingOccurrencesOfString:@"\"" withString:@""];
    return [locationString substringFromIndex:6];
}

您可以使用此函数从纬度、经度获取地址。您可以根据需要更改。我们把它作为类方法，这样我们就可以直接将它用作

NSString *strAddressFromLatLong = [CLassName getAddressFromLatLon:37.484848 withLongitude:74.48489];

编辑

请停止使用上述功能，因为它已在评论中报告停止工作(未经我测试)。我建议开始使用 SVGeocoder