在iOS中使用Geocoder类根据地址获取纬度和经度_Ios_Iphone_Objective C_Geocoding

在iOS中使用Geocoder类根据地址获取纬度和经度

ios iphone objective-c

在iOS中使用Geocoder类根据地址获取纬度和经度,ios,iphone,objective-c,geocoding,Ios,Iphone,Objective C,Geocoding,我根据经度和纬度值得到了当前位置，然后使用注释在谷歌地图上得到了多个位置。现在我想根据地址（即街道、城市和县）获得经度和纬度值。需要一些关于如何实现的指导到目前为止，我一直在尝试：- #import "ViewController.h" @interface ViewController () @end @implementation ViewController @synthesize streetField = _streetField, cityField = _cityField

我根据经度和纬度值得到了当前位置，然后使用注释在谷歌地图上得到了多个位置。现在我想根据地址（即街道、城市和县）获得经度和纬度值。需要一些关于如何实现的指导

到目前为止，我一直在尝试：-

#import "ViewController.h"

@interface ViewController ()

@end

@implementation ViewController
@synthesize streetField = _streetField, cityField = _cityField, countryField = _countryField, fetchCoordinatesButton = _fetchCoordinatesButton, nameLabel = _nameLabel, coordinatesLabel = _coordinatesLabel;
@synthesize geocoder = _geocoder;


- (void)viewDidLoad
{

    [super viewDidLoad];
    _streetField.delegate=self;
    _cityField.delegate=self;
    _countryField.delegate=self;

    // Do any additional setup after loading the view, typically from a nib.
}

- (IBAction)fetchCoordinates:(id)sender {
    NSLog(@"Fetch Coordinates");
    if (!self.geocoder) {
          NSLog(@"Geocdoing");
        self.geocoder = [[CLGeocoder alloc] init];
    }

    NSString *address = [NSString stringWithFormat:@"%@ %@ %@", self.streetField.text, self.cityField.text, self.countryField.text];
    NSLog(@"GET Addres%@",address);

    self.fetchCoordinatesButton.enabled = NO;

    [self.geocoder geocodeAddressString:address completionHandler:^(NSArray *placemarks, NSError *error) {
          NSLog(@"Fetch Gecodingaddress");
        if ([placemarks count] > 0) {
            CLPlacemark *placemark = [placemarks objectAtIndex:0];

             NSLog(@"GET placemark%@",placemark);

            CLLocation *location = placemark.location;

            NSLog(@"GET location%@",location);

            CLLocationCoordinate2D coordinate = location.coordinate;


            self.coordinatesLabel.text = [NSString stringWithFormat:@"%f, %f", coordinate.latitude, coordinate.longitude];

            NSLog(@"CoordinatesLabel%@",self.coordinatesLabel.text);


            if ([placemark.areasOfInterest count] > 0) {
                NSString *areaOfInterest = [placemark.areasOfInterest objectAtIndex:0];
                self.nameLabel.text = areaOfInterest;
                NSLog(@"NameLabe%@",self.nameLabel.text);
            }
        }

        self.fetchCoordinatesButton.enabled = YES;
    }];
}

- (void)didReceiveMemoryWarning
{
    [super didReceiveMemoryWarning];
    // Dispose of any resources that can be recreated.
}
-(BOOL)textFieldShouldReturn:(UITextField *)textField
{
    [textField resignFirstResponder];
    return YES;
}
@end

上面的代码无法为我提供经度和纬度。需要一些帮助，了解我在这里做错了什么，或者我缺少了什么

提前谢谢

这是一个非常旧的答案，请查看新的更新

- (IBAction)forwardButton:(id)sender { if([self.address.text length]) { NSString *place = self.address.text; CLGeocoder *geocoder = [[CLGeocoder alloc] init]; __unsafe_unretained RTGeoCoderViewController *weakSelf = self; [geocoder geocodeAddressString:place completionHandler:^(NSArray* placemarks, NSError* error) { NSLog(@"completed"); if ( error ) { NSLog(@"error = %@", error ); dispatch_async(dispatch_get_main_queue(), ^{ UIAlertView *alert = [[UIAlertView alloc] initWithTitle:@"Error" message:[self errorMessage:error.code] delegate:nil cancelButtonTitle:nil otherButtonTitles:@"OK", nil]; [alert show]; }); } else { NSLog(@"%@",placemarks); } }]; } }
编辑：
在将此检查与iOS8更新一起使用之前

NSLocationAlwaysUsageDescription NSLocationWhenInUseUsageDescription
这是为了获取lat和基于长的用户区域，如街道名称、州名称、国家

-(CLLocationCoordinate2D) getLocationFromAddressString: (NSString*) addressStr { double latitude = 0, longitude = 0; NSString *esc_addr = [addressStr stringByAddingPercentEscapesUsingEncoding:NSUTF8StringEncoding]; NSString *req = [NSString stringWithFormat:@"http://maps.google.com/maps/api/geocode/json?sensor=false&address=%@", esc_addr]; NSString *result = [NSString stringWithContentsOfURL:[NSURL URLWithString:req] encoding:NSUTF8StringEncoding error:NULL]; if (result) { NSScanner *scanner = [NSScanner scannerWithString:result]; if ([scanner scanUpToString:@"\"lat\" :" intoString:nil] && [scanner scanString:@"\"lat\" :" intoString:nil]) { [scanner scanDouble:&latitude]; if ([scanner scanUpToString:@"\"lng\" :" intoString:nil] && [scanner scanString:@"\"lng\" :" intoString:nil]) { [scanner scanDouble:&longitude]; } } } CLLocationCoordinate2D center; center.latitude=latitude; center.longitude = longitude; NSLog(@"View Controller get Location Logitute : %f",center.latitude); NSLog(@"View Controller get Location Latitute : %f",center.longitude); return center; }
在viewdidload方法中或根据项目的某个地方调用类似这样的方法

[self getLocationFromAddressString:@"chennai"];
只需在浏览器中传递此信息
您将使用带有lat和lon的json格式

http://maps.google.com/maps/api/geocode/json?sensor=false&address=@"your city name here" NSString *address = [NSString stringWithFormat:@"http://maps.google.com/maps/api/geocode/json?sensor=false&address=%@ %@ %@", self.streetField.text, self.cityField.text, self.countryField.text];
此方法的用法

CLLocationCoordinate2D center; center=[self getLocationFromAddressString:@"uthangarai"]; double latFrom=&center.latitude; double lonFrom=&center.longitude; NSLog(@"View Controller get Location Logitute : %f",latFrom); NSLog(@"View Controller get Location Latitute : %f",lonFrom);
试试这个

NSString *address = [NSString stringWithFormat:@"%@,%@,%@", self.streetField.text, self.cityField.text,self.countryField.text]; [self.geocoder geocodeAddressString:address completionHandler:^(NSArray *placemarks, NSError *error) { if(!error) { CLPlacemark *placemark = [placemarks objectAtIndex:0]; NSLog(@"%f",placemark.location.coordinate.latitude); NSLog(@"%f",placemark.location.coordinate.longitude); NSLog(@"%@",[NSString stringWithFormat:@"%@",[placemark description]]); } else { NSLog(@"There was a forward geocoding error\n%@",[error localizedDescription]); } } ];

寻找Swift 2.0解决方案的人可以使用以下工具：

let address = "1 Infinite Loop, CA, USA" let geocoder = CLGeocoder() geocoder.geocodeAddressString(address, completionHandler: {(placemarks, error) -> Void in if((error) != nil){ print("Error", error) } if let placemark = placemarks?.first { let coordinates:CLLocationCoordinate2D = placemark.location!.coordinate coordinates.latitude coordinates.longitude print("lat", coordinates.latitude) print("long", coordinates.longitude) } })
试试这个

- (void)getAddressFromAdrress:(NSString *)address withCompletationHandle:(void (^)(NSDictionary *))completationHandler { CLGeocoder *geoCoder = [[CLGeocoder alloc] init]; //Get the address through geoCoder [geoCoder geocodeAddressString:address completionHandler:^(NSArray *placemarks, NSError *error) { if ([placemarks count] > 0 && !error) { //get the address from placemark CLPlacemark *placemark = [placemarks objectAtIndex:0]; NSString *locatedAt = [[placemark.addressDictionary valueForKey:@"FormattedAddressLines"] componentsJoinedByString:@", "]; CLLocation *location = placemark.location; CLLocationCoordinate2D coordinate = location.coordinate; _latitudeUserLocation = coordinate.latitude; _longitudeUserLocation = coordinate.longitude; NSString *postalCode = placemark.addressDictionary[(NSString*)kABPersonAddressZIPKey]; if (postalCode == nil) postalCode = @""; if (locatedAt == nil) locatedAt = @""; NSDictionary *dictionary = [NSDictionary dictionaryWithObjectsAndKeys: postalCode ,kPostalCode, locatedAt ,kFullAddress, nil]; completationHandler(dictionary); } else { completationHandler(nil); } }]; }
Swift

public func getLocationFromAddress(address : String) -> CLLocationCoordinate2D { var lat : Double = 0.0 var lon : Double = 0.0 do { let url = String(format: "https://maps.google.com/maps/api/geocode/json?sensor=false&address=%@", (address.addingPercentEncoding(withAllowedCharacters: .urlPathAllowed)!)) let result = try Data(contentsOf: URL(string: url)!) let json = JSON(data: result) lat = json["results"][0]["geometry"]["location"]["lat"].doubleValue lon = json["results"][0]["geometry"]["location"]["lng"].doubleValue } catch let error{ print(error) } return CLLocationCoordinate2D(latitude: lat, longitude: lon) }

我使用了，但您可以根据需要解析JSON响应
使用此链接它是hope for you@Anbu.Karthik我需要根据地址获取经度和纬度值。您的代码基于Zipcode。我试过你的代码我把我的城市名和国家名传给了url，但它不起作用，请告诉我任何想法。会有一个json格式，它会显示你所在城市的格式化地址，countryi得到了，但没有得到确切的位置，例如钦奈（13.0839°N，80.2700°E），但运行此代码时会显示（13.233954 80.332361）它根据当前数据自动更改纬度和经纬度值location@karithikeyan我正在检查模拟器，它给了我正确的lat-long，但当我在地图上申请时，它将转到其他位置，请帮助…我没有获得纬度和长度我正在这里，或者尝试使用google api“”获得。通过在上述api中用+替换空格来追加地址字符串，并解析JSON响应