ios 从地址获取纬度/经度

Get latitude/longitude from address

提问by Luka

How can I get latitude and longitude from a full address (street, city, etc.) input by the user, using the iPhone SDK 3.x?

如何使用 iPhone SDK 3.x 从用户输入的完整地址（街道、城市等）获取纬度和经度？

回答by Thomas Tempelmann

Here's an updated, more compact, version of unforgiven's code, which uses the latest v3 API:

这是 unforgiven 代码的更新、更紧凑的版本，它使用最新的 v3 API：

- (CLLocationCoordinate2D) geoCodeUsingAddress:(NSString *)address
{
    double latitude = 0, longitude = 0;
    NSString *esc_addr =  [address 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;
    return center;
}

It makes the assumption that the coordinates for "location" come first, e.g. before those for "viewport", because it just takes the first coords it finds under the "lng" and "lat" keys. Feel free to use a proper JSON scanner (e.g. SBJSON) if you are worried about this simple scanning technique used here.

它假设“位置”的坐标首先出现，例如在“视口”的坐标之前，因为它只采用它在“lng”和“lat”键下找到的第一个坐标。如果您担心这里使用的这种简单的扫描技术，请随意使用适当的 JSON 扫描器（例如 SBJSON）。

回答by Valerii Hiora

You can use Google Geocoding for this. It is as simple as getting data through HTTP and parsing it (it can return JSON KML, XML, CSV).

您可以为此使用 Google 地理编码。它就像通过 HTTP 获取数据并解析它一样简单（它可以返回 JSON KML、XML、CSV）。

回答by russes

Here's a similar solution for obtaining the latitude and longitude from Google. Note: This example uses the SBJson library, which you can find on github:

这是从谷歌获取纬度和经度的类似解决方案。注意：此示例使用 SBJson 库，您可以在 github 上找到它：

+ (CLLocationCoordinate2D) geoCodeUsingAddress: (NSString *) address
{
    CLLocationCoordinate2D myLocation; 

// -- modified from the stackoverflow page - we use the SBJson parser instead of the string scanner --

        NSString       *esc_addr = [address stringByAddingPercentEscapesUsingEncoding: NSUTF8StringEncoding];
        NSString            *req = [NSString stringWithFormat: @"http://maps.google.com/maps/api/geocode/json?sensor=false&address=%@", esc_addr];
    NSDictionary *googleResponse = [[NSString stringWithContentsOfURL: [NSURL URLWithString: req] encoding: NSUTF8StringEncoding error: NULL] JSONValue];

    NSDictionary    *resultsDict = [googleResponse valueForKey:  @"results"];   // get the results dictionary
    NSDictionary   *geometryDict = [   resultsDict valueForKey: @"geometry"];   // geometry dictionary within the  results dictionary
    NSDictionary   *locationDict = [  geometryDict valueForKey: @"location"];   // location dictionary within the geometry dictionary

// -- you should be able to strip the latitude & longitude from google's location information (while understanding what the json parser returns) --

    DLog (@"-- returning latitude & longitude from google --");

    NSArray *latArray = [locationDict valueForKey: @"lat"]; NSString *latString = [latArray lastObject];     // (one element) array entries provided by the json parser
    NSArray *lngArray = [locationDict valueForKey: @"lng"]; NSString *lngString = [lngArray lastObject];     // (one element) array entries provided by the json parser

     myLocation.latitude = [latString doubleValue];     // the json parser uses NSArrays which don't support "doubleValue"
    myLocation.longitude = [lngString doubleValue];

    return myLocation;
}

回答by mrd

Update version, using iOS JSON:

更新版本，使用 iOS JSON：

- (CLLocationCoordinate2D)getLocation:(NSString *)address {

    CLLocationCoordinate2D center;
    NSString *esc_addr =  [address stringByAddingPercentEscapesUsingEncoding:NSUTF8StringEncoding];
    NSString *req = [NSString stringWithFormat:@"http://maps.google.com/maps/api/geocode/json?sensor=false&address=%@", esc_addr];
    NSData *responseData = [[NSData alloc] initWithContentsOfURL:
                        [NSURL URLWithString:req]];    NSError *error;
    NSMutableDictionary *responseDictionary = [NSJSONSerialization
                                               JSONObjectWithData:responseData
                                               options:nil
                                               error:&error];
    if( error )
    {
        NSLog(@"%@", [error localizedDescription]);
        center.latitude = 0;
        center.longitude = 0;
        return center;
    }
    else {
        NSArray *results = (NSArray *) responseDictionary[@"results"];
        NSDictionary *firstItem = (NSDictionary *) [results objectAtIndex:0];
        NSDictionary *geometry = (NSDictionary *) [firstItem objectForKey:@"geometry"];
        NSDictionary *location = (NSDictionary *) [geometry objectForKey:@"location"];
        NSNumber *lat = (NSNumber *) [location objectForKey:@"lat"];
        NSNumber *lng = (NSNumber *) [location objectForKey:@"lng"];

        center.latitude = [lat doubleValue];
        center.longitude = [lng doubleValue];
        return center;
    }
}

回答by Massimo Cafaro

The following method does what you asked for. You need to insert your Google maps key for this to work correctly.

以下方法可以满足您的要求。您需要插入您的 Google 地图密钥才能正常工作。

- (CLLocationCoordinate2D) geoCodeUsingAddress:(NSString *)address{

    int code = -1;
    int accuracy = -1;
    float latitude = 0.0f;
    float longitude = 0.0f;
    CLLocationCoordinate2D center;

    // setup maps api key
    NSString * MAPS_API_KEY = @"YOUR GOOGLE MAPS KEY HERE";

    NSString *escaped_address =  [address stringByAddingPercentEscapesUsingEncoding: NSUTF8StringEncoding];
    // Contact Google and make a geocoding request
    NSString *requestString = [NSString stringWithFormat:@"http://maps.google.com/maps/geo?q=%@&output=csv&oe=utf8&key=%@&sensor=false&gl=it", escaped_address, MAPS_API_KEY];
    NSURL *url = [NSURL URLWithString:requestString];

    NSString *result = [NSString stringWithContentsOfURL: url encoding: NSUTF8StringEncoding error:NULL];
        if(result){
            // we got a result from the server, now parse it
            NSScanner *scanner = [NSScanner scannerWithString:result];
            [scanner scanInt:&code];
            if(code == 200){
                // everything went off smoothly
                [scanner scanString:@"," intoString:nil];
                [scanner scanInt:&accuracy];

                //NSLog(@"Accuracy: %d", accuracy);

                [scanner scanString:@"," intoString:nil];
                [scanner scanFloat:&latitude];
                [scanner scanString:@"," intoString:nil];
                [scanner scanFloat:&longitude];


                center.latitude = latitude;
                center.longitude = longitude;

                return center;


            }
            else{
                // the server answer was not the one we expected
                UIAlertView *alert = [[[UIAlertView alloc] 
                                       initWithTitle: @"Warning" 
                                       message:@"Connection to Google Maps failed"
                                       delegate:nil
                                       cancelButtonTitle:nil 
                                       otherButtonTitles:@"OK", nil] autorelease];

                [alert show];

                center.latitude = 0.0f;
                center.longitude = 0.0f;

                return center;


            }

        }
        else{
            // no result back from the server
            UIAlertView *alert = [[[UIAlertView alloc] 
                                   initWithTitle: @"Warning" 
                                   message:@"Connection to Google Maps failed"
                                   delegate:nil
                                   cancelButtonTitle:nil 
                                   otherButtonTitles:@"OK", nil] autorelease];

            [alert show];

            center.latitude = 0.0f;
            center.longitude = 0.0f;

            return center;
        }

    }

        center.latitude = 0.0f;
        center.longitude = 0.0f;

        return center;

}

回答by Jablair

There's also CoreGeoLocation, which wraps up the functionality in a framework (Mac) or static library (iPhone). Supports lookups through Google or Yahoo, if you have a preference for one over the other.

还有 CoreGeoLocation，它将功能封装在框架 (Mac) 或静态库 (iPhone) 中。支持通过 Google 或 Yahoo 查找，如果您偏爱其中一个。

https://github.com/thekarladam/CoreGeoLocation

https://github.com/thekarladam/CoreGeoLocation

回答by gammapoint

For the google map key solution, as described by unforgiven above, doesn't one has to make the app free? As per google terms & conditions: 9.1 Free, Public Accessibility to Your Maps API Implementation. Your Maps API Implementation must be generally accessible to users without charge.

对于 google map key 解决方案，如上面 unforgiven 所描述的，不是必须让应用程序免费吗？根据 google 条款和条件：9.1 免费、公开访问您的 Maps API 实现。您的 Maps API 实现必须可供用户免费访问。

With map kit in sdk 3.0 this is easily done using the SDK. See apple's manuals or follow: https://developer.apple.com/documentation/mapkit

使用 sdk 3.0 中的地图工具包，这可以使用 SDK 轻松完成。请参阅苹果的手册或遵循：https: //developer.apple.com/documentation/mapkit

回答by Paresh Hirpara

- (void)viewDidLoad
{
    app=(AppDelegate *)[[UIApplication sharedApplication] delegate];
    NSLog(@"%@", app.str_address);


    NSLog(@"internet connect");

    NSString *Str_address=_txt_zipcode.text;

    double latitude1 = 0, longitude1 = 0;
    NSString *esc_addr =  [ Str_address 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:&latitude1];
            if ([scanner scanUpToString:@"\"lng\" :" intoString:nil] && [scanner scanString:@"\"lng\" :" intoString:nil])
            {
                [scanner scanDouble:&longitude1];
            }
        }
    }


    //in #.hfile
   // CLLocationCoordinate2D lat;
   // CLLocationCoordinate2D lon;
   // float address_latitude;
   // float address_longitude;


    lat.latitude=latitude1;
    lon.longitude=longitude1;

    address_latitude=lat.latitude;
    address_longitude=lon.longitude;

}

回答by Piyush Sinroja

func geoCodeUsingAddress(address: NSString) -> CLLocationCoordinate2D {
    var latitude: Double = 0
    var longitude: Double = 0
    let addressstr : NSString = "http://maps.google.com/maps/api/geocode/json?sensor=false&address=\(address)" as NSString
    let urlStr  = addressstr.addingPercentEncoding(withAllowedCharacters: .urlQueryAllowed)
    let searchURL: NSURL = NSURL(string: urlStr! as String)!
    do {
        let newdata = try Data(contentsOf: searchURL as URL)
        if let responseDictionary = try JSONSerialization.jsonObject(with: newdata, options: []) as? NSDictionary {
            print(responseDictionary)
            let array = responseDictionary.object(forKey: "results") as! NSArray
            let dic = array[0] as! NSDictionary
            let locationDic = (dic.object(forKey: "geometry") as! NSDictionary).object(forKey: "location") as! NSDictionary
            latitude = locationDic.object(forKey: "lat") as! Double
            longitude = locationDic.object(forKey: "lng") as! Double
        }} catch {
    }
    var center = CLLocationCoordinate2D()
    center.latitude = latitude
    center.longitude = longitude
    return center
}