从地址获取纬度/经度答案

【问题标题】：Get latitude/longitude from address从地址获取纬度/经度
【发布时间】：2010-12-02 22:50:18
【问题描述】：

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？

【问题讨论】：

【解决方案1】：

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;
}

它假设“位置”的坐标首先出现，例如在“视口”之前，因为它只采用在“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）。

【讨论】：

这种方法效果很好。我发现了一个错误，可能是因为谷歌改变了响应格式。 scanUpToString 和 scanString 在 : 之前应该有另一个空格。它应该看起来像：scanUpToString:@"\"lat\" :" 和 scanString:@"\"lat\" :"（对于 lat 和 lng）。
@cberkley 我进行了更改，但为了安全起见，应更改扫描仪以不介意 lat/lng 和冒号之间的空格。我们永远不知道 Google 何时会再次“修复”这种糟糕的格式。事实上，'russes' 的版本可能更干净。
我发布了我的解决方案，因为字符串扫描器在 2011 年似乎无法正常工作。让 SBJson 解析 Google 的响应对于从斯坦福 CS193 在线课程学习 iOS 编码的初学者来说很有意义。
地址参数中输入“test”会怎样？

【解决方案2】：

您可以使用谷歌地理编码for this。就像通过 HTTP 获取数据并解析一样简单（可以返回 JSON KML、XML、CSV）。

【讨论】：

【解决方案3】：

这是从 Google 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 上找到该库：

+ (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;
}

【讨论】：

【解决方案4】：

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;
    }
}

【讨论】：

【解决方案5】：

以下方法可以满足您的要求。您需要插入 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;

}

【讨论】：

这段代码现在不能工作，因为这段代码我的实时应用程序不能正常工作......！？！

【解决方案6】：

对于 google map key 解决方案，正如上面 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 实施。您的 Maps API 实施必须通常可供用户免费访问。

借助 sdk 3.0 中的地图工具包，这可以使用 SDK 轻松完成。参见苹果手册或关注：https://developer.apple.com/documentation/mapkit

【讨论】：

【解决方案7】：

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 进行查找，如果您更喜欢其中一个。

https://github.com/thekarladam/CoreGeoLocation

【讨论】：

【解决方案8】：

- (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;

}

【讨论】：

【解决方案9】：

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
}

【讨论】：

在最新的 swift 3.0 中回答