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？
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9 回答
30
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”的坐标之前，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）。
于 2010-06-09T12:15:05.390 回答
8
您可以为此使用 Google 地理编码。它就像通过 HTTP 获取数据并对其进行解析一样简单（它可以返回 JSON KML、XML、CSV）。
于 2009-10-01T12:12:05.467 回答
7
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;
}
于 2011-10-22T18:56:26.553 回答
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;
}
}
于 2014-03-08T16:05:43.760 回答
3
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;
}
于 2009-10-01T12:21:21.570 回答
1
对于 google map key 解决方案，如上面 unforgiven 所述，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。
使用 sdk 3.0 中的地图套件，这很容易使用 SDK 完成。请参阅苹果的手册或关注：https ://developer.apple.com/documentation/mapkit
于 2010-03-04T03:14:09.700 回答
1
还有CoreGeoLocation，它将功能包装在框架（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。
于 2009-10-01T14:22:35.567 回答
0
- (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;
}
于 2014-12-10T13:41:30.457 回答
0
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
}
于 2017-03-31T14:09:32.297 回答