我怎样才能得到一个完整的地址（街道，城市等）的纬度和经度由用户输入，使用iPhone SDK 3.x的？
Answer 1:
这里有一个更新的，更紧凑版本的不可饶恕的代码，它采用了最新版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;
}
它使假设为“位置”的坐标是第一位的，例如那些“视口”之前，因为它只是需要根据“液化天然气”和“纬度”键找到的第一个COORDS。 随意如果你担心这里使用这个简单的扫描技术的使用恰当的JSON扫描仪（如SBJSON）。
Answer 2:
您可以使用谷歌地理编码此 。 它是通过HTTP获取数据和分析一样简单（它可以返回JSON KML，XML，CSV）。
Answer 3:
下面是获得来自谷歌的经度和纬度类似的解决方案。 注意：此示例使用SBJson库，你可以在github：
+ (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;
}
Answer 4:
更新的版本，使用的是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;
}
}
Answer 5:
下面的方法做，你问什么。 您需要将您的谷歌地图键使其正常工作。
- (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;
}
Answer 6:
还有CoreGeoLocation，它包装起来的功能在一个框架（Mac）或静态库（iPhone）。 支持查找通过谷歌或雅虎，如果你有一个比其他的偏好。
http://github.com/thekarladam/CoreGeoLocation
Answer 7:
对于谷歌地图关键解决方案，如上面所描述杀无赦，不人们必须使应用程序免费的吗？ 按照谷歌条款及条件：9.1免费，公众了解你的地图API实现。 您的地图API实现通常必须向用户无偿使用。
随着SDK 3.0地图套件，这是使用SDK容易。 看到苹果的手册或遵循： http://www.devworld.apple.com/iphone/program/sdk/maps.html
Answer 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;
}
Answer 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
}
文章来源: Get latitude/longitude from address
