将十进制坐标转换成度，分，秒，方向

到目前为止，我还没有找到一个整洁的方法来获取方向字母，而没有一堆杂乱的if语句。有任何想法吗？理想情况下，我想用类别扩展CLLocation类来完成此操作。将十进制坐标转换成度，分，秒，方向

-(NSString *)nicePosition{ 

double latitude = [self.latitude doubleValue]; 
double longitude = [self.longitude doubleValue]; 

int latSeconds = (int)round(latitude * 3600); 
int latDegrees = latSeconds/3600; 
latSeconds = abs(latSeconds % 3600); 
int latMinutes = latSeconds/60; 
latSeconds %= 60; 

int longSeconds = (int)round(longitude * 3600); 
int longDegrees = longSeconds/3600; 
longSeconds = abs(longSeconds % 3600); 
int longMinutes = longSeconds/60; 
longSeconds %= 60; 

//TODO: Use N,E,S,W notation in lat/long 

return [NSString stringWithFormat:@"%i° %i' %i\", %i° %i' %i\"", latDegrees, latMinutes, latSeconds, longDegrees, longMinutes, longSeconds]; 
}

为了记录我做了以下。

-(NSString *)nicePosition{ 

double latitude = [self.latitude doubleValue]; 
double longitude = [self.longitude doubleValue]; 

int latSeconds = (int)round(abs(latitude * 3600)); 
int latDegrees = latSeconds/3600; 
latSeconds = latSeconds % 3600; 
int latMinutes = latSeconds/60; 
latSeconds %= 60; 

int longSeconds = (int)round(abs(longitude * 3600)); 
int longDegrees = longSeconds/3600; 
longSeconds = longSeconds % 3600; 
int longMinutes = longSeconds/60; 
longSeconds %= 60; 

char latDirection = (latitude >= 0) ? 'N' : 'S'; 
char longDirection = (longitude >= 0) ? 'E' : 'W'; 

return [NSString stringWithFormat:@"%i° %i' %i\" %c, %i° %i' %i\" %c", latDegrees, latMinutes, latSeconds, latDirection, longDegrees, longMinutes, longSeconds, longDirection]; 
}

来源

2012-01-13 trapper

我是密集的吗？它只是决定NESW的long/lat的数字符号吗？那么你只需要使用：'NSLog（@“％@％@”，（lat> 0）？@“N”:(lat <0）？@“S”：@“”，（long> 0）？@ “E”:(长<0）？@“W”）;' 仍然有一个“凌乱的ifs”元素，我想.. – 2012-01-13 14:16:22

标准方式：

char lonLetter = (lon > 0) ? 'E' : 'W'; 
char latLetter = (lat > 0) ? 'N' : 'S';

来源

2012-01-13 14:14:13

是的，这是我最终做的。你先回答。 :) – trapper 2012-01-13 14:33:55

下面是在C＃中的解决方案：

void Run(double latitude, double longitude) 
    { 
     int latSeconds = (int)Math.Round(latitude * 3600); 
     int latDegrees = latSeconds/3600; 
     latSeconds = Math.Abs(latSeconds % 3600); 
     int latMinutes = latSeconds/60; 
     latSeconds %= 60; 

     int longSeconds = (int)Math.Round(longitude * 3600); 
     int longDegrees = longSeconds/3600; 
     longSeconds = Math.Abs(longSeconds % 3600); 
     int longMinutes = longSeconds/60; 
     longSeconds %= 60; 

     Console.WriteLine("{0}° {1}' {2}\" {3}, {4}° {5}' {6}\" {7}", 
      Math.Abs(latDegrees), 
      latMinutes, 
      latSeconds, 
      latDegrees >= 0 ? "N" : "S", 
      Math.Abs(longDegrees), 
      longMinutes, 
      longSeconds, 
      latDegrees >= 0 ? "E" : "W"); 
    }

这是一个例子来看：

new Program().Run(-15.14131211, 56.345678); 
new Program().Run(15.14131211, -56.345678); 
new Program().Run(15.14131211, 56.345678);

它打印：

15° 8' 29" S, 56° 20' 44" W 
15° 8' 29" N, 56° 20' 44" E 
15° 8' 29" N, 56° 20' 44" E

希望这会有所帮助，而且它做对了。祝你好运！

来源

2012-01-13 14:23:30

int latSeconds = (int)round(abs(latitude * 3600));

这是错误！正确的是

int latSeconds = abs(round(latitude * 3600));

来源

2013-04-16 17:22:19 Rio

下面是一些Objective-C的基础之上丹尼尔的解决方案：

- (NSString*)coordinateString { 

    int latSeconds = (int)(self.latitude * 3600); 
    int latDegrees = latSeconds/3600; 
    latSeconds = ABS(latSeconds % 3600); 
    int latMinutes = latSeconds/60; 
    latSeconds %= 60; 

    int longSeconds = (int)(self.longitude * 3600); 
    int longDegrees = longSeconds/3600; 
    longSeconds = ABS(longSeconds % 3600); 
    int longMinutes = longSeconds/60; 
    longSeconds %= 60; 

    NSString* result = [NSString stringWithFormat:@"%d°%d'%d\"%@ %d°%d'%d\"%@", 
         ABS(latDegrees), 
         latMinutes, 
         latSeconds, 
         latDegrees >= 0 ? @"N" : @"S", 
         ABS(longDegrees), 
         longMinutes, 
         longSeconds, 
         longDegrees >= 0 ? @"E" : @"W"]; 

    return result;  
}

来源

2013-09-03 00:40:02 Xanfred

如何看起来迅速？ – MwcsMac 2015-01-17 04:45:19

如果你想这样做的SWIFT可以让这样的事情：

import MapKit 

extension CLLocationCoordinate2D { 

    var latitudeDegreeDescription: String { 
     return fromDecToDeg(self.latitude) + " \(self.latitude >= 0 ? "N" : "S")" 
    } 
    var longitudeDegreeDescription: String { 
     return fromDecToDeg(self.longitude) + " \(self.longitude >= 0 ? "E" : "W")" 
    } 
    private func fromDecToDeg(input: Double) -> String { 
     var inputSeconds = Int(input * 3600) 
     let inputDegrees = inputSeconds/3600 
     inputSeconds = abs(inputSeconds % 3600) 
     let inputMinutes = inputSeconds/60 
     inputSeconds %= 60 
     return "\(abs(inputDegrees))°\(inputMinutes)'\(inputSeconds)''" 
    } 
}

来源

2016-06-21 14:42:05 Axel

关于对Alex的回答，这里是Swift 3中带有输出元组的解决方案。它返回元组中的坐标。

此外，它确实扩展了CLLocationDegrees类，并且不需要额外的参数。

import MapKit 

extension CLLocationDegrees { 

    func degreeRepresentation() -> (northOrEast: Bool, degrees: Int, minutes: Int, seconds: Int) { 
     var inputSeconds = Int(self * 3600) 
     let inputDegrees = inputSeconds/3600 
     inputSeconds = abs(inputSeconds % 3600) 
     let inputMinutes = inputSeconds/60 
     inputSeconds %= 60 

     return (inputDegrees > 0, abs(inputDegrees), inputMinutes, inputSeconds) 
    } 

}

来源

2017-06-02 13:59:21 who9vy

将十进制坐标转换成度，分，秒，方向

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