Question

我正在尝试编写一个取小数度（lat或long）的函数，并将它们转换为DMS度分钟秒。我知道我的意思是将小数点数乘以60然后再次为小数。但我是一个菜鸟。我会分开号码吗？

function ConvertDDToDMS(DD) {
    eg. DD =-42.4
    D= 42;
    M= 4*60;
    S= .M * 60;
    var DMS =

    return DMS //append Direction (N, S, E, W);
}

我是在正确的轨道上吗？

Answer 1

function ConvertDDToDMS(D, lng){
    return {
        dir : D<0?lng?'W':'S':lng?'E':'N',
        deg : 0|(D<0?D=-D:D),
        min : 0|D%1*60,
        sec :(0|D*60%1*6000)/100
    };
}

以上为您提供了一个对象{deg:, min:, sec:, dir:}，其中sec被截断为两位数（例如3.14），dir是N，E，S之一，W取决于您是否将lng（经度）参数设置为true。 e.g：

ConvertDDToDMS(-18.213, true) == {
   deg : 18,
   min : 12,
   sec : 46.79,
   dir : 'W'
}

或者如果您只想要基本字符串：

function ConvertDDToDMS(D){
    return [0|D, 'd ', 0|(D<0?D=-D:D)%1*60, "' ", 0|D*60%1*60, '"'].join('');
}

ConvertDDToDMS(-18.213) == `-18d 12' 46"`

Answer 2

目前尚不清楚您需要如何输出。这是一个以字符串形式返回所有3个值的版本：

function ConvertDDToDMS(dd)
{
    var deg = dd | 0; // truncate dd to get degrees
    var frac = Math.abs(dd - deg); // get fractional part
    var min = (frac * 60) | 0; // multiply fraction by 60 and truncate
    var sec = frac * 3600 - min * 60;
    return deg + "d " + min + "' " + sec + "\"";
}

Answer 3

更新：我删除了没有任何意义的部分（感谢cwolves！）。

这里有另一个实现。它不会像以前那样短暂或高效，但希望更容易理解。

要做到正确，首先您需要了解计算的完成方式，然后才尝试实施它们。为此，伪代码是一个很好的选择，因为您用简单的英语或简单的语法写下易于理解的步骤，然后将其翻译成所选的编程语言。

我希望它有用！

/* This is the pseudocode you need to follow:
 * It's a modified version from 
 * http://en.wikipedia.org/wiki/Geographic_coordinate_conversion#Conversion_from_Decimal_Degree_to_DMS

function deg_to_dms ( degfloat )
   Compute degrees, minutes and seconds:
   deg ← integerpart ( degfloat )
   minfloat ← 60 * ( degfloat - deg )
   min ← integerpart ( minfloat )
   secfloat ← 60 * ( minfloat - min )
   Round seconds to desired accuracy:
   secfloat ← round( secfloat, digits )
   After rounding, the seconds might become 60. These two
   if-tests are not necessary if no rounding is done.
   if secfloat = 60
      min ← min + 1
      secfloat ← 0
   end if
   if min = 60
      deg ← deg + 1
      min ← 0
   end if
   Return output:
   return ( deg, min, secfloat )
end function
*/

function deg_to_dms (deg) {
   var d = Math.floor (deg);
   var minfloat = (deg-d)*60;
   var m = Math.floor(minfloat);
   var secfloat = (minfloat-m)*60;
   var s = Math.round(secfloat);
   // After rounding, the seconds might become 60. These two
   // if-tests are not necessary if no rounding is done.
   if (s==60) {
     m++;
     s=0;
   }
   if (m==60) {
     d++;
     m=0;
   }
   return ("" + d + ":" + m + ":" + s);
}

Answer 4

基于上面的回答，我把它们写成了javascript和php风格。

JS -

function convertDDToDMS(deg, lng){
    var d = parseInt(deg);
    var minfloat  = Math.abs((deg-d) * 60); 
    var m = Math.floor(minfloat);
    var secfloat = (minfloat-m)*60;
    var s = Math.round(secfloat); 
    d = Math.abs(d);

    if (s==60) {
        m++;
        s=0;
    }
    if (m==60) {
        d++;
        m=0;
    }

    return {
        dir : deg<0?lng?'W':'S':lng?'E':'N',
        deg : d,
        min : m,
        sec : s
    };
}

PHP -

function convertDDtoDMS($deg, $lng){
    $dd = intval($deg);
    $minfloat = abs(($deg - $dd) * 60);
    $mm = floor($minfloat);
    $secfloat = ($minfloat - $mm) * 60;
    $ss = round($secfloat);
    $dd = abs($dd);

    if($ss == 60){
        $mm++;
        $ss = 0;
    }

    if($mm == 60){
        $dd++;
        $mm = 0;
    }

    $dd = array(
        'dir' => $deg < 0 ? ($lng ? 'W' : 'S') : ($lng ? 'E' : 'N'),
        'deg' => abs($dd),
        'min' => $mm,
        'sec' => $ss,
    );

    return $dd;
}

Answer 5

private static DecimalFormat DecimalFormat = new DecimalFormat(".##");
public static void main(String[] args){
    double decimal_degrees = 22.4229541515;

    System.out.println(getDMS(decimal_degrees));
}
public static String getDMS(double decimal_degrees) {
    double degree =  Math.floor(decimal_degrees);
    double minutes = ((decimal_degrees - Math.floor(decimal_degrees)) * 60.0); 
    double seconds = (minutes - Math.floor(minutes)) * 60.0;
    return ((int)degree)+":"+((int)minutes)+":"+decimalFormat.format(seconds);

}

输入：22.4229541515输出：22：25：22.63

Answer 6

经过一段时间后，无法使上面的脚本正常工作；只需将dms交给脚本

function ConvertDMSToDEG(dms) {   
    var dms_Array = dms.split(/[^\d\w\.]+/); 
    var degrees = dms_Array[0];
    var minutes = dms_Array[1];
    var seconds = dms_Array[2];
    var direction = dms_Array[3];

    var deg = (Number(degrees) + Number(minutes)/60 + Number(seconds)/3600).toFixed(6);

    if (direction == "S" || direction == "W") {
        deg = deg * -1;
    } // Don't do anything for N or E
    return deg;
}

反之亦然，只是给出脚本的度数，纬度（纬度）为假

function ConvertDEGToDMS(deg, lat) {
    var absolute = Math.abs(deg);

    var degrees = Math.floor(absolute);
    var minutesNotTruncated = (absolute - degrees) * 60;
    var minutes = Math.floor(minutesNotTruncated);
    var seconds = ((minutesNotTruncated - minutes) * 60).toFixed(2);

    if (lat) {
        var direction = deg >= 0 ? "N" : "S";
    } else {
        var direction = deg >= 0 ? "E" : "W";
    }

    return degrees + "°" + minutes + "'" + seconds + "\"" + direction;
}

希望这对人们有帮助。

Answer 7

尝试这种完美的工作方式！

function truncate(n) {
    return n > 0 ? Math.floor(n) : Math.ceil(n);
}

function getDMS(dd, longOrLat) {
    let hemisphere = /^[WE]|(?:lon)/i.test(longOrLat)
    ? dd < 0
      ? "W"
      : "E"
    : dd < 0
      ? "S"
      : "N";

    const absDD = Math.abs(dd);
    const degrees = truncate(absDD);
    const minutes = truncate((absDD - degrees) * 60);
    const seconds = ((absDD - degrees - minutes / 60) * Math.pow(60, 2)).toFixed(2);

    let dmsArray = [degrees, minutes, seconds, hemisphere];
    return `${dmsArray[0]}°${dmsArray[1]}'${dmsArray[2]}" ${dmsArray[3]}`;
}

var lat = 13.041107;
var lon = 80.233232;

var latDMS = getDMS(lat, 'lat'); 
var lonDMS = getDMS(lon, 'long');
console.log('latDMS: '+ latDMS);
console.log('lonDMS: '+ lonDMS);

Output:
latDMS: 13°2'27.99" N
lonDMS: 80°13'59.64" E

Answer 8

一种解决方案，可以选择在输出秒中指定小数位，并校正由于舍入秒和分钟而导致的任何边缘情况。

// @ input {deg}     Numeric; degrees number to convert
// @ input {dplaces} Decimal places to use for output seconds
//                   Default 0 places
// @ return {DMS} string degrees (°) minutes (') seconds (")
//
function degToDMS (deg, dplaces=0) {
  var d = Math.floor (deg);          // make degrees
  var m = Math.floor((deg-d)*60);    // make minutes
  var s = Math.round(((deg-d)*60-m)*60*Math.pow(10,dplaces))/Math.pow(10,dplaces); // Make sec rounded
  s == 60 && (m++, s=0 );            // if seconds rounds to 60 then increment minutes, reset seconds
  m == 60 && (d++, m=0 );            // if minutes rounds to 60 then increment degress, reset minutes
  return (d + "° " + m + "' " + s+'"');   // create output DMS string
}

// ----- tests ------
console.log(degToDMS(55.23456));         // 55° 14' 4"
console.log(degToDMS(55.23456   ,3));    // 55° 14' 4.416"
console.log(degToDMS(4 + 20/60  ,2));    // 4° 20' 0"
console.log(degToDMS(89.64789   ,2));    // 89° 38' 52.4"
console.log(degToDMS(-23.1234567,3));    // -24° 52' 35.556"

Answer 9

此脚本在TypeScript中可运行％100：

    ConvertDDToDMS(deg: number, lng: boolean): string {

    var d = parseInt(deg.toString());
    var minfloat = Math.abs((deg - d) * 60);
    var m = Math.floor(minfloat);
    var secfloat = (minfloat - m) * 60;
    var s = Math.round((secfloat + Number.EPSILON) * 100) / 100
    d = Math.abs(d);

    if (s == 60) {
      m++;
      s = 0;
    }
    if (m == 60) {
      d++;
      m = 0;
    }

    let dms = {
      dir: deg < 0 ? lng ? 'W' : 'S' : lng ? 'E' : 'N',
      deg: d,
      min: m,
      sec: s
    };
    return `${dms.deg}\u00B0 ${dms.min}' ${dms.sec}" ${dms.dir}`
  }

Answer 10

仅供说明，答案

function ConvertDDToDMS(D){
return [0|D, 'd ', 0|(D<0?D=-D:D)%1*60, "' ", 0|D*60%1*60, '"'].join('');

}

对于-1°和0°之间的角度，不不起作用。厄运！ HC

Javascript中的十进制度到几分钟和几秒

10 个答案: