请有人发布一个SQL函数来将easting / northing转换为经度/纬度。我知道它非常复杂，但我没有找到任何在T-SQL中记录它的人。
此javascript code有效，但我无法将其转换为SQL。
我有16,000个坐标，需要将它们全部转换为lat / long。
这是我到目前为止所做的，但它没有超过while循环。
DECLARE @east real = 482353,
@north real = 213371
DECLARE @a real = 6377563.396,
@b real = 6356256.910,
@F0 real = 0.9996012717,
@lat0 real = 49*PI()/180,
@lon0 real = -2*PI()/180
DECLARE @N0 real = -100000,
@E0 real = 400000,
@e2 real = 1 - (@b*@b)/(@a*@a),
@n real = (@a-@b)/(@a+@b)
DECLARE @n2 real = @n*@n,
@n3 real = @n*@n*@n
DECLARE @lat real = @lat0,
@M real = 0
WHILE (@north-@N0-@M >= 0.00001)
BEGIN
SET @lat = ((@north-@N0-@M)/(@a*@F0)) + @lat
DECLARE @Ma real = (1 + @n + (5/4)*@n2 + (5/4)*@n3) * (@lat-@lat0),
@Mb real = (3*@n + 3*@n*@n + (21/8)*@n3) * SIN(@lat-@lat0) * COS(@lat+@lat0),
@Mc real = ((15/8)*@n2 + (15/8)*@n3) * SIN(2*(@lat-@lat0)) * COS(2*(@lat+@lat0)),
@Md real = (35/24)*@n3 * SIN(3*(@lat-@lat0)) * COS(3*(@lat+@lat0))
SET @M = @b * @F0 * (@Ma - @Mb + @Mc - @Md)
END
DECLARE @cosLat real = COS(@lat),
@sinLat real = SIN(@lat)
DECLARE @nu real = @a*@F0/sqrt(1-@e2*@sinLat*@sinLat)
DECLARE @rho real = @a*@F0*(1-@e2)/POWER(1-@e2*@sinLat*@sinLat, 1.5)
DECLARE @eta2 real = @nu/@rho-1
DECLARE @tanLat real = tan(@lat)
DECLARE @tan2lat real = @tanLat*@tanLat
DECLARE @tan4lat real = @tan2lat*@tan2lat
DECLARE @tan6lat real = @tan4lat*@tan2lat
DECLARE @secLat real = 1/@cosLat
DECLARE @nu3 real = @nu*@nu*@nu
DECLARE @nu5 real = @nu3*@nu*@nu
DECLARE @nu7 real = @nu5*@nu*@nu
DECLARE @VII real = @tanLat/(2*@rho*@nu)
DECLARE @VIII real = @tanLat/(24*@rho*@nu3)*(5+3*@tan2lat+@eta2-9*@tan2lat*@eta2)
DECLARE @IX real = @tanLat/(720*@rho*@nu5)*(61+90*@tan2lat+45*@tan4lat)
DECLARE @X real = @secLat/@nu
DECLARE @XI real = @secLat/(6*@nu3)*(@nu/@rho+2*@tan2lat)
DECLARE @XII real = @secLat/(120*@nu5)*(5+28*@tan2lat+24*@tan4lat)
DECLARE @XIIA real = @secLat/(5040*@nu7)*(61+662*@tan2lat+1320*@tan4lat+720*@tan6lat)
DECLARE @dE real = (@east-@E0)
DECLARE @dE2 real = @dE*@dE
DECLARE @dE3 real = @dE2*@dE
DECLARE @dE4 real = @dE2*@dE2,
@dE5 real = @dE3*@dE2
DECLARE @dE6 real = @dE4*@dE2,
@dE7 real = @dE5*@dE2
SET @lat = @lat - @VII*@dE2 + @VIII*@dE4 - @IX*@dE6
DECLARE @lon real = @lon0 + @X*@dE - @XI*@dE3 + @XII*@dE5 - @XIIA*@dE7
SELECT @lon, @lat
答案 0 :(得分：14)
我一直在努力解决这个问题。 我在OSGB36中有很多北向/东向点，必须定期进行转换。 请注意，下面的UDF将OSGB36（军械测量）投影中的北/东转换为WGS84投影中的纬度/经度，以便它们可以在谷歌地图中使用。
/****** Object: UserDefinedFunction [dbo].[NEtoLL] Script Date: 09/06/2012 17:06:39 ******/
SET ANSI_NULLS ON
GO
SET QUOTED_IDENTIFIER ON
GO
CREATE FUNCTION [dbo].[NEtoLL] (@East INT, @North INT, @LatOrLng VARCHAR(3)) RETURNS FLOAT AS
BEGIN
--Author: Sandy Motteram
--Date: 06 September 2012
--UDF adapted from javascript at http://www.bdcc.co.uk/LatLngToOSGB.js
--found on page http://mapki.com/wiki/Tools:Snippets
--Instructions:
--Latitude and Longitude are calculated based on BOTH the easting and northing values from the OSGB36
--This UDF takes both easting and northing values in OSGB36 projection and you must specify if a latitude or longitude co-ordinate should be returned.
--IT first converts E/N values to lat and long in OSGB36 projection, then converts those values to lat/lng in WGS84 projection
--Sample values below
--DECLARE @East INT, @North INT, @LatOrLng VARCHAR(3)
--SELECT @East = 529000, @North = 183650 --that combo should be the corner of Camden High St and Delancey St
DECLARE @Pi FLOAT
, @K0 FLOAT
, @OriginLat FLOAT
, @OriginLong FLOAT
, @OriginX FLOAT
, @OriginY FLOAT
, @a FLOAT
, @b FLOAT
, @e2 FLOAT
, @ex FLOAT
, @n1 FLOAT
, @n2 FLOAT
, @n3 FLOAT
, @OriginNorthings FLOAT
, @lat FLOAT
, @lon FLOAT
, @Northing FLOAT
, @Easting FLOAT
SELECT @Pi = 3.14159265358979323846
, @K0 = 0.9996012717 -- grid scale factor on central meridean
, @OriginLat = 49.0
, @OriginLong = -2.0
, @OriginX = 400000 -- 400 kM
, @OriginY = -100000 -- 100 kM
, @a = 6377563.396 -- Airy Spheroid
, @b = 6356256.910
/* , @e2
, @ex
, @n1
, @n2
, @n3
, @OriginNorthings*/
-- compute interim values
SELECT @a = @a * @K0
, @b = @b * @K0
SET @n1 = (@a - @b) / (@a + @b)
SET @n2 = @n1 * @n1
SET @n3 = @n2 * @n1
SET @lat = @OriginLat * @Pi / 180.0 -- to radians
SELECT @e2 = (@a * @a - @b * @b) / (@a * @a) -- first eccentricity
, @ex = (@a * @a - @b * @b) / (@b * @b) -- second eccentricity
SET @OriginNorthings = @b * @lat + @b * (@n1 * (1.0 + 5.0 * @n1 * (1.0 + @n1) / 4.0) * @lat
- 3.0 * @n1 * (1.0 + @n1 * (1.0 + 7.0 * @n1 / 8.0)) * SIN(@lat) * COS(@lat)
+ (15.0 * @n1 * (@n1 + @n2) / 8.0) * SIN(2.0 * @lat) * COS(2.0 * @lat)
- (35.0 * @n3 / 24.0) * SIN(3.0 * @lat) * COS(3.0 * @lat))
SELECT @northing = @north - @OriginY
, @easting = @east - @OriginX
DECLARE @nu FLOAT
, @phid FLOAT
, @phid2 FLOAT
, @t2 FLOAT
, @t FLOAT
, @q2 FLOAT
, @c FLOAT
, @s FLOAT
, @nphid FLOAT
, @dnphid FLOAT
, @nu2 FLOAT
, @nudivrho FLOAT
, @invnurho FLOAT
, @rho FLOAT
, @eta2 FLOAT
/* Evaluate M term: latitude of the northing on the centre meridian */
SET @northing = @northing + @OriginNorthings
SET @phid = @northing / (@b*(1.0 + @n1 + 5.0 * (@n2 + @n3) / 4.0)) - 1.0
SET @phid2 = @phid + 1.0
WHILE (ABS(@phid2 - @phid) > 0.000001)
BEGIN
SET @phid = @phid2;
SET @nphid = @b * @phid + @b * (@n1 * (1.0 + 5.0 * @n1 * (1.0 + @n1) / 4.0) * @phid
- 3.0 * @n1 * (1.0 + @n1 * (1.0 + 7.0 * @n1 / 8.0)) * SIN(@phid) * COS(@phid)
+ (15.0 * @n1 * (@n1 + @n2) / 8.0) * SIN(2.0 * @phid) * COS(2.0 * @phid)
- (35.0 * @n3 / 24.0) * SIN(3.0 * @phid) * COS(3.0 * @phid))
SET @dnphid = @b * ((1.0 + @n1 + 5.0 * (@n2 + @n3) / 4.0) - 3.0 * (@n1 + @n2 + 7.0 * @n3 / 8.0) * COS(2.0 * @phid)
+ (15.0 * (@n2 + @n3) / 4.0) * COS(4 * @phid) - (35.0 * @n3 / 8.0) * COS(6.0 * @phid))
SET @phid2 = @phid - (@nphid - @northing) / @dnphid
END
SELECT @c = COS(@phid)
, @s = SIN(@phid)
, @t = TAN(@phid)
SELECT @t2 = @t * @t
, @q2 = @easting * @easting
SET @nu2 = (@a * @a) / (1.0 - @e2 * @s * @s)
SET @nu = SQRT(@nu2)
SET @nudivrho = @a * @a * @c * @c / (@b * @b) - @c * @c + 1.0
SET @eta2 = @nudivrho - 1
SET @rho = @nu / @nudivrho;
SET @invnurho = ((1.0 - @e2 * @s * @s) * (1.0 - @e2 * @s * @s)) / (@a * @a * (1.0 - @e2))
SET @lat = @phid - @t * @q2 * @invnurho / 2.0 + (@q2 * @q2 * (@t / (24 * @rho * @nu2 * @nu) * (5 + (3 * @t2) + @eta2 - (9 * @t2 * @eta2))))
SET @lon = (@easting / (@c * @nu))
- (@easting * @q2 * ((@nudivrho + 2.0 * @t2) / (6.0 * @nu2)) / (@c * @nu))
+ (@q2 * @q2 * @easting * (5 + (28 * @t2) + (24 * @t2 * @t2)) / (120 * @nu2 * @nu2 * @nu * @c))
SELECT @lat = @lat * 180.0 / @Pi
, @lon = @lon * 180.0 / @Pi + @OriginLong
--Now convert the lat and long from OSGB36 to WGS84
DECLARE @OGlat FLOAT
, @OGlon FLOAT
, @height FLOAT
SELECT @OGlat = @lat
, @OGlon = @lon
, @height = 24 --London's mean height above sea level is 24 metres. Adjust for other locations.
DECLARE @deg2rad FLOAT
, @rad2deg FLOAT
, @radOGlat FLOAT
, @radOGlon FLOAT
SELECT @deg2rad = @Pi / 180
, @rad2deg = 180 / @Pi
--first off convert to radians
SELECT @radOGlat = @OGlat * @deg2rad
, @radOGlon = @OGlon * @deg2rad
--these are the values for WGS84(GRS80) to OSGB36(Airy)
DECLARE @a2 FLOAT
, @h FLOAT
, @xp FLOAT
, @yp FLOAT
, @zp FLOAT
, @xr FLOAT
, @yr FLOAT
, @zr FLOAT
, @sf FLOAT
, @e FLOAT
, @v FLOAT
, @x FLOAT
, @y FLOAT
, @z FLOAT
, @xrot FLOAT
, @yrot FLOAT
, @zrot FLOAT
, @hx FLOAT
, @hy FLOAT
, @hz FLOAT
, @newLon FLOAT
, @newLat FLOAT
, @p FLOAT
, @errvalue FLOAT
, @lat0 FLOAT
SELECT @a2 = 6378137 -- WGS84_AXIS
, @e2 = 0.00669438037928458 -- WGS84_ECCENTRIC
, @h = @height -- height above datum (from $GPGGA sentence)
, @a = 6377563.396 -- OSGB_AXIS
, @e = 0.0066705397616 -- OSGB_ECCENTRIC
, @xp = 446.448
, @yp = -125.157
, @zp = 542.06
, @xr = 0.1502
, @yr = 0.247
, @zr = 0.8421
, @s = -20.4894
-- convert to cartesian; lat, lon are in radians
SET @sf = @s * 0.000001
SET @v = @a / (sqrt(1 - (@e * (SIN(@radOGlat) * SIN(@radOGlat)))))
SET @x = (@v + @h) * COS(@radOGlat) * COS(@radOGlon)
SET @y = (@v + @h) * COS(@radOGlat) * SIN(@radOGlon)
SET @z = ((1 - @e) * @v + @h) * SIN(@radOGlat)
-- transform cartesian
SET @xrot = (@xr / 3600) * @deg2rad
SET @yrot = (@yr / 3600) * @deg2rad
SET @zrot = (@zr / 3600) * @deg2rad
SET @hx = @x + (@x * @sf) - (@y * @zrot) + (@z * @yrot) + @xp
SET @hy = (@x * @zrot) + @y + (@y * @sf) - (@z * @xrot) + @yp
SET @hz = (-1 * @x * @yrot) + (@y * @xrot) + @z + (@z * @sf) + @zp
-- Convert back to lat, lon
SET @newLon = ATAN(@hy / @hx)
SET @p = SQRT((@hx * @hx) + (@hy * @hy))
SET @newLat = ATAN(@hz / (@p * (1 - @e2)))
SET @v = @a2 / (SQRT(1 - @e2 * (SIN(@newLat) * SIN(@newLat))))
SET @errvalue = 1.0;
SET @lat0 = 0
WHILE (@errvalue > 0.001)
BEGIN
SET @lat0 = ATAN((@hz + @e2 * @v * SIN(@newLat)) / @p)
SET @errvalue = ABS(@lat0 - @newLat)
SET @newLat = @lat0
END
--convert back to degrees
SET @newLat = @newLat * @rad2deg
SET @newLon = @newLon * @rad2deg
DECLARE @ReturnMe FLOAT
SET @ReturnMe = 0
IF @LatOrLng = 'Lat'
SET @ReturnMe = @newLat
IF @LatOrLng = 'Lng'
SET @ReturnMe = @newLon
RETURN @ReturnMe
END
GO
答案 1 :(得分：1)
我最终使用以下javascript函数来转换值。我知道这不是一个SQL解决方案，但它为我完成了这项工作。
function OSGridToLatLong(E, N) {
var a = 6377563.396, b = 6356256.910; // Airy 1830 major & minor semi-axes
var F0 = 0.9996012717; // NatGrid scale factor on central meridian
var lat0 = 49*Math.PI/180, lon0 = -2*Math.PI/180; // NatGrid true origin
var N0 = -100000, E0 = 400000; // northing & easting of true origin, metres
var e2 = 1 - (b*b)/(a*a); // eccentricity squared
var n = (a-b)/(a+b), n2 = n*n, n3 = n*n*n;
var lat=lat0, M=0;
do {
lat = (N-N0-M)/(a*F0) + lat;
var Ma = (1 + n + (5/4)*n2 + (5/4)*n3) * (lat-lat0);
var Mb = (3*n + 3*n*n + (21/8)*n3) * Math.sin(lat-lat0) * Math.cos(lat+lat0);
var Mc = ((15/8)*n2 + (15/8)*n3) * Math.sin(2*(lat-lat0)) * Math.cos(2*(lat+lat0));
var Md = (35/24)*n3 * Math.sin(3*(lat-lat0)) * Math.cos(3*(lat+lat0));
M = b * F0 * (Ma - Mb + Mc - Md); // meridional arc
} while (N-N0-M >= 0.00001); // ie until < 0.01mm
var cosLat = Math.cos(lat), sinLat = Math.sin(lat);
var nu = a*F0/Math.sqrt(1-e2*sinLat*sinLat); // transverse radius of curvature
var rho = a*F0*(1-e2)/Math.pow(1-e2*sinLat*sinLat, 1.5); // meridional radius of curvature
var eta2 = nu/rho-1;
var tanLat = Math.tan(lat);
var tan2lat = tanLat*tanLat, tan4lat = tan2lat*tan2lat, tan6lat = tan4lat*tan2lat;
var secLat = 1/cosLat;
var nu3 = nu*nu*nu, nu5 = nu3*nu*nu, nu7 = nu5*nu*nu;
var VII = tanLat/(2*rho*nu);
var VIII = tanLat/(24*rho*nu3)*(5+3*tan2lat+eta2-9*tan2lat*eta2);
var IX = tanLat/(720*rho*nu5)*(61+90*tan2lat+45*tan4lat);
var X = secLat/nu;
var XI = secLat/(6*nu3)*(nu/rho+2*tan2lat);
var XII = secLat/(120*nu5)*(5+28*tan2lat+24*tan4lat);
var XIIA = secLat/(5040*nu7)*(61+662*tan2lat+1320*tan4lat+720*tan6lat);
var dE = (E-E0), dE2 = dE*dE, dE3 = dE2*dE, dE4 = dE2*dE2, dE5 = dE3*dE2, dE6 = dE4*dE2, dE7 = dE5*dE2;
lat = lat - VII*dE2 + VIII*dE4 - IX*dE6;
var lon = lon0 + X*dE - XI*dE3 + XII*dE5 - XIIA*dE7;
return {
longitude: lon.toDeg(),
latitude: lat.toDeg()
};
}
Number.prototype.toRad = function() { // convert degrees to radians
return this * Math.PI / 180;
}
Number.prototype.toDeg = function() { // convert radians to degrees (signed)
return this * 180 / Math.PI;
}
Number.prototype.padLZ = function(w) {
var n = this.toString();
for (var i=0; i<w-n.length; i++) n = '0' + n;
return n;
}
答案 2 :(得分：1)
我需要相同的功能，并且javascript使得难以与DB交互。我已将您的JS转换为PHP，这在更新数据库时可能更有用 - 即：查询表，循环结果集，调用函数，更新表。
function OSGridToLatLong($E, $N) {
$a = 6377563.396;
$b = 6356256.910; // Airy 1830 major & minor semi-axes
$F0 = 0.9996012717; // NatGrid scale factor on central meridian
$lat0 = 49*M_PI/180;
$lon0 = -2*M_PI/180; // NatGrid true origin
$N0 = -100000;
$E0 = 400000; // northing & easting of true origin, metres
$e2 = 1 - ($b*$b)/($a*$a); // eccentricity squared
$n = ($a-$b)/($a+$b);
$n2 = $n*$n;
$n3 = $n*$n*$n;
$lat=$lat0;
$M=0;
do {
$lat = ($N-$N0-$M)/($a*$F0) + $lat;
$Ma = (1 + $n + (5/4)*$n2 + (5/4)*$n3) * ($lat-$lat0);
$Mb = (3*$n + 3*$n*$n + (21/8)*$n3) * sin($lat-$lat0) * cos($lat+$lat0);
$Mc = ((15/8)*$n2 + (15/8)*$n3) * sin(2*($lat-$lat0)) * cos(2*($lat+$lat0));
$Md = (35/24)*$n3 * sin(3*($lat-$lat0)) * cos(3*($lat+$lat0));
$M = $b * $F0 * ($Ma - $Mb + $Mc - $Md); // meridional arc
} while ($N-$N0-$M >= 0.00001); // ie until < 0.01mm
$cosLat = cos($lat);
$sinLat = sin($lat);
$nu = $a*$F0/sqrt(1-$e2*$sinLat*$sinLat); // transverse radius of curvature
$rho = $a*$F0*(1-$e2)/pow(1-$e2*$sinLat*$sinLat, 1.5); // meridional radius of curvature
$eta2 = $nu/$rho-1;
$tanLat = tan($lat);
$tan2lat = $tanLat*$tanLat;
$tan4lat = $tan2lat*$tan2lat;
$tan6lat = $tan4lat*$tan2lat;
$secLat = 1/$cosLat;
$nu3 = $nu*$nu*$nu;
$nu5 = $nu3*$nu*$nu;
$nu7 = $nu5*$nu*$nu;
$VII = $tanLat/(2*$rho*$nu);
$VIII = $tanLat/(24*$rho*$nu3)*(5+3*$tan2lat+$eta2-9*$tan2lat*$eta2);
$IX = $tanLat/(720*$rho*$nu5)*(61+90*$tan2lat+45*$tan4lat);
$X = $secLat/$nu;
$XI = $secLat/(6*$nu3)*($nu/$rho+2*$tan2lat);
$XII = $secLat/(120*$nu5)*(5+28*$tan2lat+24*$tan4lat);
$XIIA = $secLat/(5040*$nu7)*(61+662*$tan2lat+1320*$tan4lat+720*$tan6lat);
$dE = ($E-$E0);
$dE2 = $dE*$dE;
$dE3 = $dE2*$dE;
$dE4 = $dE2*$dE2;
$dE5 = $dE3*$dE2;
$dE6 = $dE4*$dE2;
$dE7 = $dE5*$dE2;
$lat = $lat - $VII*$dE2 + $VIII*$dE4 - $IX*$dE6;
$lon = $lon0 + $X*$dE - $XI*$dE3 + $XII*$dE5 - $XIIA*$dE7;
return array(
'longitude' => $lon * 180 / M_PI,
'latitude' => $lat * 180 / M_PI
);
}
答案 3 :(得分：0)
如果有人对非SQL解决方案感兴趣，我强烈建议您使用此http://www.howtocreate.co.uk/php/gridref.php PHP / JavaScript类。
这里要提到的一件重要事情是图书馆支持Helmert transformation。
PHP
$grutoolbox = Grid_Ref_Utils::toolbox();
$source_coords = Array(54.607720,-6.411990);
//get the ellipsoids that will be used
$Airy_1830 = $grutoolbox->get_ellipsoid('Airy_1830');
$WGS84 = $grutoolbox->get_ellipsoid('WGS84');
$Airy_1830_mod = $grutoolbox->get_ellipsoid('Airy_1830_mod');
//get the transform parameters that will be used
$UK_to_GPS = $grutoolbox->get_transformation('OSGB36_to_WGS84');
$GPS_to_Ireland = $grutoolbox->get_transformation('WGS84_to_Ireland65');
//convert to GPS coordinates
$gps_coords = $grutoolbox->Helmert_transform($source_coords,$Airy_1830,$UK_to_GPS,$WGS84);
//convert to destination coordinates
print $grutoolbox->Helmert_transform($source_coords,$WGS84,$GPS_to_Ireland,$Airy_1830_mod,$grutoolbox->HTML);
的JavaScript
var grutoolbox = gridRefUtilsToolbox();
var sourceCoords = [54.607720,-6.411990];
//get the ellipsoids that will be used
var Airy1830 = grutoolbox.getEllipsoid('Airy_1830');
var WGS84 = grutoolbox.getEllipsoid('WGS84');
var Airy1830Mod = grutoolbox.getEllipsoid('Airy_1830_mod');
//get the transform parameters that will be used
var UKToGPS = grutoolbox.getTransformation('OSGB36_to_WGS84');
var GPSToIreland = grutoolbox.getTransformation('WGS84_to_Ireland65');
//convert to GPS coordinates
var gpsCoords = grutoolbox.HelmertTransform(sourceCoords,Airy1830,UKToGPS,WGS84);
//convert to destination coordinates
element.innerHTML = grutoolbox.HelmertTransform(sourceCoords,WGS84,GPSToIreland,Airy1830Mod,grutoolbox.HTML);
答案 4 :(得分：0)
我已经在.NET中开发了一个库，可以从事务sql调用该库，将WGS84 / UTM坐标转换为纬度和经度
它的作用恰好相反，但是由于它使用CoordinateSharp，因此您可以下载代码并轻松对其进行更改，以将其从lat / long转换为wgs84。
您可以从github下载它：
https://github.com/j-v-garcia/UTM2LATITUDE
usage:
SELECT dbo.UTM2LATITUDE(723399.51,4373328.5,'S',30) AS Latitude, dbo.UTM2LONGITUDE(723399.51,4373328.5,'S',30) AS Longitude
result:
39,4805657453054 -0,402592727245112
<param name="XUTM">pos UTM X</param>
<param name="YUTM">pos UTM Y</param>
<param name="LatBand">Latitude band grid zone designation letter (see http://www.dmap.co.uk/utmworld.htm) </param>
<param name="LongBand">Longitude band grid zone designation number (see http://www.dmap.co.uk/utmworld.htm) </param>
答案 5 :(得分：-1)
我为此感到挣扎。 对我来说最好的选择是使用这个python库 https://pypi.org/project/convertbng/ 希望能帮助到你。 问任何我想帮忙的事情。