我正在制作一个涉及谷歌地图的 PHP 应用程序。地图只接受lat&lng对，而我要显示的数据仅带有 UTM 样式坐标。是否有一个开源 PHP 函数可以从一个转换为另一个？
这样的事情会很棒：
$UTM_ZONE = '32';
$UTMX = '60329834,34';
$UTMY = '67382984,9';
$latlng = convert($UTM_ZONE, $UTMX, $UTMY);
// $latlng = now looks like
// array('lat' => '59.4472917501', 'lng' => '5.3928572425')
这是PHP中的代码，非常感谢，很好用！
<?php
function ToLL($north, $east, $utmZone)
{
// This is the lambda knot value in the reference
$LngOrigin = Deg2Rad($utmZone * 6 - 183);
// The following set of class constants define characteristics of the
// ellipsoid, as defined my the WGS84 datum. These values need to be
// changed if a different dataum is used.
$FalseNorth = 0; // South or North?
//if (lat < 0.) FalseNorth = 10000000. // South or North?
//else FalseNorth = 0.
$Ecc = 0.081819190842622; // Eccentricity
$EccSq = $Ecc * $Ecc;
$Ecc2Sq = $EccSq / (1. - $EccSq);
$Ecc2 = sqrt($Ecc2Sq); // Secondary eccentricity
$E1 = ( 1 - sqrt(1-$EccSq) ) / ( 1 + sqrt(1-$EccSq) );
$E12 = $E1 * $E1;
$E13 = $E12 * $E1;
$E14 = $E13 * $E1;
$SemiMajor = 6378137.0; // Ellipsoidal semi-major axis (Meters)
$FalseEast = 500000.0; // UTM East bias (Meters)
$ScaleFactor = 0.9996; // Scale at natural origin
// Calculate the Cassini projection parameters
$M1 = ($north - $FalseNorth) / $ScaleFactor;
$Mu1 = $M1 / ( $SemiMajor * (1 - $EccSq/4.0 - 3.0*$EccSq*$EccSq/64.0 - 5.0*$EccSq*$EccSq*$EccSq/256.0) );
$Phi1 = $Mu1 + (3.0*$E1/2.0 - 27.0*$E13/32.0) * sin(2.0*$Mu1);
+ (21.0*$E12/16.0 - 55.0*$E14/32.0) * sin(4.0*$Mu1);
+ (151.0*$E13/96.0) * sin(6.0*$Mu1);
+ (1097.0*$E14/512.0) * sin(8.0*$Mu1);
$sin2phi1 = sin($Phi1) * sin($Phi1);
$Rho1 = ($SemiMajor * (1.0-$EccSq) ) / pow(1.0-$EccSq*$sin2phi1,1.5);
$Nu1 = $SemiMajor / sqrt(1.0-$EccSq*$sin2phi1);
// Compute parameters as defined in the POSC specification. T, C and D
$T1 = tan($Phi1) * tan($Phi1);
$T12 = $T1 * $T1;
$C1 = $Ecc2Sq * cos($Phi1) * cos($Phi1);
$C12 = $C1 * $C1;
$D = ($east - $FalseEast) / ($ScaleFactor * $Nu1);
$D2 = $D * $D;
$D3 = $D2 * $D;
$D4 = $D3 * $D;
$D5 = $D4 * $D;
$D6 = $D5 * $D;
// Compute the Latitude and Longitude and convert to degrees
$lat = $Phi1 - $Nu1*tan($Phi1)/$Rho1 * ( $D2/2.0 - (5.0 + 3.0*$T1 + 10.0*$C1 - 4.0*$C12 - 9.0*$Ecc2Sq)*$D4/24.0 + (61.0 + 90.0*$T1 + 298.0*$C1 + 45.0*$T12 - 252.0*$Ecc2Sq - 3.0*$C12)*$D6/720.0 );
$lat = Rad2Deg($lat);
$lon = $LngOrigin + ($D - (1.0 + 2.0*$T1 + $C1)*$D3/6.0 + (5.0 - 2.0*$C1 + 28.0*$T1 - 3.0*$C12 + 8.0*$Ecc2Sq + 24.0*$T12)*$D5/120.0) / cos($Phi1);
$lon = Rad2Deg($lon);
// Create a object to store the calculated Latitude and Longitude values
$PC_LatLon['lat'] = $lat;
$PC_LatLon['lon'] = $lon;
// Returns a PC_LatLon object
return $PC_LatLon;
}
?>
我发现了一种可以完成这项工作的肮脏课程。脏我的意思是函数名称很奇怪，代码的格式也不是很漂亮，但它确实可以完成这项工作。
如果我找到更好的课程，我一定会更新这个答案
我知道现在回答这个问题已经很晚了，但是由于我无法使用上述任何代码，所以我编写了自己的版本，它实际上非常易于使用。这是地址： https ://github.com/maroofi/coordinates 将 UTM 转换为 LatLong：
utm2ll(729286.9550018794,4021544.8279992654,40,true);
输出：
{"success":true,"attr":{"lat":36.311665575271,"lon":59.553858137274}}
要将 LatLong 转换为 UTM：
ll2utm(36.311665575277935,59.55385813725379);
输出：
{"success":true,"attr":{"x":729286.95500188,"y":4021544.8279993,"zone":40,"aboveEquator":true}}
希望能帮助到你。
您要求使用 PHP，但这里是 javascript。只需输入一些'$'，你应该会很好;)。这将返回 WGS84 中的纬度/经度。无担保，使用风险自负。
////////////////////////////////////////////////////////////////////////////////////////////
//
// ToLL - function to compute Latitude and Longitude given UTM Northing and Easting in meters
//
// Description:
// This function converts input north and east coordinates (meters)
// to the corresponding WGS84 Lat/Lon values relative to the defined
// UTM zone.
//
// Parameters:
// north - (i) Northing (meters)
// east - (i) Easting (meters)
// utmZone - (i) UTM Zone of the North and East parameters
// lat - (o) Latitude in degrees
// lon - (o) Longitude in degrees
//
function ToLL(north,east,utmZone)
{
// This is the lambda knot value in the reference
var LngOrigin = DegToRad(utmZone * 6 - 183)
// The following set of class constants define characteristics of the
// ellipsoid, as defined my the WGS84 datum. These values need to be
// changed if a different dataum is used.
var FalseNorth = 0. // South or North?
//if (lat < 0.) FalseNorth = 10000000. // South or North?
//else FalseNorth = 0.
var Ecc = 0.081819190842622 // Eccentricity
var EccSq = Ecc * Ecc
var Ecc2Sq = EccSq / (1. - EccSq)
var Ecc2 = Math.sqrt(Ecc2Sq) // Secondary eccentricity
var E1 = ( 1 - Math.sqrt(1-EccSq) ) / ( 1 + Math.sqrt(1-EccSq) )
var E12 = E1 * E1
var E13 = E12 * E1
var E14 = E13 * E1
var SemiMajor = 6378137.0 // Ellipsoidal semi-major axis (Meters)
var FalseEast = 500000.0 // UTM East bias (Meters)
var ScaleFactor = 0.9996 // Scale at natural origin
// Calculate the Cassini projection parameters
var M1 = (north - FalseNorth) / ScaleFactor
var Mu1 = M1 / ( SemiMajor * (1 - EccSq/4.0 - 3.0*EccSq*EccSq/64.0 -
5.0*EccSq*EccSq*EccSq/256.0) )
var Phi1 = Mu1 + (3.0*E1/2.0 - 27.0*E13/32.0) * Math.sin(2.0*Mu1)
+ (21.0*E12/16.0 - 55.0*E14/32.0) * Math.sin(4.0*Mu1)
+ (151.0*E13/96.0) * Math.sin(6.0*Mu1)
+ (1097.0*E14/512.0) * Math.sin(8.0*Mu1)
var sin2phi1 = Math.sin(Phi1) * Math.sin(Phi1)
var Rho1 = (SemiMajor * (1.0-EccSq) ) / Math.pow(1.0-EccSq*sin2phi1,1.5)
var Nu1 = SemiMajor / Math.sqrt(1.0-EccSq*sin2phi1)
// Compute parameters as defined in the POSC specification. T, C and D
var T1 = Math.tan(Phi1) * Math.tan(Phi1)
var T12 = T1 * T1
var C1 = Ecc2Sq * Math.cos(Phi1) * Math.cos(Phi1)
var C12 = C1 * C1
var D = (east - FalseEast) / (ScaleFactor * Nu1)
var D2 = D * D
var D3 = D2 * D
var D4 = D3 * D
var D5 = D4 * D
var D6 = D5 * D
// Compute the Latitude and Longitude and convert to degrees
var lat = Phi1 - Nu1*Math.tan(Phi1)/Rho1 *
( D2/2.0 - (5.0 + 3.0*T1 + 10.0*C1 - 4.0*C12 - 9.0*Ecc2Sq)*D4/24.0
+ (61.0 + 90.0*T1 + 298.0*C1 + 45.0*T12 - 252.0*Ecc2Sq - 3.0*C12)*D6/720.0 )
lat = RadToDeg(lat)
var lon = LngOrigin +
( D - (1.0 + 2.0*T1 + C1)*D3/6.0
+ (5.0 - 2.0*C1 + 28.0*T1 - 3.0*C12 + 8.0*Ecc2Sq + 24.0*T12)*D5/120.0) / Math.cos(Phi1)
lon = RadToDeg(lon)
// Create a object to store the calculated Latitude and Longitude values
var sendLatLon = new PC_LatLon(lat,lon)
// Returns a PC_LatLon object
return sendLatLon
}
<?php
function LatLonPointUTMtoLL($f, $f1, $j = 32) {
$d = 0.99960000000000004;
$d1 = 6378137;
$d2 = 0.0066943799999999998;
$d4 = (1 - sqrt(1 - $d2)) / (1 + sqrt(1 - $d2));
$d15 = $f1 - 500000;
$d16 = $f;
$d11 = (($j - 1) * 6 - 180) + 3;
$d3 = $d2 / (1 - $d2);
$d10 = $d16 / $d;
$d12 = $d10 / ($d1 * (1 - $d2 / 4 - (3 * $d2 * $d2) / 64 - (5 * pow($d2, 3)) / 256));
$d14 = $d12 + ((3 * $d4) / 2 - (27 * pow($d4, 3)) / 32) * sin(2 * $d12) + ((21 * $d4 * $d4) / 16 - (55 * pow($d4, 4)) / 32) * sin(4 * $d12) + ((151 * pow($d4, 3)) / 96) * sin(6 * $d12);
$d5 = $d1 / sqrt(1 - $d2 * sin($d14) * sin($d14));
$d6 = tan($d14) * tan($d14);
$d7 = $d3 * cos($d14) * cos($d14);
$d8 = ($d1 * (1 - $d2)) / pow(1 - $d2 * sin($d14) * sin($d14) , 1.5);
$d9 = $d15 / ($d5 * $d);
$d17 = $d14 - (($d5 * tan($d14)) / $d8) * ((($d9 * $d9) / 2 - (((5 + 3 * $d6 + 10 * $d7) - 4 * $d7 * $d7 - 9 * $d3) * pow($d9, 4)) / 24) + (((61 + 90 * $d6 + 298 * $d7 + 45 * $d6 * $d6) - 252 * $d3 - 3 * $d7 * $d7) * pow($d9, 6)) / 720);
$d17 = rad2deg($d17);
$d18 = (($d9 - ((1 + 2 * $d6 + $d7) * pow($d9, 3)) / 6) + (((((5 - 2 * $d7) + 28 * $d6) - 3 * $d7 * $d7) + 8 * $d3 + 24 * $d6 * $d6) * pow($d9, 5)) / 120) / cos($d14);
$d18 = $d11 + rad2deg($d18);
return array(
'lat' => $d17,
'lng' => $d18
);
}
?>
为了完整起见，在 Composer 中打包的人们的另一个选择是https://packagist.org/packages/php-coord/php-coord
它的文档记录不是很好，但看起来通过结合 Easting、Northing 和 Zone 可以返回纬度和经度。例如：
$easting = 505716.941;
$northing = 6961780.872;
$zone = 56;
$UTMRef = new PHPCoord\UTMRef($easting, $northing, NULL, $zone, $zone);
$LatLng = $UTMRef->toLatLng();
print "Lat/Lng:" . $LatLng->getLat() . ", " . $LatLng->getLng() . "\n";
请注意，它似乎不如上面提到的 gPoint 准确。
发布的脚本的结果有点出乎我的意料，但我在http://www.uwgb.edu/dutchs/usefuldata/ConvertUTMNoOZ.HTM上找到了一个工具，它给了我预期的结果。我在这里做了一个 PHP 版本：https ://gist.github.com/datagutten/8083549
另一种可能性是包 proj4php ( https://github.com/proj4php/proj4php )。非常强大，它允许在许多不同的坐标系之间进行转换。