Php 基于多个位置（纬度、经度）的距离计算，包括起点/终点

我想计算多个位置（环路）之间行驶的总距离，包括距离（起点（车库）-第一个位置停车点）和（最后一个位置终点-终点（车库））。 例子： （车库+D1）+（D1+D2）+（D2+E1）+（E1+E2）+E2+车库）

我对正确的循环有问题。以下是我的简化代码：

<?
$driver = 5;

     $result2 = mysql_query("SELECT * FROM test WHERE id='$driver' LIMIT 1") or die(mysql_error());
     while($row2 = mysql_fetch_array( $result2 )) {
         $lon=$row2['lon'];
         $lat=$row2['lat'];
    echo "$lon, $lat";
     }

   $result = mysql_query("SELECT * FROM test1 WHERE driver='$driver'") or die(mysql_error());  
    while($row = mysql_fetch_array( $result )) {

         $lon1=$row['lon1'];
         $lat1=$row['lat1'];
         $lon2=$row['lon2'];
         $lat2=$row['lat2'];

        //////////  distance between driver address and starting address    
        $distancecalc = (3958*3.1415926*sqrt(($lat-$lat1)*($lat-$lat1) + cos($lat/57.29578)*cos($lat1/57.29578)*($lon-$lon1)*($lon-$lon1))/180);
        //////////  distance between statring address and finishing address  - multiple adsresses
        $distancecalc1 = $distancecalc1 + (3958*3.1415926*sqrt(($lat2-$lat1)*($lat2-$lat1) + cos($lat2/57.29578)*cos($lat1/57.29578)*($lon2-$lon1)*($lon2-$lon1))/180);
        //////////  distance between finishing address and driver address
        $distancecalc2 = (3958*3.1415926*sqrt(($lat2-$lat)*($lat2-$lat) + cos($lat2/57.29578)*cos($lat/57.29578)*($lon2-$lon)*($lon2-$lon))/180);

        $distancetotal = $distancecalc + $distancecalc1 +$distancecalc2;

        echo "$distancecalc<br>
        $distancecalc1<br>
        $distancecalc2<br>";
    }
    echo "$distancetotal";
  ?>

---

以下是您需要的函数：

function distance ($lat1, $lon1, $lat2, $lon2) {
    return (3958*3.1415926*sqrt(($lat2-$lat1)*($lat2-$lat1) + cos($lat2/57.29578)*cos($lat1/57.29578)*($lon2-$lon1)*($lon2-$lon1))/180);
}

函数体与您在线使用的公式完全相同，因此我不明白您为什么需要帮助。
数字6367-是地球的半径，单位为公里

    DELIMITER $$
DROP FUNCTION IF EXISTS geodist $$
CREATE FUNCTION geodist (
  src_lat DECIMAL(9,6), src_lon DECIMAL(9,6),
  dst_lat DECIMAL(9,6), dst_lon DECIMAL(9,6)
) RETURNS DECIMAL(6,2) DETERMINISTIC
BEGIN
 SET @dist := 6367 * 2 * ASIN(SQRT(
      POWER(SIN((src_lat - ABS(dst_lat)) * PI()/180 / 2), 2) +
      COS(src_lat * PI()/180) *
      COS(ABS(dst_lat) * PI()/180) *
      POWER(SIN((src_lon - dst_lon) * PI()/180 / 2), 2)
    ));
 RETURN @dist;
END $$
DELIMITER ;

嗯。我在mac_gyver（php怪胎）的帮助下解决了这个问题。所有的计算都按我的意愿进行。我的代码如下：

<?
include "connectdb.php";
$driver = 5;
$datestamp = '2013/05/07';
$result2 = mysql_query("SELECT * FROM drivers WHERE id='$driver' LIMIT 1") or die(mysql_error());
while($row2 = mysql_fetch_array( $result2 )) {
         $lon=$row2['lon'];
         $lat=$row2['lat'];
     }

$result = mysql_query("SELECT * FROM quotedb WHERE moveday='$datestamp' AND driver='$driver' AND cleared='Not Cleared' AND status='Done' ORDER BY moveday, timeday") or die(mysql_error());  
    $distance = 0; // accumulate the distance
    $first_pass = true;  // flag to detect the first row inside the loop
while($row = mysql_fetch_assoc( $result )) {
         $lon2a=$lon2;
         $lat2a=$lat2;
         $lon1=$row['lon1'];
         $lat1=$row['lat1'];
         $lon2=$row['lon2'];
         $lat2=$row['lat2'];
         // calculate the distance from the Garage to the first point of the first row
    if($first_pass){
        $distance += (3958*3.1415926*sqrt(($lat-$lat1)*($lat-$lat1) + cos($lat/57.29578)*cos($lat1/57.29578)*($lon-$lon1)*($lon-$lon1))/180);
        $first_pass = false;
    }
    // calculate the distance for each row (segment) in the route
    $distance += (3958*3.1415926*sqrt(($lat2-$lat1)*($lat2-$lat1) + cos($lat2/57.29578)*cos($lat1/57.29578)*($lon2-$lon1)*($lon2-$lon1))/180);
    if ( $lon2a == "" or $lat2a =="" ) {            
        } else {
        // calculate the distance from the second point of the first row to the first point of the next row 
    $distance += (3958*3.1415926*sqrt(($lat2a-$lat1)*($lat2a-$lat1) + cos($lat2a/57.29578)*cos($lat1/57.29578)*($lon2a-$lon1)*($lon2a-$lon1))/180);
        }
    }
    // calculate the distance from the second point of the last row to the Garage
    $distance += (3958*3.1415926*sqrt(($lat2-$lat)*($lat2-$lat) + cos($lat2/57.29578)*cos($lat/57.29578)*($lon2-$lon)*($lon2-$lon))/180);
echo "$distance<br>
";
  ?>

改进版本：

<?
include "connectdb.php";
$driver = 5;
$datestamp = '2013/05/07';
$result2 = mysql_query("SELECT * FROM drivers WHERE id='$driver' LIMIT 1") or die(mysql_error());
while($row2 = mysql_fetch_array( $result2 )) {
         $garage_lon=$row2['lon'];
         $garage_lat=$row2['lat'];
     }

$result = mysql_query("SELECT * FROM quotedb WHERE moveday='$datestamp' AND driver='$driver' AND cleared='Not Cleared' AND status='Done' ORDER BY moveday, timeday") or die(mysql_error());  

    function calculate_distance($lon1, $lat1, $lon2, $lat2) {
    return (3958 * 3.1415926 * sqrt(($lat2 - $lat1) * ($lat2 - $lat1) + cos($lat2 / 57.29578) * cos($lat1 / 57.29578) * ($lon2 - $lon1) * ($lon2 - $lon1)) / 180);}  

$previous_lon = $garage_lon;
$previous_lat = $garage_lat;
$distance = 0; // accumulate the distance

while($row = mysql_fetch_assoc( $result ))
{
    $lon1 = $row['lon1'];
    $lat1 = $row['lat1'];
    $lon2 = $row['lon2'];
    $lat2 = $row['lat2'];
    if ( $previous_lon && $previous_lat )
    {
        // calculate the distance from the second point of the first row to the first point of the next row 
        $distance += calculate_distance($lon1, $lat1, $previous_lon, $previous_lat);
    }
    // calculate the distance for each row (segment) in the route
    $distance += calculate_distance($lon1, $lat1, $lon2, $lat2);
    $previous_lon = $lon2;
    $previous_lat = $lat2;
}
// calculate the distance from the second point of the last row to the Garage
$distance += calculate_distance($garage_lon, $garage_lat, $lon2, $lat2);  
    $distance = round($distance,0);
echo "$distance<br>
";
  ?>

您可以编写一个函数
距离（$lat1，$lon1，$lat2，$lon2）
。您还可以通过
SELECT
语句中的计算，在两个表之间的一个SQL联接中完成整个过程。您可以在SQL中创建一个用户定义的
DISTANCE（）
函数来缩短它。这已经是一个很好的起点。。但是把它带到生活中是另一个故事，如果有一些代码示例会很好。。。我会调查的你的总体计算似乎不正确。您每次通过循环累积
$distancecalc1
和
$distancecalc2
，但
$distancecalc
没有累积。您不需要每次分配
$distancetotal
，因为您没有在循环中使用它。如果其他三个变量是累积的，你可以在最后把它们加在一起。正如我在会议上提到的，需要计算的是起点（具有恒定lat，lon的车库-也是终点）到第一个目的地（起点）之间的距离。同时有多个地点可供选择。然后司机必须从最后一个目的地回到车库。举两个例子吧。（车库+D1）+（D1+D2）+（D2+E1）+（E1+E2）+E2+车库）。我必须循环它。我还是不知道怎么办？？