Question

我试图找到两点之间的距离，但每个纬度和经度我从数据库中获取基于的邮编

我试图将查询作为变量，但我不知道如何使这个想法最终起作用。

任务：我有10个商店，每个地址和邮政编码，然后我有1000个用户地址和邮政编码，我需要为每个用户找到3个最近的商店，理想情况下提取为电子表格，这样每个用户将是一个额外的3列将关闭商店。

我下载完整的邮政编码数据库，并尝试将用户/商店邮政编码与邮政编码数据库连接起来，获取每个用户/商店的经纬度，但是从这里我现在确定我需要做什么以及如何处理它进一步。

P.S。请不要宣传谷歌地图API，我知道，但对我来说太贵了。

任何有关此任务的建议或任何帮助都会让我们感激不尽。

谢谢。

这是我的PHP代码：

$sqlTescLat = "SELECT lat FROM postcodelatlng, tesco WHERE tesco.postcode = postcodelatlng.postcode";
$sqlTescLng = "SELECT lng FROM postcodelatlng, tesco WHERE tesco.postcode = postcodelatlng.postcode";

$sqlShopLat = "SELECT lat FROM postcodelatlng, shops WHERE shops.postcode = postcodelatlng.postcode";
$sqlShopLng = "SELECT lng FROM postcodelatlng, shops WHERE shops.postcode = postcodelatlng.postcode";

$tescoLat = mysqli_query($con,$sqlTescLat);
$tescoLng = mysqli_query($con,$sqlTescLng);
$shopsLat = mysqli_query($con,$sqlShopLat);
$shopsLng = mysqli_query($con,$sqlShopLng);

function distance($lat1, $lon1, $lat2, $lon2, $unit) {
    $theta = $lon1 - $lon2;
    $dist = sin(deg2rad($lat1)) * sin(deg2rad($lat2)) +  cos(deg2rad($lat1)) * cos(deg2rad($lat2)) * cos(deg2rad($theta));
    $dist = acos($dist);
    $dist = rad2deg($dist);
    $miles = $dist * 60 * 1.1515;
    $unit = strtoupper($unit);

    if ($unit == "K") {
        return ($miles * 1.609344);
    } else if ($unit == "N") {
        return ($miles * 0.8684);
    } else {
        return $miles;
    }
}

echo distance($tescoLat, $tescoLng, $shopsLat, $shopsLng, "M") . " Miles<br>";

这就是数据库的样子：

shops
ID  NAME    ADDRESS             POSTCODE
1   Home1   63 Simmonds Close   BH15 3EB


tesco
ID  NAME            ADDRESS                             POSTCODE
1   Tesco Poole     Waterloo Rd, Poole BH17 7EJ, UK     BH17 7EJ


postcodelatlng(this is big post codes database) as e.g.:
ID  POSTCODE    lat                     lng
1   AB10 1XG    57.144165160000000      -2.114847768000000

更新

$sqlDestLat = "SELECT lat FROM postcodelatlng, tesco WHERE tesco.postcode = postcodelatlng.postcode";
$sqlDestLng = "SELECT lng FROM postcodelatlng, tesco WHERE tesco.postcode = postcodelatlng.postcode";
$sqlOrigLat = "SELECT lat FROM postcodelatlng, shops WHERE shops.postcode = postcodelatlng.postcode";
$sqlOrigLng = "SELECT lng FROM postcodelatlng, shops WHERE shops.postcode = postcodelatlng.postcode";


$sqlDISTANCE="SELECT *, 3956 * 2 * ASIN(SQRT( POWER(SIN(('$sqlOrigLat' - abs('$sqlDestLat')) * pi()/180 / 2),2) + COS('$sqlOrigLat' * pi()/180 ) * COS(abs('$sqlDestLat') *  pi()/180) * POWER(SIN(('$sqlOrigLng' – '$sqlDestLng') * pi()/180 / 2), 2) )) as distance FROM tesco, shops, postcodelatlng having distance < 20 BY distance limit 10;"

MySQLi使用经度/纬度将查询用作查找距离的变量

0 个答案: