MySQL大圆距离(Haversine公式)

我有一个工作的PHP脚本,该脚本获取经度和纬度值,然后将它们输入到MySQL查询中。我只想将其制作成MySQL。这是我当前的PHP代码:

if ($distance != "Any" && $customer_zip != "") { //get the great circle distance

//get the origin zip code info

$zip_sql = "SELECT * FROM zip_code WHERE zip_code = '$customer_zip'";

$result = mysql_query($zip_sql);

$row = mysql_fetch_array($result);

$origin_lat = $row['lat'];

$origin_lon = $row['lon'];

//get the range

$lat_range = $distance/69.172;

$lon_range = abs($distance/(cos($details[0]) * 69.172));

$min_lat = number_format($origin_lat - $lat_range, "4", ".", "");

$max_lat = number_format($origin_lat + $lat_range, "4", ".", "");

$min_lon = number_format($origin_lon - $lon_range, "4", ".", "");

$max_lon = number_format($origin_lon + $lon_range, "4", ".", "");

$sql .= "lat BETWEEN '$min_lat' AND '$max_lat' AND lon BETWEEN '$min_lon' AND '$max_lon' AND ";

}

有谁知道如何完全使用MySQL?我浏览了一下互联网,但是关于它的大多数文献都令人困惑。

回答:

来自Google Code常见问题解答-使用PHP,MySQL和Google

Maps创建商店定位器:

这是一条SQL语句,它将找到距离37,-122坐标25英里范围内的最近20个位置。它根据该行的纬度/经度和目标纬度/经度计算距离,然后仅查询距离值小于25的行,按距离对整个查询进行排序,并将其限制为20个结果。要按公里而不是英里进行搜索,请将3959替换为6371。

SELECT id, ( 3959 * acos( cos( radians(37) ) * cos( radians( lat ) ) 

* cos( radians( lng ) - radians(-122) ) + sin( radians(37) ) * sin(radians(lat)) ) ) AS distance

FROM markers

HAVING distance < 25

ORDER BY distance

LIMIT 0 , 20;

以上是 MySQL大圆距离(Haversine公式) 的全部内容, 来源链接: utcz.com/qa/418202.html

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