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。
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;