Haversine公式中的MySQL語法錯誤

Question

我在下面的代碼中收到令人沮喪的MySql語法錯誤。實際的錯誤是：

SQLSTATE[42000]: Syntax error or access violation: 1064 You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near ' JOIN storetable ON pricelist.storecode = storetable.storecode JOIN itemlist ON' at line 9"

附加信息：基於谷歌地圖的PHP/MySQL例如這段代碼：http://code.google.com/apis/maps/articles/phpsqlsearch.html

我能得到這個代碼工作正常之前，我嘗試添加的JOIN語句。

$query = sprintf("SELECT storetable.storeaddress, 
         storetable.storename, 
         storetable.lat, 
         storetable.lng, 
         (3959 * acos(cos(radians('%s')) * 
         cos(radians(storetable.lat)) *        cos(radians(storetable.lng) - radians('%s')) + 
         sin(radians('%s')) * sin(radians(storetable.lat)))) AS distance 
        FROM pricelist 
        HAVING distance < 25 
       ORDER BY distance 
        LIMIT 0 , 20, 
        JOIN storetable ON pricelist.storecode = storetable.storecode 
        JOIN itemlist ON pricelist.upccode = itemlist.upccode", 
       mysql_real_escape_string($latitude), 
       mysql_real_escape_string($longitude), 
       mysql_real_escape_string($latitude)); 

Answer 1

加入是在錯誤的地方 - 他們來了之後FROM子句中，前WHERE子句 - 使用：

$query = sprintf("SELECT storetable.storeaddress, 
         storetable.storename, 
         storetable.lat, 
         storetable.lng, 
         (3959 * acos(cos(radians('%s')) * 
         cos(radians(lat)) * cos(radians(lng) - radians('%s')) + 
         sin(radians('%s')) * sin(radians(lat)))) AS distance 
        FROM pricelist 
        JOIN storetable ON pricelist.storecode = storetable.storecode 
        JOIN itemlist ON pricelist.upccode = itemlist.upccode 
        HAVING distance < 25 
       ORDER BY distance 
        LIMIT 0, 20", 
       mysql_real_escape_string($latitude), 
       mysql_real_escape_string($longitude), 
       mysql_real_escape_string($latitude));