如何改進SQL查詢

查看下面的SQL查詢，它似乎運行良好。它允許我通過郵編或城鎮搜索記錄。如何改進SQL查詢

用戶輸入郵編或城市從文本框中。

從uk_postcodes表中有郵編列表（BB1，BB2，BB3等）。

如果用戶輸入BB2或BB2 XXX，它將顯示一個在BB2區域發貨的店鋪記錄列表。（我用PHP來獲得郵政編碼的第一部分，它進入SQL查詢之前）

如果Blackburn用戶類型，那麼它會顯示與默認送達地區的商店名單（D.shop_id = S.shop_id AND D.postcode_id = S.postcode_id）

$SQL = "(SELECT DISTINCT S.*, D.delivery_cost, D.postcode AS DeliveryAreaPostcode FROM shops AS S 
       JOIN shop_delivery_area AS D ON D.shop_id = S.shop_id 
       JOIN uk_postcodes AS P ON P.postcode_id = D.postcode_id 
        WHERE P.postcode = '$search' ORDER BY D.postcode) 
       UNION 
    (SELECT DISTINCT S.* , D.delivery_cost, D.postcode AS DeliveryAreaPostcode FROM shops AS S 
       JOIN shop_delivery_area AS D ON D.shop_id = S.shop_id AND D.postcode_id = S.postcode_id 
       JOIN uk_postcodes AS P ON P.postcode_id = D.postcode_id 
        WHERE S.town = '$search')";

是這個SQL查詢很好？或者什麼可以改進或使其變小？

來源

2011-11-16 I' ll-Be-Back

使用UNION ALL是因爲不需要完成獨特的檢查，所以比使用UNION更快。在這種情況下，除非有人住在同樣也是郵編的小鎮，否則你不需要那些開銷！

來源

2011-11-16 16:12:42

您可以使用一個控制變量，一些

$SQL = "SELECT DISTINCT S.*, D.delivery_cost, D.postcode AS DeliveryAreaPostcode 
     FROM shops AS S 
      JOIN shop_delivery_area AS D ON D.shop_id = S.shop_id 
      JOIN uk_postcodes AS P ON P.postcode_id = D.postcode_id 
     WHERE (P.postcode = '{0}' and '{1}' = 'postcode') 
     or (S.town = '{0}' and D.postcode_id = S.postcode_id and '{1}' = 'town') 
      ORDER BY D.postcode".format($search, $myflagcontrol);

$ myflagcontrol是如何確定的當前用戶選擇

格式是這裏的虛擬函數代碼

// Add format extension to string class 
String.prototype.format= function(){  
var str = this; 
for (var i = 0, l = arguments.length; i < l; i++) 
    str = str.replace("{" + i + "}", arguments[i]); 
return str; 
}

來源

2011-11-16 16:07:28

如何改進SQL查詢

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