轉換查詢處理關鍵值列

Question

我需要幫助來更新這個查詢：

$query = sprintf(
    "SELECT 
     lat, 
     lng, 
     ( 
      3959 * acos(cos(radians('%s')) * 
         cos(radians(lat)) * 
         cos(radians(lng) - radians('%s')) + 
         sin(radians('%s')) * 
         sin(radians(lat)) 
         ) 
     ) AS distance 
    FROM 
     markers 
    HAVING 
     distance < '%s' 
    ORDER BY 
     distance LIMIT 0 , 20" 

與表結構，如：

lat  | lng 
23.0324234 | -103.0324234 

要查詢，我可以從一個搶shipping_lat和shipping_long鍵值列結構如下：

meta_key  | meta_value 
shipping_lat | 23.0324234 
shipping_long | -103.0324234 

Answer 1

使用以下內容作爲子查詢你加入你的主要查詢（或類似）？

SELECT t1.meta_value AS lat, t2.meta_value AS lng 
FROM meta_table t1 JOIN meta_table t2 
WHERE t1.meta_key = 'shipping_lat' 
    AND t2.meta_key = 'shipping_long' 

即：

SELECT 
    lat, 
    lng, 
    ( 
     3959 * acos(cos(radians('%s')) * 
        cos(radians(lat)) * 
        cos(radians(lng) - radians('%s')) + 
        sin(radians('%s')) * 
        sin(radians(lat)) 
        ) 
    ) AS distance 
FROM 
    markers JOIN (
     SELECT t1.meta_value AS lat, t2.meta_value AS lng 
     FROM meta_table t1 JOIN meta_table t2 
     WHERE t1.meta_key = 'shipping_lat' 
     AND t2.meta_key = 'shipping_long' 
    ) AS t 
HAVING 
    distance < '%s' 
ORDER BY 
    distance LIMIT 0 , 20 

人們可以通過直接接合meta_table到markers兩次（在相關meta_key S）然後引用避免子查詢中的meta_value小號，而不是直接lat和lng ，但我認爲上述內容更具可讀性且更易於遵循。但是，爲了完整性：

SELECT 
    lat.meta_value, 
    lng.meta_value, 
    ( 
     3959 * acos(cos(radians('%s')) * 
        cos(radians(lat.meta_value)) * 
        cos(radians(lng.meta_value) - radians('%s')) + 
        sin(radians('%s')) * 
        sin(radians(lat.meta_value)) 
        ) 
    ) AS distance 
FROM 
    markers 
     JOIN meta_table lat ON lat.meta_key = 'shipping_lat' 
     JOIN meta_table lng ON lng.meta_key = 'shipping_long' 
HAVING 
    distance < '%s' 
ORDER BY 
    distance LIMIT 0 , 20