查詢中有兩個相同的子查詢 - 如何使它們成爲1？

Question

我有表的地方

place_id | city | country_code | zipcode | lat | lon 

現在我想說明的地方A. A地點25英里範圍內的地方已place_id 1.

SELECT * 
FROM `places` 
WHERE (3959 * acos(cos(radians((SELECT lat FROM places WHERE place_id=1))) * cos(radians(lat)) * cos(radians(lon) - radians((SELECT lon FROM places WHERE place_id=1))) + sin(radians((SELECT lat FROM places WHERE place_id=1))) * sin(radians(lat)))) < 25; 

該工程確定，但有兩個同樣的子查詢

SELECT lat FROM places WHERE place_id=1 

是否可以優化該查詢不能有兩個相同的子查詢，但讓他們變成1？

Answer 1

您的三個子查詢將針對外部表中的每一行執行。

你可以做的是那些子查詢轉換爲單個JOIN這將僅適用於整個查詢執行一次，以便找到的place_id 1的緯度和經度：

SELECT a.* 
FROM places a 
JOIN (SELECT lat, lon FROM places WHERE place_id = 1) b ON 
     (3959 * acos(cos(radians(b.lat)) * cos(radians(a.lat)) * cos(radians(a.lon) - radians(b.lon)) + sin(radians(b.lat)) * sin(radians(a.lat)))) < 25; 

Answer 2

從子查詢派生一個新表，並與原有的表連接，像

SELECT * 
FROM places p, (SELECT lat AS new_lat FROM places WHERE place_id = 1) l 
WHERE blah blah 

現在你可以替換列「new_lat」子查詢。

Answer 3

可以使用可變@var_lon緩存

SELECT * 
FROM `places` 
WHERE (3959 * acos(cos(radians((SELECT lat FROM places WHERE place_id=1))) * cos(radians(lat)) * cos(radians(lon) - radians((SELECT @var_lon := lon FROM places WHERE place_id=1))) + sin(radians(@var_lon)) * sin(radians(lat)))) < 25;