如何編寫此SQL子查詢？

Question

我有此查詢運行良好

從此查詢中，我從我的位置中選擇所有餐廳3 KM，這是我的第一張表。

SELECT foodjoint_id,foodjoint_name,open_hours,cont_no,address_line,city ( 3959 * acos( cos( radians('".$userLatitude."') ) * cos( radians( foodjoint_latitude) ) * cos( radians( foodjoint_longitude) - radians('".$userLongitude."') ) + sin( radians('".$userLatitude."') ) * sin( radians( foodjoint_latitude) ) ) ) AS distance
FROM provider_food_joints
HAVING distance < '3' ORDER BY distance LIMIT 0 , 20

但是我需要從3Km中的食品接頭中選擇AVG等級。

該查詢也運行完美：

select AVG(customer_ratings) from customer_review where foodjoint_id=".$foodjoint_id 

但是我需要添加這兩個查詢，通過它們我可以選擇所有這些食品接頭及其等級AVG。

Answer 1

只需放置子查詢，您將得到結果：

`SELECT foodjoint_id,foodjoint_name,open_hours,cont_no,address_line,city ( 3959 * acos( cos( radians('".$userLatitude."') ) * cos( radians( foodjoint_latitude) ) * cos( radians( foodjoint_longitude) - radians('".$userLongitude."') ) + sin( radians('".$userLatitude."') ) * sin( radians( foodjoint_latitude) ) ) ) AS distance,

(select AVG(customer_ratings) from customer_review where customer_review.foodjoint_id=provider_food_joints.foodjoint_id) as Customer_Reviews

 FROM provider_food_joints 

HAVING distance < '3' ORDER BY distance LIMIT 0 , 20`