的Sql子查询需要添加两个查询

Question

我有一个完美的最后一天运行谢赫Farooque回答查询 Link of that question

现在我还有一个问题，我需要过滤那些foodjoint_id详细哪都一样cuisine_id下。 用户要提交纬度长，cuisine_id我需要过滤那些FoodJoint

正如我告诉你，我已经通过，现在正在运行，我需要补充的美食过滤Lat Long网寻找食物联合。

这是运行查询是

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 foodjoint_id=provider_food_joints.foodjoint_id) AS customer_rating 
FROM provider_food_joints 
HAVING distance < '3' ORDER BY distance 

，我已经添加了它：

SELECT foodjoint_id FROM menu_item WHERE cuisine_id=''.$userGivenCuisineId.'' 

我很抱歉地说，这个问题仍然没有解决

Answer 1

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('".$userLatitude."')) * 
    sin(radians(foodjoint_latitude)))) AS distance, 
(select AVG(customer_ratings) from customer_review where 
foodjoint_id=provider_food_joints.foodjoint_id) as customer_rating 
FROM provider_food_joints 
where foodjoint_id in 
(SELECT foodjoint_id FROM menu_item WHERE cuisine_id='".$userGivenCuisineId."') 
HAVING distance < '3' ORDER BY distance