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Home/ Questions/Q 7879603
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Editorial Team
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Editorial Team
Asked: June 3, 20262026-06-03T03:56:04+00:00 2026-06-03T03:56:04+00:00

I have this query which is running perfectly From this query I am selecting

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I have this query which is running perfectly

From this query I am selecting all restaurant 3 KM from my location this is my 1st table.

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

But I need to select the AVG rating from those food joint which are with in this 3Km.

The query is also running perfectly:

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

but I need to add this two query through which I can select all those food joint and their rating AVG.

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  1. Editorial Team
    Editorial Team
    2026-06-03T03:56:05+00:00Added an answer on June 3, 2026 at 3:56 am

    Just place the subquery and you will get your result:

    `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`
    
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