Sql Sub query need to add two query

Question

I have a query which is running perfectly last day answered by Shaikh Farooque Link of that question

Now I have another problem I need to filter those foodjoint_id detail which are under the same cuisine_id. User going to submit lat long and cuisine_id I need to filter those FoodJoint

As I told you I already search Food Joint by the Lat Long which is running now I need to add cuisine filter.

The query which is running is

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

and I have added it with :

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

I am sorry to say that the problem is still unsolved

Can you more clearify your question

Muhammad Raheel
– Muhammad Raheel

2012-05-07 09:52:53 +00:00
Commented May 7, 2012 at 9:52 — Muhammad Raheel
– Muhammad Raheel, Commented May 7, 2012 at 9:52
Kindly try the following answer.

Shaikh Farooque
– Shaikh Farooque

2012-05-07 10:04:08 +00:00
Commented May 7, 2012 at 10:04 — Shaikh Farooque
– Shaikh Farooque, Commented May 7, 2012 at 10:04

Shaikh Farooque · Accepted Answer · 2012-05-07 09:59:47Z

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

answered May 7, 2012 at 9:59

Shaikh Farooque

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