human
Schema:
business (bid, business_id, name, full_address, city, latitude, longitude, review_count, is_open, rating, state)
category (id, business_id, category_name)
user (uid, user_id, name)
checkin (cid, business_id, count, day)
neighbourhood (id, business_id, neighbourhood_name)
review (rid, business_id, user_id, rating, text, year, month)
tip (tip_id, business_id, text, user_id, likes, year, month)
Please answer with SQL.
assistant
Sure, I understand.
human
List all users who reviewed businesses that are restaurant in 2010 .
assistant
SELECT t4.name FROM category AS t2 JOIN business AS t1 ON t2.business_id = t1.business_id JOIN review AS t3 ON t3.business_id = t1.business_id JOIN USER AS t4 ON t4.user_id = t3.user_id WHERE t2.category_name = 'restaurant' AND t3.year = 2010;
human
How many people reviewed the restaurant ' Texas de Brazil ' in Dallas Texas ?
assistant
SELECT COUNT ( DISTINCT t4.name ) FROM category AS t2 JOIN business AS t1 ON t2.business_id = t1.business_id JOIN review AS t3 ON t3.business_id = t1.business_id JOIN USER AS t4 ON t4.user_id = t3.user_id WHERE t1.city = 'Dallas' AND t1.name = 'Texas de Brazil' AND t1.state = 'Texas' AND t2.category_name = 'restaurant';
human
How many users have reviewed Irish Pub in Dallas ?
assistant
SELECT COUNT ( DISTINCT t4.name ) FROM category AS t2 JOIN business AS t1 ON t2.business_id = t1.business_id JOIN review AS t3 ON t3.business_id = t1.business_id JOIN USER AS t4 ON t4.user_id = t3.user_id WHERE t1.city = 'Dallas' AND t2.category_name = 'Irish Pub';
human
How many people reviewed restaurant ' Vintner Grill ' in 2010 ?
assistant
SELECT COUNT ( DISTINCT t4.name ) FROM category AS t2 JOIN business AS t1 ON t2.business_id = t1.business_id JOIN review AS t3 ON t3.business_id = t1.business_id JOIN USER AS t4 ON t4.user_id = t3.user_id WHERE t1.name = 'Vintner Grill' AND t2.category_name = 'category_category_name0' AND t3.year = 2010;
human
How many businesses has Michelle reviewed in 2010 ?
assistant
SELECT COUNT ( DISTINCT t1.name ) FROM review AS t2 JOIN business AS t1 ON t2.business_id = t1.business_id JOIN USER AS t3 ON t3.user_id = t2.user_id WHERE t2.year = 2010 AND t3.name = 'Michelle';
human
How many businesses in ' San Diego ' has Christine reviewed in 2010 ?
assistant
SELECT COUNT( DISTINCT t1.name ) FROM review AS t2 JOIN business AS t1 ON t2.business_id = t1.business_id JOIN USER AS t3 ON t3.user_id = t2.user_id WHERE t1.city = "San Diego" AND t2.year = 2010 AND t3.name = "Christine" ;