1sf/bird-sql-train-with-schema huggingface dataset:
Database: public_review_platform
Tables: Attributes, Categories, Compliments, Days, Years, Business_Attributes, Business_Categories, Business_Hours, Checkins, Elite, Reviews, Tips, Users_Compliments, Business, Users
Columns (table index, column name): -1, *, 0, attribute_id, 0, attribute_name, 1, category_id, 1, category_name, 2, compliment_id, 2, compliment_type, 3, day_id, 3, day_of_week, 4, year_id, 4, actual_year, 5, attribute_id, 5, business_id, 5, attribute_value, 6, business_id, 6, category_id, 7, business_id, 7, day_id, 7, opening_time, 7, closing_time, 8, business_id, 8, day_id, 8, label_time_0, 8, label_time_1, 8, label_time_2, 8, label_time_3, 8, label_time_4, 8, label_time_5, 8, label_time_6, 8, label_time_7, 8, label_time_8, 8, label_time_9, 8, label_time_10, 8, label_time_11, 8, label_time_12, 8, label_time_13, 8, label_time_14, 8, label_time_15, 8, label_time_16, 8, label_time_17, 8, label_time_18, 8, label_time_19, 8, label_time_20, 8, label_time_21, 8, label_time_22, 8, label_time_23, 9, user_id, 9, year_id, 10, business_id, 10, user_id, 10, review_stars, 10, review_votes_funny, 10, review_votes_useful, 10, review_votes_cool, 10, review_length, 11, business_id, 11, user_id, 11, likes, 11, tip_length, 12, compliment_id, 12, user_id, 12, number_of_compliments, 13, business_id, 13, active, 13, city, 13, state, 13, stars, 13, review_count, 14, user_id, 14, user_yelping_since_year, 14, user_average_stars, 14, user_votes_funny, 14, user_votes_useful, 14, user_votes_cool, 14, user_review_count, 14, user_fans
Question: List the closing time and day of week of active businesses in Scottsdale with stars greater than the 60% of average age of star rating.
Evidence: active business refers to active = 'true';  'Scottsdale' is the name of city; stars greater than the 60% of average age of star rating refers to stars  > avg(stars) * 60%
SQL: SELECT T2.closing_time, T3.day_of_week FROM Business AS T1 INNER JOIN Business_Hours AS T2 ON T1.business_id = T2.business_id INNER JOIN Days AS T3 ON T2.day_id = T3.day_id WHERE T1.city LIKE 'Scottsdale' AND T1.active LIKE 'TRUE' AND T1.stars > 0.6 * ( SELECT AVG(T1.stars) FROM Business AS T1 INNER JOIN Business_Hours AS T2 ON T1.business_id = T2.business_id INNER JOIN Days AS T3 ON T2.day_id = T3.day_id WHERE T1.city LIKE 'Scottsdale' AND T1.active LIKE 'TRUE' )
Database: public_review_platform
Tables: Attributes, Categories, Compliments, Days, Years, Business_Attributes, Business_Categories, Business_Hours, Checkins, Elite, Reviews, Tips, Users_Compliments, Business, Users
Columns (table index, column name): -1, *, 0, attribute_id, 0, attribute_name, 1, category_id, 1, category_name, 2, compliment_id, 2, compliment_type, 3, day_id, 3, day_of_week, 4, year_id, 4, actual_year, 5, attribute_id, 5, business_id, 5, attribute_value, 6, business_id, 6, category_id, 7, business_id, 7, day_id, 7, opening_time, 7, closing_time, 8, business_id, 8, day_id, 8, label_time_0, 8, label_time_1, 8, label_time_2, 8, label_time_3, 8, label_time_4, 8, label_time_5, 8, label_time_6, 8, label_time_7, 8, label_time_8, 8, label_time_9, 8, label_time_10, 8, label_time_11, 8, label_time_12, 8, label_time_13, 8, label_time_14, 8, label_time_15, 8, label_time_16, 8, label_time_17, 8, label_time_18, 8, label_time_19, 8, label_time_20, 8, label_time_21, 8, label_time_22, 8, label_time_23, 9, user_id, 9, year_id, 10, business_id, 10, user_id, 10, review_stars, 10, review_votes_funny, 10, review_votes_useful, 10, review_votes_cool, 10, review_length, 11, business_id, 11, user_id, 11, likes, 11, tip_length, 12, compliment_id, 12, user_id, 12, number_of_compliments, 13, business_id, 13, active, 13, city, 13, state, 13, stars, 13, review_count, 14, user_id, 14, user_yelping_since_year, 14, user_average_stars, 14, user_votes_funny, 14, user_votes_useful, 14, user_votes_cool, 14, user_review_count, 14, user_fans
Question: How many users have no followers in 2014?
Evidence: in 2004 refers to user_yelping_since_year = 2004; no follower refers to user_fans = 'None'
SQL: SELECT COUNT(user_id) FROM Users WHERE user_yelping_since_year = 2004 AND user_fans LIKE 'None'
Database: public_review_platform
Tables: Attributes, Categories, Compliments, Days, Years, Business_Attributes, Business_Categories, Business_Hours, Checkins, Elite, Reviews, Tips, Users_Compliments, Business, Users
Columns (table index, column name): -1, *, 0, attribute_id, 0, attribute_name, 1, category_id, 1, category_name, 2, compliment_id, 2, compliment_type, 3, day_id, 3, day_of_week, 4, year_id, 4, actual_year, 5, attribute_id, 5, business_id, 5, attribute_value, 6, business_id, 6, category_id, 7, business_id, 7, day_id, 7, opening_time, 7, closing_time, 8, business_id, 8, day_id, 8, label_time_0, 8, label_time_1, 8, label_time_2, 8, label_time_3, 8, label_time_4, 8, label_time_5, 8, label_time_6, 8, label_time_7, 8, label_time_8, 8, label_time_9, 8, label_time_10, 8, label_time_11, 8, label_time_12, 8, label_time_13, 8, label_time_14, 8, label_time_15, 8, label_time_16, 8, label_time_17, 8, label_time_18, 8, label_time_19, 8, label_time_20, 8, label_time_21, 8, label_time_22, 8, label_time_23, 9, user_id, 9, year_id, 10, business_id, 10, user_id, 10, review_stars, 10, review_votes_funny, 10, review_votes_useful, 10, review_votes_cool, 10, review_length, 11, business_id, 11, user_id, 11, likes, 11, tip_length, 12, compliment_id, 12, user_id, 12, number_of_compliments, 13, business_id, 13, active, 13, city, 13, state, 13, stars, 13, review_count, 14, user_id, 14, user_yelping_since_year, 14, user_average_stars, 14, user_votes_funny, 14, user_votes_useful, 14, user_votes_cool, 14, user_review_count, 14, user_fans
Question: List at least 5 users that has received less than 5 low compliments from
other users.
Evidence: less than 5 low compliment refers to number_of_compliments < 5
SQL: SELECT user_id FROM Users_Compliments WHERE number_of_compliments LIKE 'Low' GROUP BY user_id ORDER BY COUNT(number_of_compliments) > 5 LIMIT 5
Database: public_review_platform
Tables: Attributes, Categories, Compliments, Days, Years, Business_Attributes, Business_Categories, Business_Hours, Checkins, Elite, Reviews, Tips, Users_Compliments, Business, Users
Columns (table index, column name): -1, *, 0, attribute_id, 0, attribute_name, 1, category_id, 1, category_name, 2, compliment_id, 2, compliment_type, 3, day_id, 3, day_of_week, 4, year_id, 4, actual_year, 5, attribute_id, 5, business_id, 5, attribute_value, 6, business_id, 6, category_id, 7, business_id, 7, day_id, 7, opening_time, 7, closing_time, 8, business_id, 8, day_id, 8, label_time_0, 8, label_time_1, 8, label_time_2, 8, label_time_3, 8, label_time_4, 8, label_time_5, 8, label_time_6, 8, label_time_7, 8, label_time_8, 8, label_time_9, 8, label_time_10, 8, label_time_11, 8, label_time_12, 8, label_time_13, 8, label_time_14, 8, label_time_15, 8, label_time_16, 8, label_time_17, 8, label_time_18, 8, label_time_19, 8, label_time_20, 8, label_time_21, 8, label_time_22, 8, label_time_23, 9, user_id, 9, year_id, 10, business_id, 10, user_id, 10, review_stars, 10, review_votes_funny, 10, review_votes_useful, 10, review_votes_cool, 10, review_length, 11, business_id, 11, user_id, 11, likes, 11, tip_length, 12, compliment_id, 12, user_id, 12, number_of_compliments, 13, business_id, 13, active, 13, city, 13, state, 13, stars, 13, review_count, 14, user_id, 14, user_yelping_since_year, 14, user_average_stars, 14, user_votes_funny, 14, user_votes_useful, 14, user_votes_cool, 14, user_review_count, 14, user_fans
Question: List at least 10 users ID that has 4 as an average ratings of all reviews sent.
Evidence: 4 as an average rating refers to user_average_stars = 4
SQL: SELECT COUNT(user_id) FROM Users WHERE user_average_stars = 4 LIMIT 10
Database: public_review_platform
Tables: Attributes, Categories, Compliments, Days, Years, Business_Attributes, Business_Categories, Business_Hours, Checkins, Elite, Reviews, Tips, Users_Compliments, Business, Users
Columns (table index, column name): -1, *, 0, attribute_id, 0, attribute_name, 1, category_id, 1, category_name, 2, compliment_id, 2, compliment_type, 3, day_id, 3, day_of_week, 4, year_id, 4, actual_year, 5, attribute_id, 5, business_id, 5, attribute_value, 6, business_id, 6, category_id, 7, business_id, 7, day_id, 7, opening_time, 7, closing_time, 8, business_id, 8, day_id, 8, label_time_0, 8, label_time_1, 8, label_time_2, 8, label_time_3, 8, label_time_4, 8, label_time_5, 8, label_time_6, 8, label_time_7, 8, label_time_8, 8, label_time_9, 8, label_time_10, 8, label_time_11, 8, label_time_12, 8, label_time_13, 8, label_time_14, 8, label_time_15, 8, label_time_16, 8, label_time_17, 8, label_time_18, 8, label_time_19, 8, label_time_20, 8, label_time_21, 8, label_time_22, 8, label_time_23, 9, user_id, 9, year_id, 10, business_id, 10, user_id, 10, review_stars, 10, review_votes_funny, 10, review_votes_useful, 10, review_votes_cool, 10, review_length, 11, business_id, 11, user_id, 11, likes, 11, tip_length, 12, compliment_id, 12, user_id, 12, number_of_compliments, 13, business_id, 13, active, 13, city, 13, state, 13, stars, 13, review_count, 14, user_id, 14, user_yelping_since_year, 14, user_average_stars, 14, user_votes_funny, 14, user_votes_useful, 14, user_votes_cool, 14, user_review_count, 14, user_fans
Question: What city does the business have a business hour from 10 am to 12 pm on Sunday?
Evidence: 10 am refers to opening_time = '10AM'; 12 pm refers to closing_time = '12PM'; on Sunday refers to day_of_week = 'Sunday'
SQL: SELECT T1.city FROM Business AS T1 INNER JOIN Business_Hours AS T2 ON T1.business_id = T2.business_id INNER JOIN Days AS T3 ON T2.day_id = T3.day_id WHERE T2.opening_time LIKE '10AM' AND T2.closing_time LIKE '12PM' AND T3.day_of_week LIKE 'Sunday'
Database: public_review_platform
Tables: Attributes, Categories, Compliments, Days, Years, Business_Attributes, Business_Categories, Business_Hours, Checkins, Elite, Reviews, Tips, Users_Compliments, Business, Users
Columns (table index, column name): -1, *, 0, attribute_id, 0, attribute_name, 1, category_id, 1, category_name, 2, compliment_id, 2, compliment_type, 3, day_id, 3, day_of_week, 4, year_id, 4, actual_year, 5, attribute_id, 5, business_id, 5, attribute_value, 6, business_id, 6, category_id, 7, business_id, 7, day_id, 7, opening_time, 7, closing_time, 8, business_id, 8, day_id, 8, label_time_0, 8, label_time_1, 8, label_time_2, 8, label_time_3, 8, label_time_4, 8, label_time_5, 8, label_time_6, 8, label_time_7, 8, label_time_8, 8, label_time_9, 8, label_time_10, 8, label_time_11, 8, label_time_12, 8, label_time_13, 8, label_time_14, 8, label_time_15, 8, label_time_16, 8, label_time_17, 8, label_time_18, 8, label_time_19, 8, label_time_20, 8, label_time_21, 8, label_time_22, 8, label_time_23, 9, user_id, 9, year_id, 10, business_id, 10, user_id, 10, review_stars, 10, review_votes_funny, 10, review_votes_useful, 10, review_votes_cool, 10, review_length, 11, business_id, 11, user_id, 11, likes, 11, tip_length, 12, compliment_id, 12, user_id, 12, number_of_compliments, 13, business_id, 13, active, 13, city, 13, state, 13, stars, 13, review_count, 14, user_id, 14, user_yelping_since_year, 14, user_average_stars, 14, user_votes_funny, 14, user_votes_useful, 14, user_votes_cool, 14, user_review_count, 14, user_fans
Question: How many businesses are opened for 24 hours?
Evidence: opened for 24 hours refers to attribute_name = 'Open 24 Hours' AND attribute_value = 'true'
SQL: SELECT COUNT(T2.business_id) FROM Attributes AS T1 INNER JOIN Business_Attributes AS T2 ON T1.attribute_id = T2.attribute_id WHERE T2.attribute_value LIKE 'TRUE' AND T1.attribute_name LIKE 'Open 24 Hours'
Database: public_review_platform
Tables: Attributes, Categories, Compliments, Days, Years, Business_Attributes, Business_Categories, Business_Hours, Checkins, Elite, Reviews, Tips, Users_Compliments, Business, Users
Columns (table index, column name): -1, *, 0, attribute_id, 0, attribute_name, 1, category_id, 1, category_name, 2, compliment_id, 2, compliment_type, 3, day_id, 3, day_of_week, 4, year_id, 4, actual_year, 5, attribute_id, 5, business_id, 5, attribute_value, 6, business_id, 6, category_id, 7, business_id, 7, day_id, 7, opening_time, 7, closing_time, 8, business_id, 8, day_id, 8, label_time_0, 8, label_time_1, 8, label_time_2, 8, label_time_3, 8, label_time_4, 8, label_time_5, 8, label_time_6, 8, label_time_7, 8, label_time_8, 8, label_time_9, 8, label_time_10, 8, label_time_11, 8, label_time_12, 8, label_time_13, 8, label_time_14, 8, label_time_15, 8, label_time_16, 8, label_time_17, 8, label_time_18, 8, label_time_19, 8, label_time_20, 8, label_time_21, 8, label_time_22, 8, label_time_23, 9, user_id, 9, year_id, 10, business_id, 10, user_id, 10, review_stars, 10, review_votes_funny, 10, review_votes_useful, 10, review_votes_cool, 10, review_length, 11, business_id, 11, user_id, 11, likes, 11, tip_length, 12, compliment_id, 12, user_id, 12, number_of_compliments, 13, business_id, 13, active, 13, city, 13, state, 13, stars, 13, review_count, 14, user_id, 14, user_yelping_since_year, 14, user_average_stars, 14, user_votes_funny, 14, user_votes_useful, 14, user_votes_cool, 14, user_review_count, 14, user_fans
Question: List at least 5 active business ID that are good for groups and dancing.
Evidence: "Good for Groups" and "Good for Dancing" are attribute_name; active business refers to active = true'
SQL: SELECT T2.business_id FROM Attributes AS T1 INNER JOIN Business_Attributes AS T2 ON T1.attribute_id = T2.attribute_id INNER JOIN Business AS T3 ON T2.business_id = T3.business_id WHERE T2.attribute_value LIKE 'TRUE' AND T1.attribute_name LIKE 'Good for Dancing' AND T1.attribute_name LIKE 'Good for Groups' LIMIT 5
Database: public_review_platform
Tables: Attributes, Categories, Compliments, Days, Years, Business_Attributes, Business_Categories, Business_Hours, Checkins, Elite, Reviews, Tips, Users_Compliments, Business, Users
Columns (table index, column name): -1, *, 0, attribute_id, 0, attribute_name, 1, category_id, 1, category_name, 2, compliment_id, 2, compliment_type, 3, day_id, 3, day_of_week, 4, year_id, 4, actual_year, 5, attribute_id, 5, business_id, 5, attribute_value, 6, business_id, 6, category_id, 7, business_id, 7, day_id, 7, opening_time, 7, closing_time, 8, business_id, 8, day_id, 8, label_time_0, 8, label_time_1, 8, label_time_2, 8, label_time_3, 8, label_time_4, 8, label_time_5, 8, label_time_6, 8, label_time_7, 8, label_time_8, 8, label_time_9, 8, label_time_10, 8, label_time_11, 8, label_time_12, 8, label_time_13, 8, label_time_14, 8, label_time_15, 8, label_time_16, 8, label_time_17, 8, label_time_18, 8, label_time_19, 8, label_time_20, 8, label_time_21, 8, label_time_22, 8, label_time_23, 9, user_id, 9, year_id, 10, business_id, 10, user_id, 10, review_stars, 10, review_votes_funny, 10, review_votes_useful, 10, review_votes_cool, 10, review_length, 11, business_id, 11, user_id, 11, likes, 11, tip_length, 12, compliment_id, 12, user_id, 12, number_of_compliments, 13, business_id, 13, active, 13, city, 13, state, 13, stars, 13, review_count, 14, user_id, 14, user_yelping_since_year, 14, user_average_stars, 14, user_votes_funny, 14, user_votes_useful, 14, user_votes_cool, 14, user_review_count, 14, user_fans
Question: Among the active businesses in Ahwatukee, which of them are still open in Sunday?
Evidence: active business refers to active = 'true'; 'Ahwatukee' is the name of city; open in Sunday refers to day_of_week = 'Sunday'
SQL: SELECT T1.business_id FROM Business AS T1 INNER JOIN Business_Hours AS T2 ON T1.business_id = T2.business_id INNER JOIN Days AS T3 ON T2.day_id = T3.day_id WHERE T1.city LIKE 'Ahwatukee' AND T1.active LIKE 'TRUE' AND T3.day_of_week LIKE 'Sunday'
Database: public_review_platform
Tables: Attributes, Categories, Compliments, Days, Years, Business_Attributes, Business_Categories, Business_Hours, Checkins, Elite, Reviews, Tips, Users_Compliments, Business, Users
Columns (table index, column name): -1, *, 0, attribute_id, 0, attribute_name, 1, category_id, 1, category_name, 2, compliment_id, 2, compliment_type, 3, day_id, 3, day_of_week, 4, year_id, 4, actual_year, 5, attribute_id, 5, business_id, 5, attribute_value, 6, business_id, 6, category_id, 7, business_id, 7, day_id, 7, opening_time, 7, closing_time, 8, business_id, 8, day_id, 8, label_time_0, 8, label_time_1, 8, label_time_2, 8, label_time_3, 8, label_time_4, 8, label_time_5, 8, label_time_6, 8, label_time_7, 8, label_time_8, 8, label_time_9, 8, label_time_10, 8, label_time_11, 8, label_time_12, 8, label_time_13, 8, label_time_14, 8, label_time_15, 8, label_time_16, 8, label_time_17, 8, label_time_18, 8, label_time_19, 8, label_time_20, 8, label_time_21, 8, label_time_22, 8, label_time_23, 9, user_id, 9, year_id, 10, business_id, 10, user_id, 10, review_stars, 10, review_votes_funny, 10, review_votes_useful, 10, review_votes_cool, 10, review_length, 11, business_id, 11, user_id, 11, likes, 11, tip_length, 12, compliment_id, 12, user_id, 12, number_of_compliments, 13, business_id, 13, active, 13, city, 13, state, 13, stars, 13, review_count, 14, user_id, 14, user_yelping_since_year, 14, user_average_stars, 14, user_votes_funny, 14, user_votes_useful, 14, user_votes_cool, 14, user_review_count, 14, user_fans
Question: List the categories of all active businesses that were not in Arizona.
Evidence: active business refers to active = 'true'; not in Arizona refers to state ! = 'AZ'; category refers to category_name
SQL: SELECT T3.category_name FROM Business AS T1 INNER JOIN Business_Categories ON T1.business_id = Business_Categories.business_id INNER JOIN Categories AS T3 ON Business_Categories.category_id = T3.category_id WHERE T1.active LIKE 'TRUE' AND T1.state NOT LIKE 'AZ'
Database: public_review_platform
Tables: Attributes, Categories, Compliments, Days, Years, Business_Attributes, Business_Categories, Business_Hours, Checkins, Elite, Reviews, Tips, Users_Compliments, Business, Users
Columns (table index, column name): -1, *, 0, attribute_id, 0, attribute_name, 1, category_id, 1, category_name, 2, compliment_id, 2, compliment_type, 3, day_id, 3, day_of_week, 4, year_id, 4, actual_year, 5, attribute_id, 5, business_id, 5, attribute_value, 6, business_id, 6, category_id, 7, business_id, 7, day_id, 7, opening_time, 7, closing_time, 8, business_id, 8, day_id, 8, label_time_0, 8, label_time_1, 8, label_time_2, 8, label_time_3, 8, label_time_4, 8, label_time_5, 8, label_time_6, 8, label_time_7, 8, label_time_8, 8, label_time_9, 8, label_time_10, 8, label_time_11, 8, label_time_12, 8, label_time_13, 8, label_time_14, 8, label_time_15, 8, label_time_16, 8, label_time_17, 8, label_time_18, 8, label_time_19, 8, label_time_20, 8, label_time_21, 8, label_time_22, 8, label_time_23, 9, user_id, 9, year_id, 10, business_id, 10, user_id, 10, review_stars, 10, review_votes_funny, 10, review_votes_useful, 10, review_votes_cool, 10, review_length, 11, business_id, 11, user_id, 11, likes, 11, tip_length, 12, compliment_id, 12, user_id, 12, number_of_compliments, 13, business_id, 13, active, 13, city, 13, state, 13, stars, 13, review_count, 14, user_id, 14, user_yelping_since_year, 14, user_average_stars, 14, user_votes_funny, 14, user_votes_useful, 14, user_votes_cool, 14, user_review_count, 14, user_fans
Question: List the category of the business with high review count but received 2 stars.
Evidence: high review count refers to review_count = 'High'; received 2 stars refers to stars = 2; category refers to category_name
SQL: SELECT T3.category_name FROM Business AS T1 INNER JOIN Business_Categories ON T1.business_id = Business_Categories.business_id INNER JOIN Categories AS T3 ON Business_Categories.category_id = T3.category_id WHERE T1.stars = 2 AND T1.review_count LIKE 'High'
Database: public_review_platform
Tables: Attributes, Categories, Compliments, Days, Years, Business_Attributes, Business_Categories, Business_Hours, Checkins, Elite, Reviews, Tips, Users_Compliments, Business, Users
Columns (table index, column name): -1, *, 0, attribute_id, 0, attribute_name, 1, category_id, 1, category_name, 2, compliment_id, 2, compliment_type, 3, day_id, 3, day_of_week, 4, year_id, 4, actual_year, 5, attribute_id, 5, business_id, 5, attribute_value, 6, business_id, 6, category_id, 7, business_id, 7, day_id, 7, opening_time, 7, closing_time, 8, business_id, 8, day_id, 8, label_time_0, 8, label_time_1, 8, label_time_2, 8, label_time_3, 8, label_time_4, 8, label_time_5, 8, label_time_6, 8, label_time_7, 8, label_time_8, 8, label_time_9, 8, label_time_10, 8, label_time_11, 8, label_time_12, 8, label_time_13, 8, label_time_14, 8, label_time_15, 8, label_time_16, 8, label_time_17, 8, label_time_18, 8, label_time_19, 8, label_time_20, 8, label_time_21, 8, label_time_22, 8, label_time_23, 9, user_id, 9, year_id, 10, business_id, 10, user_id, 10, review_stars, 10, review_votes_funny, 10, review_votes_useful, 10, review_votes_cool, 10, review_length, 11, business_id, 11, user_id, 11, likes, 11, tip_length, 12, compliment_id, 12, user_id, 12, number_of_compliments, 13, business_id, 13, active, 13, city, 13, state, 13, stars, 13, review_count, 14, user_id, 14, user_yelping_since_year, 14, user_average_stars, 14, user_votes_funny, 14, user_votes_useful, 14, user_votes_cool, 14, user_review_count, 14, user_fans
Question: How many businesses have a romantic ambiance?
Evidence: romantic ambiance refers to attribute_name = 'ambience_romantic' AND attribute_value = 'true'
SQL: SELECT COUNT(T2.business_id) FROM Attributes AS T1 INNER JOIN Business_Attributes AS T2 ON T1.attribute_id = T2.attribute_id WHERE T2.attribute_value = 'true' AND T1.attribute_name = 'ambience_romantic'
Database: public_review_platform
Tables: Attributes, Categories, Compliments, Days, Years, Business_Attributes, Business_Categories, Business_Hours, Checkins, Elite, Reviews, Tips, Users_Compliments, Business, Users
Columns (table index, column name): -1, *, 0, attribute_id, 0, attribute_name, 1, category_id, 1, category_name, 2, compliment_id, 2, compliment_type, 3, day_id, 3, day_of_week, 4, year_id, 4, actual_year, 5, attribute_id, 5, business_id, 5, attribute_value, 6, business_id, 6, category_id, 7, business_id, 7, day_id, 7, opening_time, 7, closing_time, 8, business_id, 8, day_id, 8, label_time_0, 8, label_time_1, 8, label_time_2, 8, label_time_3, 8, label_time_4, 8, label_time_5, 8, label_time_6, 8, label_time_7, 8, label_time_8, 8, label_time_9, 8, label_time_10, 8, label_time_11, 8, label_time_12, 8, label_time_13, 8, label_time_14, 8, label_time_15, 8, label_time_16, 8, label_time_17, 8, label_time_18, 8, label_time_19, 8, label_time_20, 8, label_time_21, 8, label_time_22, 8, label_time_23, 9, user_id, 9, year_id, 10, business_id, 10, user_id, 10, review_stars, 10, review_votes_funny, 10, review_votes_useful, 10, review_votes_cool, 10, review_length, 11, business_id, 11, user_id, 11, likes, 11, tip_length, 12, compliment_id, 12, user_id, 12, number_of_compliments, 13, business_id, 13, active, 13, city, 13, state, 13, stars, 13, review_count, 14, user_id, 14, user_yelping_since_year, 14, user_average_stars, 14, user_votes_funny, 14, user_votes_useful, 14, user_votes_cool, 14, user_review_count, 14, user_fans
Question: List the city of the business where they open from 1 pm to 6 pm on Saturday.
Evidence: 1 pm refers to opening_time = '1PM'; 6 pm refers to closing_time = '6PM'; on Saturday refers to day_of_week = 'Saturday'
SQL: SELECT T1.city FROM Business AS T1 INNER JOIN Business_Hours AS T2 ON T1.business_id = T2.business_id INNER JOIN Days AS T3 ON T2.day_id = T3.day_id WHERE T2.closing_time LIKE '6PM' AND T2.opening_time LIKE '1PM' AND T3.day_of_week LIKE 'Saturday'