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: Who has got the most number of "funny" type of compliments? Give the user ID.
Evidence: type of compliments refers to compliment_type; most number of funny type of compliments refers to MAX(COUNT(number of compliments = 'high' WHERE compliment_type = 'funny'));
SQL: SELECT user_id FROM Users_Compliments WHERE compliment_id IN ( SELECT compliment_id FROM Compliments WHERE compliment_type LIKE 'funny' )
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: Give the number of "drive-thru" businesses in "Scottsdale" with business ID number less than "1000".
Evidence: drive-thru refers to attribute_name = 'Drive-Thru' AND attribute_value = 'true'; Scottsdale refers to city = 'Scottsdale'; business_id < 1000;
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 T3.business_id < 1000 AND T3.city LIKE 'Scottsdale' AND T1.attribute_name LIKE 'Drive-Thru' AND T2.attribute_value 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: What is the average rating for the all Yelp businesses that open 24 hours?
Evidence: open 24 hours refers to attribute_name = 'Open 24 Hours' AND attribute_value = 'true'; rating refers to stars; average rating = AVG(stars);
SQL: SELECT CAST(SUM(T3.stars) AS REAL) / COUNT(T2.business_id) AS "avg" 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 T1.attribute_name LIKE 'Open 24 Hours' AND T2.attribute_value 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: For all the Yelp businesses that allow customers bring their own beer, what percentage of them are in "Phoenix"?
Evidence: bring their own beer refers to attribute_name = 'BYOB' AND attribute_value = 'TRUE'; Phoenix refers to city = 'Phoenix'; percentage = MULTIPLY(DIVIDE(SUM(city = 'Phoenix'), COUNT(business_id)), 1.0)
SQL: SELECT CAST(SUM(CASE WHEN T3.city LIKE 'Phoenix' THEN 1 ELSE 0 END) AS REAL) * 100 / COUNT(T2.business_id) AS "percentage" 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 T1.attribute_name LIKE 'BYOB' AND T2.attribute_value 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: List the names of business in AZ with a rating of 5.
Evidence: AZ refers to state = 'AZ'; rating refers to stars;
SQL: SELECT business_id FROM Business WHERE state LIKE 'AZ' AND stars = 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: How many active businesses of city are underrated?
Evidence: active businesses refers to active = 'true'; underrated refers to review_count = 'Low';
SQL: SELECT COUNT(business_id) FROM Business WHERE review_count LIKE 'Low' AND 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 user ids from 1 to 20 have no fan users and have low ratings?
Evidence: user_id BETWEEN 1 AND 20; no fan users refers to user_fans = 'None'; low ratings refers to user_review_count = 'Low';
SQL: SELECT COUNT(user_id) FROM Users WHERE user_id BETWEEN 1 AND 20 AND user_fans LIKE 'None' AND user_review_count LIKE 'Low'
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: Indicate the opening hours of businesses are with category in fashion.
Evidence: opening hours refers to opening_time; category refers to category_name;
SQL: SELECT T4.opening_time FROM Categories AS T1 INNER JOIN Business_Categories AS T2 ON T1.category_id = T2.category_id INNER JOIN Business AS T3 ON T2.business_id = T3.business_id INNER JOIN Business_Hours AS T4 ON T3.business_id = T4.business_id WHERE T1.category_name LIKE 'Fashion'
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 operating in the shopping business have opening times before 8AM?
Evidence: shopping business refers to category_name = 'Shopping'; opening time before 8AM refers to opening_time < '8AM';
SQL: SELECT COUNT(T3.business_id) FROM Categories AS T1 INNER JOIN Business_Categories AS T2 ON T1.category_id = T2.category_id INNER JOIN Business AS T3 ON T2.business_id = T3.business_id INNER JOIN Business_Hours AS T4 ON T3.business_id = T4.business_id WHERE T4.opening_time < '8AM' AND T1.category_name LIKE 'Shopping'
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: Provide name of businesses whose category is pets and are still opened after 9PM.
Evidence: category refers to category_name; open after 9pm refers to closing_time > '9PM';
SQL: SELECT T3.business_id FROM Categories AS T1 INNER JOIN Business_Categories AS T2 ON T1.category_id = T2.category_id INNER JOIN Business AS T3 ON T2.business_id = T3.business_id INNER JOIN Business_Hours AS T4 ON T3.business_id = T4.business_id WHERE T4.closing_time > '9PM' AND T1.category_name LIKE 'Pets'
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 with the category are open from Monday to Thursday?
Evidence: open from Monday to Thursday refers to day_of_week BETWEEN Monday AND Thursday and day_id BETWEEN 2 AND 5;
SQL: SELECT COUNT(T2.business_id) FROM Categories AS T1 INNER JOIN Business_Categories AS T2 ON T1.category_id = T2.category_id INNER JOIN Business AS T3 ON T2.business_id = T3.business_id INNER JOIN Business_Hours AS T4 ON T3.business_id = T4.business_id INNER JOIN Days AS T5 ON T4.day_id = T5.day_id WHERE T5.day_of_week LIKE 'Monday' OR T5.day_of_week LIKE 'Tuesday' OR T5.day_of_week LIKE 'Wednesday' OR T5.day_of_week LIKE 'Thursday'
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: Please indicate the review count of the "active life" businesses in Phoenix.
Evidence: active life refers to category_name = 'Active Life'; Phoenix refers to city = 'Phoenix';
SQL: SELECT COUNT(*) FROM Categories AS T1 INNER JOIN Business_Categories AS T2 ON T1.category_id = T2.category_id INNER JOIN Business AS T3 ON T2.business_id = T3.business_id WHERE T1.category_name = 'Active Life' AND T3.city = 'Phoenix'