difficulty
int64
1
4
nl_query
stringlengths
23
90
sql_query
stringlengths
84
540
db_id
stringclasses
1 value
3
How many reviews has Adrienne written ?
SELECT COUNT( DISTINCT ( REVIEWalias0.TEXT ) ) FROM REVIEW AS REVIEWalias0 , USER AS USERalias0 WHERE USERalias0.NAME = "Adrienne" AND USERalias0.USER_ID = REVIEWalias0.USER_ID ;
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3
How many reviews has Michelle written in March 2014 ?
SELECT COUNT( DISTINCT ( REVIEWalias0.TEXT ) ) FROM REVIEW AS REVIEWalias0 , USER AS USERalias0 WHERE REVIEWalias0.MONTH = "March" AND REVIEWalias0.YEAR = 2014 AND USERalias0.NAME = "Michelle" AND USERalias0.USER_ID = REVIEWalias0.USER_ID ;
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3
How many businesses has Michelle reviewed in 2010 ?
SELECT COUNT( DISTINCT ( BUSINESSalias0.NAME ) ) FROM BUSINESS AS BUSINESSalias0 , REVIEW AS REVIEWalias0 , USER AS USERalias0 WHERE REVIEWalias0.BUSINESS_ID = BUSINESSalias0.BUSINESS_ID AND REVIEWalias0.YEAR = 2010 AND USERalias0.NAME = "Michelle" AND USERalias0.USER_ID = REVIEWalias0.USER_ID ;
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3
How many businesses in " San Diego " has Christine reviewed in 2010 ?
SELECT COUNT( DISTINCT ( BUSINESSalias0.NAME ) ) FROM BUSINESS AS BUSINESSalias0 , REVIEW AS REVIEWalias0 , USER AS USERalias0 WHERE BUSINESSalias0.CITY = "San Diego" AND REVIEWalias0.BUSINESS_ID = BUSINESSalias0.BUSINESS_ID AND REVIEWalias0.YEAR = 2010 AND USERalias0.NAME = "Christine" AND USERalias0.USER_ID = REVIEWa...
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3
How many Target are there in " Los Angeles " ?
SELECT COUNT( DISTINCT ( BUSINESSalias0.BUSINESS_ID ) ) FROM BUSINESS AS BUSINESSalias0 WHERE BUSINESSalias0.CITY = "Los Angeles" AND BUSINESSalias0.NAME = "Target" ;
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3
How many users have reviewed Irish Pub in Dallas ?
SELECT COUNT( DISTINCT ( USERalias0.NAME ) ) FROM BUSINESS AS BUSINESSalias0 , CATEGORY AS CATEGORYalias0 , REVIEW AS REVIEWalias0 , USER AS USERalias0 WHERE BUSINESSalias0.CITY = "Dallas" AND CATEGORYalias0.BUSINESS_ID = BUSINESSalias0.BUSINESS_ID AND CATEGORYalias0.CATEGORY_NAME = "Irish Pub" AND REVIEWalias0.BUSINES...
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3
What is the average rating of reviews written in year 2014 ?
SELECT AVG ( REVIEWalias0.RATING ) FROM REVIEW AS REVIEWalias0 WHERE REVIEWalias0.YEAR = 2014 ;
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3
How many people reviewed restaurant " Vintner Grill " in 2010 ?
SELECT COUNT( DISTINCT ( USERalias0.NAME ) ) FROM BUSINESS AS BUSINESSalias0 , REVIEW AS REVIEWalias0 , USER AS USERalias0 WHERE BUSINESSalias0.NAME = "Vintner Grill" AND REVIEWalias0.BUSINESS_ID = BUSINESSalias0.BUSINESS_ID AND REVIEWalias0.YEAR = 2010 AND USERalias0.USER_ID = REVIEWalias0.USER_ID ;
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3
How many people reviewed restaurant " Vintner Grill " in 2010 ?
SELECT COUNT( DISTINCT ( USERalias0.NAME ) ) FROM BUSINESS AS BUSINESSalias0 , CATEGORY AS CATEGORYalias0 , REVIEW AS REVIEWalias0 , USER AS USERalias0 WHERE BUSINESSalias0.NAME = "Vintner Grill" AND CATEGORYalias0.BUSINESS_ID = BUSINESSalias0.BUSINESS_ID AND CATEGORYalias0.CATEGORY_NAME = "category_name0" AND REVIEWal...
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3
Find the number of reviews on businesses located in " South Summerlin " neighborhood
SELECT SUM( BUSINESSalias0.REVIEW_COUNT ) FROM BUSINESS AS BUSINESSalias0 , NEIGHBORHOOD AS NEIGHBORHOODalias0 WHERE NEIGHBORHOODalias0.BUSINESS_ID = BUSINESSalias0.BUSINESS_ID AND NEIGHBORHOODalias0.NEIGHBORHOOD_NAME = "South Summerlin" ;
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3
Find the number of reviews on businesses located in " South Summerlin " neighborhood
SELECT COUNT( DISTINCT ( REVIEWalias0.TEXT ) ) FROM BUSINESS AS BUSINESSalias0 , NEIGHBORHOOD AS NEIGHBORHOODalias0 , REVIEW AS REVIEWalias0 WHERE NEIGHBORHOODalias0.BUSINESS_ID = BUSINESSalias0.BUSINESS_ID AND NEIGHBORHOODalias0.NEIGHBORHOOD_NAME = "South Summerlin" AND REVIEWalias0.BUSINESS_ID = BUSINESSalias0.BUSINE...
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3
Find the number of users called Michelle
SELECT COUNT( DISTINCT ( USERalias0.NAME ) ) FROM USER AS USERalias0 WHERE USERalias0.NAME = "Michelle" ;
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3
Return me the number of businesses that are restaurant .
SELECT COUNT( DISTINCT ( BUSINESSalias0.NAME ) ) FROM BUSINESS AS BUSINESSalias0 , CATEGORY AS CATEGORYalias0 WHERE CATEGORYalias0.BUSINESS_ID = BUSINESSalias0.BUSINESS_ID AND CATEGORYalias0.CATEGORY_NAME = "restaurant" ;
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3
Return me the number of cities that has " Panda Express " .
SELECT COUNT( DISTINCT ( BUSINESSalias0.CITY ) ) FROM BUSINESS AS BUSINESSalias0 WHERE BUSINESSalias0.NAME = "Panda Express" ;
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3
Return me the number of tips that are written by Michelle .
SELECT COUNT( DISTINCT ( TIPalias0.TEXT ) ) FROM TIP AS TIPalias0 , USER AS USERalias0 WHERE USERalias0.NAME = "Michelle" AND USERalias0.USER_ID = TIPalias0.USER_ID ;
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3
Find the total checkins in " Brighton Heights " neighborhood
SELECT SUM( CHECKINalias0.COUNT ) FROM BUSINESS AS BUSINESSalias0 , CHECKIN AS CHECKINalias0 , NEIGHBORHOOD AS NEIGHBORHOODalias0 WHERE CHECKINalias0.BUSINESS_ID = BUSINESSalias0.BUSINESS_ID AND NEIGHBORHOODalias0.BUSINESS_ID = BUSINESSalias0.BUSINESS_ID AND NEIGHBORHOODalias0.NEIGHBORHOOD_NAME = "Brighton Heights" ;
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3
Find the total checkins in Sunday
SELECT SUM( CHECKINalias0.COUNT ) FROM CHECKIN AS CHECKINalias0 WHERE CHECKINalias0.DAY = "Sunday" ;
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3
Find the total number of reviews written in March
SELECT COUNT( DISTINCT ( REVIEWalias0.TEXT ) ) FROM REVIEW AS REVIEWalias0 WHERE REVIEWalias0.MONTH = "March" ;
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3
Find the number of tips written in each month
SELECT COUNT( DISTINCT ( TIPalias0.TEXT ) ) , TIPalias0.MONTH FROM TIP AS TIPalias0 GROUP BY TIPalias0.MONTH ;
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3
How many neighborhoods have a business with rating 5 in Madison ?
SELECT COUNT( DISTINCT ( NEIGHBORHOODalias0.NEIGHBORHOOD_NAME ) ) FROM BUSINESS AS BUSINESSalias0 , NEIGHBORHOOD AS NEIGHBORHOODalias0 WHERE BUSINESSalias0.CITY = "Madison" AND BUSINESSalias0.RATING = 5 AND NEIGHBORHOODalias0.BUSINESS_ID = BUSINESSalias0.BUSINESS_ID ;
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4
Give me all the Moroccan restaurant in Texas
SELECT BUSINESSalias0.NAME FROM BUSINESS AS BUSINESSalias0 , CATEGORY AS CATEGORYalias0 , CATEGORY AS CATEGORYalias1 WHERE BUSINESSalias0.STATE = "Texas" AND CATEGORYalias0.BUSINESS_ID = BUSINESSalias0.BUSINESS_ID AND CATEGORYalias0.CATEGORY_NAME = "Moroccan" AND CATEGORYalias1.BUSINESS_ID = BUSINESSalias0.BUSINESS_ID ...
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4
Which Thai restaurant has the most number of reviews
SELECT BUSINESSalias0.NAME FROM BUSINESS AS BUSINESSalias0 , CATEGORY AS CATEGORYalias0 , CATEGORY AS CATEGORYalias1 WHERE CATEGORYalias0.BUSINESS_ID = BUSINESSalias0.BUSINESS_ID AND CATEGORYalias0.CATEGORY_NAME = "restaurant" AND CATEGORYalias1.BUSINESS_ID = BUSINESSalias0.BUSINESS_ID AND CATEGORYalias1.CATEGORY_NAME ...
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4
Which Thai restaurant has the most number of reviews
SELECT BUSINESSalias0.NAME FROM BUSINESS AS BUSINESSalias0 , CATEGORY AS CATEGORYalias0 , CATEGORY AS CATEGORYalias1 , REVIEW AS REVIEWalias0 WHERE CATEGORYalias0.BUSINESS_ID = BUSINESSalias0.BUSINESS_ID AND CATEGORYalias0.CATEGORY_NAME = "restaurant" AND CATEGORYalias1.BUSINESS_ID = BUSINESSalias0.BUSINESS_ID AND CATE...
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4
which business has the most number of checkins
SELECT BUSINESSalias0.NAME FROM BUSINESS AS BUSINESSalias0 , CHECKIN AS CHECKINalias0 WHERE CHECKINalias0.BUSINESS_ID = BUSINESSalias0.BUSINESS_ID GROUP BY BUSINESSalias0.NAME ORDER BY SUM( CHECKINalias0.COUNT ) DESC LIMIT 1 ;
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4
find the user with the most number of reviews
SELECT USERalias0.NAME FROM REVIEW AS REVIEWalias0 , USER AS USERalias0 WHERE USERalias0.USER_ID = REVIEWalias0.USER_ID GROUP BY USERalias0.NAME ORDER BY COUNT( DISTINCT ( REVIEWalias0.TEXT ) ) DESC LIMIT 1 ;
yelp
4
find the user with the most number of reviews
SELECT USERalias0.USER_ID FROM REVIEW AS REVIEWalias0 , USER AS USERalias0 WHERE USERalias0.USER_ID = REVIEWalias0.USER_ID GROUP BY USERalias0.USER_ID ORDER BY COUNT( DISTINCT ( REVIEWalias0.TEXT ) ) DESC LIMIT 1 ;
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4
which neighborhood has the most number of businesses in Madison
SELECT NEIGHBORHOODalias0.NEIGHBORHOOD_NAME FROM BUSINESS AS BUSINESSalias0 , NEIGHBORHOOD AS NEIGHBORHOODalias0 WHERE BUSINESSalias0.CITY = "Madison" AND NEIGHBORHOODalias0.BUSINESS_ID = BUSINESSalias0.BUSINESS_ID GROUP BY NEIGHBORHOODalias0.NEIGHBORHOOD_NAME ORDER BY COUNT( DISTINCT ( BUSINESSalias0.NAME ) ) DESC LIM...
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4
Which neighborhood in Madison has the Italian restaurant with the highest stars
SELECT NEIGHBORHOODalias0.NEIGHBORHOOD_NAME FROM BUSINESS AS BUSINESSalias0 , CATEGORY AS CATEGORYalias0 , CATEGORY AS CATEGORYalias1 , NEIGHBORHOOD AS NEIGHBORHOODalias0 WHERE BUSINESSalias0.CITY = "Madison" AND CATEGORYalias0.BUSINESS_ID = BUSINESSalias0.BUSINESS_ID AND CATEGORYalias0.CATEGORY_NAME = "restaurant" AND...
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4
Find all Chinese restaurant in Dallas with at least 4 stars
SELECT BUSINESSalias0.NAME FROM BUSINESS AS BUSINESSalias0 , CATEGORY AS CATEGORYalias0 , CATEGORY AS CATEGORYalias1 WHERE BUSINESSalias0.CITY = "Dallas" AND BUSINESSalias0.RATING > 4 AND CATEGORYalias0.BUSINESS_ID = BUSINESSalias0.BUSINESS_ID AND CATEGORYalias0.CATEGORY_NAME = "Chinese" AND CATEGORYalias1.BUSINESS_ID ...
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4
Find all Mexican restaurant in Dallas with a rating above 3.5
SELECT BUSINESSalias0.NAME FROM BUSINESS AS BUSINESSalias0 , CATEGORY AS CATEGORYalias0 , CATEGORY AS CATEGORYalias1 WHERE BUSINESSalias0.CITY = "Dallas" AND BUSINESSalias0.RATING > 3.5 AND CATEGORYalias0.BUSINESS_ID = BUSINESSalias0.BUSINESS_ID AND CATEGORYalias0.CATEGORY_NAME = "Mexican" AND CATEGORYalias1.BUSINESS_I...
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4
Find all restaurant with Valet Service in Dallas Texas
SELECT BUSINESSalias0.NAME FROM BUSINESS AS BUSINESSalias0 , CATEGORY AS CATEGORYalias0 , CATEGORY AS CATEGORYalias1 WHERE BUSINESSalias0.CITY = "Dallas" AND BUSINESSalias0.STATE = "Texas" AND CATEGORYalias0.BUSINESS_ID = BUSINESSalias0.BUSINESS_ID AND CATEGORYalias0.CATEGORY_NAME = "Valet Service" AND CATEGORYalias1.B...
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4
Find all Italian restaurant in the Meadowood neighborhood of Madison
SELECT BUSINESSalias0.NAME FROM BUSINESS AS BUSINESSalias0 , CATEGORY AS CATEGORYalias0 , CATEGORY AS CATEGORYalias1 , NEIGHBORHOOD AS NEIGHBORHOODalias0 WHERE BUSINESSalias0.CITY = "Madison" AND CATEGORYalias0.BUSINESS_ID = BUSINESSalias0.BUSINESS_ID AND CATEGORYalias0.CATEGORY_NAME = "Italian" AND CATEGORYalias1.BUSI...
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4
Find all Bars in " Los Angeles " with at least 30 reviews and average rating above 3 stars
SELECT BUSINESSalias0.NAME FROM BUSINESS AS BUSINESSalias0 , CATEGORY AS CATEGORYalias0 WHERE BUSINESSalias0.CITY = "Los Angeles" AND BUSINESSalias0.RATING > 3 AND BUSINESSalias0.REVIEW_COUNT > 30 AND CATEGORYalias0.BUSINESS_ID = BUSINESSalias0.BUSINESS_ID AND CATEGORYalias0.CATEGORY_NAME = "Bars" ;
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4
How many Egyptian restaurant are there in Edinburgh ?
SELECT COUNT( DISTINCT ( BUSINESSalias0.NAME ) ) FROM BUSINESS AS BUSINESSalias0 , CATEGORY AS CATEGORYalias0 , CATEGORY AS CATEGORYalias1 WHERE BUSINESSalias0.CITY = "Edinburgh" AND CATEGORYalias0.BUSINESS_ID = BUSINESSalias0.BUSINESS_ID AND CATEGORYalias0.CATEGORY_NAME = "restaurant" AND CATEGORYalias1.BUSINESS_ID = ...
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4
Which Indian restaurant in Dallas has the highest rating ?
SELECT BUSINESSalias0.NAME FROM BUSINESS AS BUSINESSalias0 , CATEGORY AS CATEGORYalias0 , CATEGORY AS CATEGORYalias1 WHERE BUSINESSalias0.CITY = "Dallas" AND CATEGORYalias0.BUSINESS_ID = BUSINESSalias0.BUSINESS_ID AND CATEGORYalias0.CATEGORY_NAME = "Indian" AND CATEGORYalias1.BUSINESS_ID = BUSINESSalias0.BUSINESS_ID AN...
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4
Which Italian restaurant in Dallas has the highest rating ?
SELECT BUSINESSalias0.NAME FROM BUSINESS AS BUSINESSalias0 , CATEGORY AS CATEGORYalias0 , CATEGORY AS CATEGORYalias1 WHERE BUSINESSalias0.CITY = "Dallas" AND CATEGORYalias0.BUSINESS_ID = BUSINESSalias0.BUSINESS_ID AND CATEGORYalias0.CATEGORY_NAME = "restaurant" AND CATEGORYalias1.BUSINESS_ID = BUSINESSalias0.BUSINESS_I...
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4
Which Indian restaurant in Dallas has the most number of reviews ?
SELECT BUSINESSalias0.NAME FROM BUSINESS AS BUSINESSalias0 , CATEGORY AS CATEGORYalias0 , CATEGORY AS CATEGORYalias1 WHERE BUSINESSalias0.CITY = "Dallas" AND CATEGORYalias0.BUSINESS_ID = BUSINESSalias0.BUSINESS_ID AND CATEGORYalias0.CATEGORY_NAME = "restaurant" AND CATEGORYalias1.BUSINESS_ID = BUSINESSalias0.BUSINESS_I...
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4
Which Indian restaurant in Dallas has the most number of reviews ?
SELECT BUSINESSalias0.NAME FROM BUSINESS AS BUSINESSalias0 , CATEGORY AS CATEGORYalias0 , CATEGORY AS CATEGORYalias1 , REVIEW AS REVIEWalias0 WHERE BUSINESSalias0.CITY = "Dallas" AND CATEGORYalias0.BUSINESS_ID = BUSINESSalias0.BUSINESS_ID AND CATEGORYalias0.CATEGORY_NAME = "restaurant" AND CATEGORYalias1.BUSINESS_ID = ...
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4
Find users whose average review rating is below 3
SELECT USERalias0.NAME FROM REVIEW AS REVIEWalias0 , USER AS USERalias0 WHERE USERalias0.USER_ID = REVIEWalias0.USER_ID GROUP BY USERalias0.NAME HAVING AVG ( REVIEWalias0.RATING ) < 3 ;
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4
Find the business with the most number of reviews in April
SELECT BUSINESSalias0.NAME FROM BUSINESS AS BUSINESSalias0 , REVIEW AS REVIEWalias0 WHERE REVIEWalias0.BUSINESS_ID = BUSINESSalias0.BUSINESS_ID AND REVIEWalias0.MONTH = "April" GROUP BY BUSINESSalias0.NAME ORDER BY COUNT( DISTINCT ( REVIEWalias0.TEXT ) ) DESC LIMIT 1 ;
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4
Find the business which has the most number of categories
SELECT BUSINESSalias0.NAME FROM BUSINESS AS BUSINESSalias0 , CATEGORY AS CATEGORYalias0 WHERE CATEGORYalias0.BUSINESS_ID = BUSINESSalias0.BUSINESS_ID GROUP BY BUSINESSalias0.NAME ORDER BY COUNT( DISTINCT ( CATEGORYalias0.CATEGORY_NAME ) ) DESC LIMIT 1 ;
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