query_id
int64 0
1.03k
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stringclasses 20
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sequencelengths 1
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stringlengths 18
174
| answer
stringlengths 20
422
| difficulty
stringclasses 4
values |
|---|---|---|---|---|---|
900 | network_1 | [
"Likes",
"Friend",
"Highschooler"
] | Show name of all students who have some friends and also are liked by someone else. | SELECT T2.name FROM Friend AS T1 JOIN Highschooler AS T2 ON T1.student_id = T2.id INTERSECT SELECT T2.name FROM Likes AS T1 JOIN Highschooler AS T2 ON T1.liked_id = T2.id | hard |
901 | network_1 | [
"Likes",
"Friend",
"Highschooler"
] | What are the names of high schoolers who both have friends and are liked? | SELECT T2.name FROM Friend AS T1 JOIN Highschooler AS T2 ON T1.student_id = T2.id INTERSECT SELECT T2.name FROM Likes AS T1 JOIN Highschooler AS T2 ON T1.liked_id = T2.id | hard |
902 | network_1 | [
"Likes"
] | Count the number of likes for each student id. | SELECT student_id , count(*) FROM Likes GROUP BY student_id | medium |
903 | network_1 | [
"Likes"
] | How many likes correspond to each student id? | SELECT student_id , count(*) FROM Likes GROUP BY student_id | medium |
904 | network_1 | [
"Likes",
"Highschooler"
] | Show the names of high schoolers who have likes, and numbers of likes for each. | SELECT T2.name , count(*) FROM Likes AS T1 JOIN Highschooler AS T2 ON T1.student_id = T2.id GROUP BY T1.student_id | medium |
905 | network_1 | [
"Likes",
"Highschooler"
] | What are the names of high schoolers who have likes, and how many likes does each have? | SELECT T2.name , count(*) FROM Likes AS T1 JOIN Highschooler AS T2 ON T1.student_id = T2.id GROUP BY T1.student_id | medium |
906 | network_1 | [
"Likes",
"Highschooler"
] | What is the name of the high schooler who has the greatest number of likes? | SELECT T2.name FROM Likes AS T1 JOIN Highschooler AS T2 ON T1.student_id = T2.id GROUP BY T1.student_id ORDER BY count(*) DESC LIMIT 1 | extra |
907 | network_1 | [
"Likes",
"Highschooler"
] | Give the name of the student with the most likes. | SELECT T2.name FROM Likes AS T1 JOIN Highschooler AS T2 ON T1.student_id = T2.id GROUP BY T1.student_id ORDER BY count(*) DESC LIMIT 1 | extra |
908 | network_1 | [
"Likes",
"Highschooler"
] | Show the names of students who have at least 2 likes. | SELECT T2.name FROM Likes AS T1 JOIN Highschooler AS T2 ON T1.student_id = T2.id GROUP BY T1.student_id HAVING count(*) >= 2 | medium |
909 | network_1 | [
"Likes",
"Highschooler"
] | What are the names of students who have 2 or more likes? | SELECT T2.name FROM Likes AS T1 JOIN Highschooler AS T2 ON T1.student_id = T2.id GROUP BY T1.student_id HAVING count(*) >= 2 | medium |
910 | network_1 | [
"Friend",
"Highschooler"
] | Show the names of students who have a grade higher than 5 and have at least 2 friends. | SELECT T2.name FROM Friend AS T1 JOIN Highschooler AS T2 ON T1.student_id = T2.id WHERE T2.grade > 5 GROUP BY T1.student_id HAVING count(*) >= 2 | hard |
911 | network_1 | [
"Friend",
"Highschooler"
] | What are the names of high schoolers who have a grade of over 5 and have 2 or more friends? | SELECT T2.name FROM Friend AS T1 JOIN Highschooler AS T2 ON T1.student_id = T2.id WHERE T2.grade > 5 GROUP BY T1.student_id HAVING count(*) >= 2 | hard |
912 | network_1 | [
"Likes",
"Highschooler"
] | How many likes does Kyle have? | SELECT count(*) FROM Likes AS T1 JOIN Highschooler AS T2 ON T1.student_id = T2.id WHERE T2.name = "Kyle" | medium |
913 | network_1 | [
"Likes",
"Highschooler"
] | Return the number of likes that the high schooler named Kyle has. | SELECT count(*) FROM Likes AS T1 JOIN Highschooler AS T2 ON T1.student_id = T2.id WHERE T2.name = "Kyle" | medium |
914 | network_1 | [
"Friend",
"Highschooler"
] | Find the average grade of all students who have some friends. | SELECT avg(grade) FROM Highschooler WHERE id IN (SELECT T1.student_id FROM Friend AS T1 JOIN Highschooler AS T2 ON T1.student_id = T2.id) | hard |
915 | network_1 | [
"Friend",
"Highschooler"
] | What is the average grade of students who have friends? | SELECT avg(grade) FROM Highschooler WHERE id IN (SELECT T1.student_id FROM Friend AS T1 JOIN Highschooler AS T2 ON T1.student_id = T2.id) | hard |
916 | network_1 | [
"Friend",
"Highschooler"
] | Find the minimum grade of students who have no friends. | SELECT min(grade) FROM Highschooler WHERE id NOT IN (SELECT T1.student_id FROM Friend AS T1 JOIN Highschooler AS T2 ON T1.student_id = T2.id) | extra |
917 | network_1 | [
"Friend",
"Highschooler"
] | What is the lowest grade of students who do not have any friends? | SELECT min(grade) FROM Highschooler WHERE id NOT IN (SELECT T1.student_id FROM Friend AS T1 JOIN Highschooler AS T2 ON T1.student_id = T2.id) | extra |
918 | dog_kennels | [
"Owners",
"Professionals"
] | Which states have both owners and professionals living there? | SELECT state FROM Owners INTERSECT SELECT state FROM Professionals | hard |
919 | dog_kennels | [
"Owners",
"Professionals"
] | Find the states where both owners and professionals live. | SELECT state FROM Owners INTERSECT SELECT state FROM Professionals | hard |
920 | dog_kennels | [
"Dogs",
"Treatments"
] | What is the average age of the dogs who have gone through any treatments? | SELECT avg(age) FROM Dogs WHERE dog_id IN ( SELECT dog_id FROM Treatments ) | hard |
921 | dog_kennels | [
"Dogs",
"Treatments"
] | Find the average age of the dogs who went through treatments. | SELECT avg(age) FROM Dogs WHERE dog_id IN ( SELECT dog_id FROM Treatments ) | hard |
922 | dog_kennels | [
"Professionals",
"Treatments"
] | Which professionals live in the state of Indiana or have done treatment on more than 2 treatments? List his or her id, last name and cell phone. | SELECT professional_id , last_name , cell_number FROM Professionals WHERE state = 'Indiana' UNION SELECT T1.professional_id , T1.last_name , T1.cell_number FROM Professionals AS T1 JOIN Treatments AS T2 ON T1.professional_id = T2.professional_id GROUP BY T1.professional_id HAVING count(*) > 2 | extra |
923 | dog_kennels | [
"Professionals",
"Treatments"
] | Find the id, last name and cell phone of the professionals who live in the state of Indiana or have performed more than two treatments. | SELECT professional_id , last_name , cell_number FROM Professionals WHERE state = 'Indiana' UNION SELECT T1.professional_id , T1.last_name , T1.cell_number FROM Professionals AS T1 JOIN Treatments AS T2 ON T1.professional_id = T2.professional_id GROUP BY T1.professional_id HAVING count(*) > 2 | extra |
924 | dog_kennels | [
"treatments",
"dogs"
] | Which dogs have not cost their owner more than 1000 for treatment ? List the dog names . | select name from dogs where dog_id not in ( select dog_id from treatments group by dog_id having sum(cost_of_treatment) > 1000 ) | hard |
925 | dog_kennels | [
"treatments",
"dogs"
] | What are the names of the dogs for which the owner has not spend more than 1000 for treatment ? | select name from dogs where dog_id not in ( select dog_id from treatments group by dog_id having sum(cost_of_treatment) > 1000 ) | hard |
926 | dog_kennels | [
"Owners",
"Dogs",
"Professionals"
] | Which first names are used for professionals or owners but are not used as dog names? | SELECT first_name FROM Professionals UNION SELECT first_name FROM Owners EXCEPT SELECT name FROM Dogs | hard |
927 | dog_kennels | [
"Owners",
"Dogs",
"Professionals"
] | Find the first names that are used for professionals or owners but are not used as dog names. | SELECT first_name FROM Professionals UNION SELECT first_name FROM Owners EXCEPT SELECT name FROM Dogs | hard |
928 | dog_kennels | [
"Professionals",
"Treatments"
] | Which professional did not operate any treatment on dogs? List the professional's id, role and email. | SELECT professional_id , role_code , email_address FROM Professionals EXCEPT SELECT T1.professional_id , T1.role_code , T1.email_address FROM Professionals AS T1 JOIN Treatments AS T2 ON T1.professional_id = T2.professional_id | extra |
929 | dog_kennels | [
"Professionals",
"Treatments"
] | Give me the id, role and email of the professionals who did not perform any treatment on dogs. | SELECT professional_id , role_code , email_address FROM Professionals EXCEPT SELECT T1.professional_id , T1.role_code , T1.email_address FROM Professionals AS T1 JOIN Treatments AS T2 ON T1.professional_id = T2.professional_id | extra |
930 | dog_kennels | [
"Owners",
"Dogs"
] | Which owner owns the most dogs? List the owner id, first name and last name. | SELECT T1.owner_id , T2.first_name , T2.last_name FROM Dogs AS T1 JOIN Owners AS T2 ON T1.owner_id = T2.owner_id GROUP BY T1.owner_id ORDER BY count(*) DESC LIMIT 1 | extra |
931 | dog_kennels | [
"Owners",
"Dogs"
] | Return the owner id, first name and last name of the owner who has the most dogs. | SELECT T1.owner_id , T2.first_name , T2.last_name FROM Dogs AS T1 JOIN Owners AS T2 ON T1.owner_id = T2.owner_id GROUP BY T1.owner_id ORDER BY count(*) DESC LIMIT 1 | extra |
932 | dog_kennels | [
"Professionals",
"Treatments"
] | Which professionals have done at least two treatments? List the professional's id, role, and first name. | SELECT T1.professional_id , T1.role_code , T1.first_name FROM Professionals AS T1 JOIN Treatments AS T2 ON T1.professional_id = T2.professional_id GROUP BY T1.professional_id HAVING count(*) >= 2 | medium |
933 | dog_kennels | [
"Professionals",
"Treatments"
] | What are the id, role, and first name of the professionals who have performed two or more treatments? | SELECT T1.professional_id , T1.role_code , T1.first_name FROM Professionals AS T1 JOIN Treatments AS T2 ON T1.professional_id = T2.professional_id GROUP BY T1.professional_id HAVING count(*) >= 2 | medium |
934 | dog_kennels | [
"Dogs",
"Breeds"
] | What is the name of the breed with the most dogs? | SELECT T1.breed_name FROM Breeds AS T1 JOIN Dogs AS T2 ON T1.breed_code = T2.breed_code GROUP BY T1.breed_name ORDER BY count(*) DESC LIMIT 1 | extra |
935 | dog_kennels | [
"Dogs",
"Breeds"
] | Which breed do the most dogs have? Give me the breed name. | SELECT T1.breed_name FROM Breeds AS T1 JOIN Dogs AS T2 ON T1.breed_code = T2.breed_code GROUP BY T1.breed_name ORDER BY count(*) DESC LIMIT 1 | extra |
936 | dog_kennels | [
"Owners",
"Dogs",
"Treatments"
] | Which owner has paid for the most treatments on his or her dogs? List the owner id and last name. | SELECT T1.owner_id , T1.last_name FROM Owners AS T1 JOIN Dogs AS T2 ON T1.owner_id = T2.owner_id JOIN Treatments AS T3 ON T2.dog_id = T3.dog_id GROUP BY T1.owner_id ORDER BY count(*) DESC LIMIT 1 | extra |
937 | dog_kennels | [
"Owners",
"Dogs",
"Treatments"
] | Tell me the owner id and last name of the owner who spent the most on treatments of his or her dogs. | SELECT T1.owner_id , T1.last_name FROM Owners AS T1 JOIN Dogs AS T2 ON T1.owner_id = T2.owner_id JOIN Treatments AS T3 ON T2.dog_id = T3.dog_id GROUP BY T1.owner_id ORDER BY count(*) DESC LIMIT 1 | extra |
938 | dog_kennels | [
"Treatment_types",
"Treatments"
] | What is the description of the treatment type that costs the least money in total? | SELECT T1.treatment_type_description FROM Treatment_types AS T1 JOIN Treatments AS T2 ON T1.treatment_type_code = T2.treatment_type_code GROUP BY T1.treatment_type_code ORDER BY sum(cost_of_treatment) ASC LIMIT 1 | extra |
939 | dog_kennels | [
"Treatment_types",
"Treatments"
] | Give me the description of the treatment type whose total cost is the lowest. | SELECT T1.treatment_type_description FROM Treatment_types AS T1 JOIN Treatments AS T2 ON T1.treatment_type_code = T2.treatment_type_code GROUP BY T1.treatment_type_code ORDER BY sum(cost_of_treatment) ASC LIMIT 1 | extra |
940 | dog_kennels | [
"Owners",
"Dogs",
"Treatments"
] | Which owner has paid the largest amount of money in total for their dogs? Show the owner id and zip code. | SELECT T1.owner_id , T1.zip_code FROM Owners AS T1 JOIN Dogs AS T2 ON T1.owner_id = T2.owner_id JOIN Treatments AS T3 ON T2.dog_id = T3.dog_id GROUP BY T1.owner_id ORDER BY sum(T3.cost_of_treatment) DESC LIMIT 1 | extra |
941 | dog_kennels | [
"Owners",
"Dogs",
"Treatments"
] | Find the owner id and zip code of the owner who spent the most money in total for his or her dogs. | SELECT T1.owner_id , T1.zip_code FROM Owners AS T1 JOIN Dogs AS T2 ON T1.owner_id = T2.owner_id JOIN Treatments AS T3 ON T2.dog_id = T3.dog_id GROUP BY T1.owner_id ORDER BY sum(T3.cost_of_treatment) DESC LIMIT 1 | extra |
942 | dog_kennels | [
"Professionals",
"Treatments"
] | Which professionals have done at least two types of treatments? List the professional id and cell phone. | SELECT T1.professional_id , T1.cell_number FROM Professionals AS T1 JOIN Treatments AS T2 ON T1.professional_id = T2.professional_id GROUP BY T1.professional_id HAVING count(*) >= 2 | medium |
943 | dog_kennels | [
"Professionals",
"Treatments"
] | Find the id and cell phone of the professionals who operate two or more types of treatments. | SELECT T1.professional_id , T1.cell_number FROM Professionals AS T1 JOIN Treatments AS T2 ON T1.professional_id = T2.professional_id GROUP BY T1.professional_id HAVING count(*) >= 2 | medium |
944 | dog_kennels | [
"Professionals",
"Treatments"
] | What are the first name and last name of the professionals who have done treatment with cost below average? | SELECT DISTINCT T1.first_name , T1.last_name FROM Professionals AS T1 JOIN Treatments AS T2 WHERE cost_of_treatment < ( SELECT avg(cost_of_treatment) FROM Treatments ) | extra |
945 | dog_kennels | [
"Professionals",
"Treatments"
] | Which professionals have operated a treatment that costs less than the average? Give me theor first names and last names. | SELECT DISTINCT T1.first_name , T1.last_name FROM Professionals AS T1 JOIN Treatments AS T2 WHERE cost_of_treatment < ( SELECT avg(cost_of_treatment) FROM Treatments ) | extra |
946 | dog_kennels | [
"Professionals",
"Treatments"
] | List the date of each treatment, together with the first name of the professional who operated it. | SELECT T1.date_of_treatment , T2.first_name FROM Treatments AS T1 JOIN Professionals AS T2 ON T1.professional_id = T2.professional_id | medium |
947 | dog_kennels | [
"Professionals",
"Treatments"
] | What are the date and the operating professional's first name of each treatment? | SELECT T1.date_of_treatment , T2.first_name FROM Treatments AS T1 JOIN Professionals AS T2 ON T1.professional_id = T2.professional_id | medium |
948 | dog_kennels | [
"Treatments",
"treatment_types"
] | List the cost of each treatment and the corresponding treatment type description. | SELECT T1.cost_of_treatment , T2.treatment_type_description FROM Treatments AS T1 JOIN treatment_types AS T2 ON T1.treatment_type_code = T2.treatment_type_code | medium |
949 | dog_kennels | [
"Treatments",
"treatment_types"
] | What are the cost and treatment type description of each treatment? | SELECT T1.cost_of_treatment , T2.treatment_type_description FROM Treatments AS T1 JOIN treatment_types AS T2 ON T1.treatment_type_code = T2.treatment_type_code | medium |
950 | dog_kennels | [
"Owners",
"Dogs"
] | List each owner's first name, last name, and the size of his for her dog. | SELECT T1.first_name , T1.last_name , T2.size_code FROM Owners AS T1 JOIN Dogs AS T2 ON T1.owner_id = T2.owner_id | medium |
951 | dog_kennels | [
"Owners",
"Dogs"
] | What are each owner's first name, last name, and the size of their dog? | SELECT T1.first_name , T1.last_name , T2.size_code FROM Owners AS T1 JOIN Dogs AS T2 ON T1.owner_id = T2.owner_id | medium |
952 | dog_kennels | [
"Owners",
"Dogs"
] | List pairs of the owner's first name and the dogs's name. | SELECT T1.first_name , T2.name FROM Owners AS T1 JOIN Dogs AS T2 ON T1.owner_id = T2.owner_id | medium |
953 | dog_kennels | [
"Owners",
"Dogs"
] | What are each owner's first name and their dogs's name? | SELECT T1.first_name , T2.name FROM Owners AS T1 JOIN Dogs AS T2 ON T1.owner_id = T2.owner_id | medium |
954 | dog_kennels | [
"Dogs",
"Treatments"
] | List the names of the dogs of the rarest breed and the treatment dates of them. | SELECT T1.name , T2.date_of_treatment FROM Dogs AS T1 JOIN Treatments AS T2 ON T1.dog_id = T2.dog_id WHERE T1.breed_code = ( SELECT breed_code FROM Dogs GROUP BY breed_code ORDER BY count(*) ASC LIMIT 1 ) | extra |
955 | dog_kennels | [
"Dogs",
"Treatments"
] | Which dogs are of the rarest breed? Show their names and treatment dates. | SELECT T1.name , T2.date_of_treatment FROM Dogs AS T1 JOIN Treatments AS T2 ON T1.dog_id = T2.dog_id WHERE T1.breed_code = ( SELECT breed_code FROM Dogs GROUP BY breed_code ORDER BY count(*) ASC LIMIT 1 ) | extra |
956 | dog_kennels | [
"Owners",
"Dogs"
] | Which dogs are owned by someone who lives in Virginia? List the owner's first name and the dog's name. | SELECT T1.first_name , T2.name FROM Owners AS T1 JOIN Dogs AS T2 ON T1.owner_id = T2.owner_id WHERE T1.state = 'Virginia' | medium |
957 | dog_kennels | [
"Owners",
"Dogs"
] | Find the first names of owners living in Virginia and the names of dogs they own. | SELECT T1.first_name , T2.name FROM Owners AS T1 JOIN Dogs AS T2 ON T1.owner_id = T2.owner_id WHERE T1.state = 'Virginia' | medium |
958 | dog_kennels | [
"Dogs",
"Treatments"
] | What are the arriving date and the departing date of the dogs who have gone through a treatment? | SELECT DISTINCT T1.date_arrived , T1.date_departed FROM Dogs AS T1 JOIN Treatments AS T2 ON T1.dog_id = T2.dog_id | medium |
959 | dog_kennels | [
"Dogs",
"Treatments"
] | Find the arriving date and the departing date of the dogs that received a treatment. | SELECT DISTINCT T1.date_arrived , T1.date_departed FROM Dogs AS T1 JOIN Treatments AS T2 ON T1.dog_id = T2.dog_id | medium |
960 | dog_kennels | [
"Owners",
"Dogs"
] | List the last name of the owner owning the youngest dog. | SELECT T1.last_name FROM Owners AS T1 JOIN Dogs AS T2 ON T1.owner_id = T2.owner_id WHERE T2.age = ( SELECT max(age) FROM Dogs ) | extra |
961 | dog_kennels | [
"Owners",
"Dogs"
] | Who owns the youngest dog? Give me his or her last name. | SELECT T1.last_name FROM Owners AS T1 JOIN Dogs AS T2 ON T1.owner_id = T2.owner_id WHERE T2.age = ( SELECT max(age) FROM Dogs ) | extra |
962 | dog_kennels | [
"Professionals"
] | List the emails of the professionals who live in the state of Hawaii or the state of Wisconsin. | SELECT email_address FROM Professionals WHERE state = 'Hawaii' OR state = 'Wisconsin' | medium |
963 | dog_kennels | [
"Professionals"
] | What are the emails of the professionals living in either the state of Hawaii or the state of Wisconsin? | SELECT email_address FROM Professionals WHERE state = 'Hawaii' OR state = 'Wisconsin' | medium |
964 | dog_kennels | [
"Dogs"
] | What are the arriving date and the departing date of all the dogs? | SELECT date_arrived , date_departed FROM Dogs | medium |
965 | dog_kennels | [
"Dogs"
] | List the arrival date and the departure date for all the dogs. | SELECT date_arrived , date_departed FROM Dogs | medium |
966 | dog_kennels | [
"Treatments"
] | How many dogs went through any treatments? | SELECT count(DISTINCT dog_id) FROM Treatments | easy |
967 | dog_kennels | [
"Treatments"
] | Count the number of dogs that went through a treatment. | SELECT count(DISTINCT dog_id) FROM Treatments | easy |
968 | dog_kennels | [
"Treatments"
] | How many professionals have performed any treatment to dogs? | SELECT count(DISTINCT professional_id) FROM Treatments | easy |
969 | dog_kennels | [
"Treatments"
] | Find the number of professionals who have ever treated dogs. | SELECT count(DISTINCT professional_id) FROM Treatments | easy |
970 | dog_kennels | [
"professionals"
] | Which professionals live in a city containing the substring 'West'? List his or her role, street, city and state. | SELECT role_code , street , city , state FROM professionals WHERE city LIKE '%West%' | medium |
971 | dog_kennels | [
"professionals"
] | Find the role, street, city and state of the professionals living in a city that contains the substring 'West'. | SELECT role_code , street , city , state FROM professionals WHERE city LIKE '%West%' | medium |
972 | dog_kennels | [
"Owners"
] | Which owners live in the state whose name contains the substring 'North'? List his first name, last name and email. | SELECT first_name , last_name , email_address FROM Owners WHERE state LIKE '%North%' | medium |
973 | dog_kennels | [
"Owners"
] | Return the first name, last name and email of the owners living in a state whose name contains the substring 'North'. | SELECT first_name , last_name , email_address FROM Owners WHERE state LIKE '%North%' | medium |
974 | dog_kennels | [
"Dogs"
] | How many dogs have an age below the average? | SELECT count(*) FROM Dogs WHERE age < ( SELECT avg(age) FROM Dogs ) | hard |
975 | dog_kennels | [
"Dogs"
] | Count the number of dogs of an age below the average. | SELECT count(*) FROM Dogs WHERE age < ( SELECT avg(age) FROM Dogs ) | hard |
976 | dog_kennels | [
"Treatments"
] | How much does the most recent treatment cost? | SELECT cost_of_treatment FROM Treatments ORDER BY date_of_treatment DESC LIMIT 1 | medium |
977 | dog_kennels | [
"Treatments"
] | Show me the cost of the most recently performed treatment. | SELECT cost_of_treatment FROM Treatments ORDER BY date_of_treatment DESC LIMIT 1 | medium |
978 | dog_kennels | [
"Dogs",
"Treatments"
] | How many dogs have not gone through any treatment? | SELECT count(*) FROM Dogs WHERE dog_id NOT IN ( SELECT dog_id FROM Treatments ) | extra |
979 | dog_kennels | [
"treatments",
"dogs"
] | Tell me the number of dogs that have not received any treatment . | select count(*) from dogs where dog_id not in ( select dog_id from treatments ) | extra |
980 | dog_kennels | [
"Owners",
"Dogs"
] | How many owners temporarily do not have any dogs? | SELECT count(*) FROM Owners WHERE owner_id NOT IN ( SELECT owner_id FROM Dogs ) | extra |
981 | dog_kennels | [
"Owners",
"Dogs"
] | Find the number of owners who do not own any dogs at this moment. | SELECT count(*) FROM Owners WHERE owner_id NOT IN ( SELECT owner_id FROM Dogs ) | extra |
982 | dog_kennels | [
"Professionals",
"Treatments"
] | How many professionals did not operate any treatment on dogs? | SELECT count(*) FROM Professionals WHERE professional_id NOT IN ( SELECT professional_id FROM Treatments ) | extra |
983 | dog_kennels | [
"Professionals",
"Treatments"
] | Find the number of professionals who have not treated any dogs. | SELECT count(*) FROM Professionals WHERE professional_id NOT IN ( SELECT professional_id FROM Treatments ) | extra |
984 | dog_kennels | [
"Dogs"
] | List the dog name, age and weight of the dogs who have been abandoned? 1 stands for yes, and 0 stands for no. | SELECT name , age , weight FROM Dogs WHERE abandoned_yn = 1 | medium |
985 | dog_kennels | [
"Dogs"
] | What are the dog name, age and weight of the dogs that were abandoned? Note that 1 stands for yes, and 0 stands for no in the tables. | SELECT name , age , weight FROM Dogs WHERE abandoned_yn = 1 | medium |
986 | dog_kennels | [
"Dogs"
] | What is the average age of all the dogs? | SELECT avg(age) FROM Dogs | easy |
987 | dog_kennels | [
"Dogs"
] | Compute the average age of all the dogs. | SELECT avg(age) FROM Dogs | easy |
988 | dog_kennels | [
"Dogs"
] | What is the age of the oldest dog? | SELECT max(age) FROM Dogs | easy |
989 | dog_kennels | [
"Dogs"
] | Tell me the age of the oldest dog. | SELECT max(age) FROM Dogs | easy |
990 | dog_kennels | [
"Charges"
] | How much does each charge type costs? List both charge type and amount. | SELECT charge_type , charge_amount FROM Charges | medium |
991 | dog_kennels | [
"Charges"
] | List each charge type and its amount. | SELECT charge_type , charge_amount FROM Charges | medium |
992 | dog_kennels | [
"Charges"
] | How much does the most expensive charge type costs? | SELECT max(charge_amount) FROM Charges | easy |
993 | dog_kennels | [
"Charges"
] | What is the charge amount of the most expensive charge type? | SELECT max(charge_amount) FROM Charges | easy |
994 | dog_kennels | [
"professionals"
] | List the email, cell phone and home phone of all the professionals. | SELECT email_address , cell_number , home_phone FROM professionals | medium |
995 | dog_kennels | [
"professionals"
] | What are the email, cell phone and home phone of each professional? | SELECT email_address , cell_number , home_phone FROM professionals | medium |
996 | dog_kennels | [
"dogs"
] | What are all the possible breed type and size type combinations? | SELECT DISTINCT breed_code , size_code FROM dogs | medium |
997 | dog_kennels | [
"dogs"
] | Find the distinct breed type and size type combinations for dogs. | SELECT DISTINCT breed_code , size_code FROM dogs | medium |
998 | dog_kennels | [
"Treatment_types",
"Treatments",
"professionals"
] | List the first name of all the professionals along with the description of the treatment they have done. | SELECT DISTINCT T1.first_name , T3.treatment_type_description FROM professionals AS T1 JOIN Treatments AS T2 ON T1.professional_id = T2.professional_id JOIN Treatment_types AS T3 ON T2.treatment_type_code = T3.treatment_type_code | medium |
999 | dog_kennels | [
"Treatment_types",
"Treatments",
"professionals"
] | What are each professional's first name and description of the treatment they have performed? | SELECT DISTINCT T1.first_name , T3.treatment_type_description FROM professionals AS T1 JOIN Treatments AS T2 ON T1.professional_id = T2.professional_id JOIN Treatment_types AS T3 ON T2.treatment_type_code = T3.treatment_type_code | medium |
Subsets and Splits