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network_1 | 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 | Show name of all students who have some friends and also are liked by someone else. | [
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network_1 | 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 | What are the names of high schoolers who both have friends and are liked? | [
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network_1 | SELECT student_id , count(*) FROM Likes GROUP BY student_id | Count the number of likes for each student id. | [
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network_1 | SELECT T2.name , count(*) FROM Likes AS T1 JOIN Highschooler AS T2 ON T1.student_id = T2.id GROUP BY T1.student_id | Show the names of high schoolers who have likes, and numbers of likes for each. | [
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network_1 | SELECT T2.name , count(*) FROM Likes AS T1 JOIN Highschooler AS T2 ON T1.student_id = T2.id GROUP BY T1.student_id | What are the names of high schoolers who have likes, and how many likes does each have? | [
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network_1 | 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 | What is the name of the high schooler who has the greatest number of likes? | [
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network_1 | 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 | Give the name of the student with the most likes. | [
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network_1 | 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 | Show the names of students who have at least 2 likes. | [
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network_1 | 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 | What are the names of students who have 2 or more likes? | [
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network_1 | 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 | Show the names of students who have a grade higher than 5 and have at least 2 friends. | [
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network_1 | 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 | What are the names of high schoolers who have a grade of over 5 and have 2 or more friends? | [
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network_1 | SELECT count(*) FROM Likes AS T1 JOIN Highschooler AS T2 ON T1.student_id = T2.id WHERE T2.name = "Kyle" | How many likes does Kyle have? | [
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network_1 | SELECT count(*) FROM Likes AS T1 JOIN Highschooler AS T2 ON T1.student_id = T2.id WHERE T2.name = "Kyle" | Return the number of likes that the high schooler named Kyle has. | [
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network_1 | 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) | Find the average grade of all students who have some friends. | [
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dog_kennels | SELECT state FROM Owners INTERSECT SELECT state FROM Professionals | Which states have both owners and professionals living there? | [
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dog_kennels | SELECT avg(age) FROM Dogs WHERE dog_id IN ( SELECT dog_id FROM Treatments ) | What is the average age of the dogs who have gone through any treatments? | [
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dog_kennels | 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 | 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. | [
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dog_kennels | 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 | 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. | [
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dog_kennels | select name from dogs where dog_id not in ( select dog_id from treatments group by dog_id having sum(cost_of_treatment) > 1000 ) | Which dogs have not cost their owner more than 1000 for treatment ? List the dog names . | [
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dog_kennels | SELECT first_name FROM Professionals UNION SELECT first_name FROM Owners EXCEPT SELECT name FROM Dogs | Which first names are used for professionals or owners but are not used as dog names? | [
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dog_kennels | 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 | Which professional did not operate any treatment on dogs? List the professional's id, role and email. | [
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dog_kennels | 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 | Give me the id, role and email of the professionals who did not perform any treatment on dogs. | [
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dog_kennels | 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 | Which owner owns the most dogs? List the owner id, first name and last name. | [
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dog_kennels | 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 | Which professionals have done at least two treatments? List the professional's id, role, and first name. | [
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dog_kennels | 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 | What are the id, role, and first name of the professionals who have performed two or more treatments? | [
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dog_kennels | 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 | What is the name of the breed with the most dogs? | [
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dog_kennels | 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 | Which breed do the most dogs have? Give me the breed name. | [
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dog_kennels | 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 | Which owner has paid for the most treatments on his or her dogs? List the owner id and last name. | [
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dog_kennels | 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 | Tell me the owner id and last name of the owner who spent the most on treatments of his or her dogs. | [
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dog_kennels | 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 | What is the description of the treatment type that costs the least money in total? | [
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dog_kennels | 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 | Give me the description of the treatment type whose total cost is the lowest. | [
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dog_kennels | 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 | Which owner has paid the largest amount of money in total for their dogs? Show the owner id and zip code. | [
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dog_kennels | 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 | Find the owner id and zip code of the owner who spent the most money in total for his or her dogs. | [
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dog_kennels | 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 | Which professionals have done at least two types of treatments? List the professional id and cell phone. | [
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dog_kennels | 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 | Find the id and cell phone of the professionals who operate two or more types of treatments. | [
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dog_kennels | 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 ) | What are the first name and last name of the professionals who have done treatment with cost below average? | [
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dog_kennels | 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 ) | Which professionals have operated a treatment that costs less than the average? Give me theor first names and last names. | [
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dog_kennels | SELECT T1.date_of_treatment , T2.first_name FROM Treatments AS T1 JOIN Professionals AS T2 ON T1.professional_id = T2.professional_id | List the date of each treatment, together with the first name of the professional who operated it. | [
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dog_kennels | SELECT T1.date_of_treatment , T2.first_name FROM Treatments AS T1 JOIN Professionals AS T2 ON T1.professional_id = T2.professional_id | What are the date and the operating professional's first name of each treatment? | [
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dog_kennels | 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 | List the cost of each treatment and the corresponding treatment type description. | [
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dog_kennels | 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 | List each owner's first name, last name, and the size of his for her dog. | [
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dog_kennels | 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 | What are each owner's first name, last name, and the size of their dog? | [
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dog_kennels | SELECT T1.first_name , T2.name FROM Owners AS T1 JOIN Dogs AS T2 ON T1.owner_id = T2.owner_id | What are each owner's first name and their dogs's name? | [
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dog_kennels | 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 ) | List the names of the dogs of the rarest breed and the treatment dates of them. | [
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dog_kennels | 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 ) | Which dogs are of the rarest breed? Show their names and treatment dates. | [
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dog_kennels | 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' | Which dogs are owned by someone who lives in Virginia? List the owner's first name and the dog's name. | [
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dog_kennels | 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' | Find the first names of owners living in Virginia and the names of dogs they own. | [
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dog_kennels | SELECT DISTINCT T1.date_arrived , T1.date_departed FROM Dogs AS T1 JOIN Treatments AS T2 ON T1.dog_id = T2.dog_id | What are the arriving date and the departing date of the dogs who have gone through a treatment? | [
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dog_kennels | 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 ) | Who owns the youngest dog? Give me his or her last name. | [
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dog_kennels | SELECT email_address FROM Professionals WHERE state = 'Hawaii' OR state = 'Wisconsin' | List the emails of the professionals who live in the state of Hawaii or the state of Wisconsin. | [
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dog_kennels | SELECT count(DISTINCT dog_id) FROM Treatments | How many dogs went through any treatments? | [
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dog_kennels | SELECT count(DISTINCT professional_id) FROM Treatments | How many professionals have performed any treatment to dogs? | [
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CREATE TABLE Treatment_Types (
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dog_kennels | SELECT role_code , street , city , state FROM professionals WHERE city LIKE '%West%' | Which professionals live in a city containing the substring 'West'? List his or her role, street, city and state. | [
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dog_kennels | SELECT cost_of_treatment FROM Treatments ORDER BY date_of_treatment DESC LIMIT 1 | Show me the cost of the most recently performed treatment. | [
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dog_kennels | SELECT email_address , cell_number , home_phone FROM professionals | List the email, cell phone and home phone of all the professionals. | [
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dog_kennels | SELECT email_address , cell_number , home_phone FROM professionals | What are the email, cell phone and home phone of each professional? | [
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dog_kennels | SELECT DISTINCT breed_code , size_code FROM dogs | What are all the possible breed type and size type combinations? | [
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dog_kennels | 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 | List the first name of all the professionals along with the description of the treatment they have done. | [
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dog_kennels | 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 | What are each professional's first name and description of the treatment they have performed? | [
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"of",
"the",
"treatment",
"they",
"have",
"performed",
"?"
] | CREATE TABLE Breeds (
breed_code TEXT PRIMARY KEY,
breed_name TEXT
);
CREATE TABLE Charges (
charge_id NUMBER PRIMARY KEY,
charge_type TEXT,
charge_amount NUMBER
);
CREATE TABLE Sizes (
size_code TEXT PRIMARY KEY,
size_description TEXT
);
CREATE TABLE Treatment_Types (
treatment_type_code TEXT PRIMAR... |
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