Dludora/Bird-minidev-sqlserver-oracle huggingface dataset:
Database: superhero

Question: What is the percentage of superheroes who act in their own self-interest or make decisions based on their own moral code? Indicate how many of the said superheroes were published by Marvel Comics.
Evidence: published by Marvel Comics refers to publisher_name = 'Marvel Comics'; superheroes who act in their own self-interest or make decisions based on their own moral code refers to alignment = 'Bad'; calculation = MULTIPLY(DIVIDE(SUM(alignment = 'Bad); count(id)), 100)
SQL (SQLite): SELECT (CAST(COUNT(*) AS REAL) * 100 / (SELECT COUNT(*) FROM superhero)), CAST(SUM(CASE WHEN T2.publisher_name = 'Marvel Comics' THEN 1 ELSE 0 END) AS REAL) FROM superhero AS T1 INNER JOIN publisher AS T2 ON T1.publisher_id = T2.id INNER JOIN alignment AS T3 ON T3.id = T1.alignment_id WHERE T3.alignment = 'Bad'
SQL (SQL Server): SELECT (TRY_CAST(COUNT(*) AS FLOAT) * 100 / (SELECT COUNT(*) AS _agg1 FROM superhero)), TRY_CAST(SUM(CASE WHEN T2.publisher_name = 'Marvel Comics' THEN 1 ELSE 0 END) AS FLOAT) FROM superhero AS T1 INNER JOIN publisher AS T2 ON T1.publisher_id = T2.id INNER JOIN alignment AS T3 ON T3.id = T1.alignment_id WHERE T3.alignment = 'Bad'
SQL (Oracle): SELECT (S.BAD_CNT * 100 / NULLIF(T.TOTAL_CNT, 0)), S.MARVEL_CNT FROM (  SELECT COUNT(*) BAD_CNT,          SUM(CASE WHEN T2.publisher_name = 'Marvel Comics' THEN 1 ELSE 0 END) MARVEL_CNT   FROM superhero T1   INNER JOIN publisher T2 ON T1.publisher_id = T2.id   INNER JOIN alignment T3 ON T3.id = T1.alignment_id   WHERE T3.alignment = 'Bad') S CROSS JOIN (SELECT COUNT(*) TOTAL_CNT FROM superhero) T

Question: Between DC and Marvel Comics, which publisher has published more superheroes? Find the difference in the number of superheroes they have published.
Evidence: DC refers to publisher_name = 'DC Comics'; Marvel Comics refers to publisher_name = 'Marvel Comics'; calculation = SUBTRACT(SUM(publisher_name = 'Marvel Comics'), SUM(publisher_name = 'DC Comics'))
SQL (SQLite): SELECT SUM(CASE WHEN T2.publisher_name = 'Marvel Comics' THEN 1 ELSE 0 END) - SUM(CASE WHEN T2.publisher_name = 'DC Comics' THEN 1 ELSE 0 END) FROM superhero AS T1 INNER JOIN publisher AS T2 ON T1.publisher_id = T2.id
SQL (SQL Server): SELECT SUM(CASE WHEN T2.publisher_name = 'Marvel Comics' THEN 1 ELSE 0 END) - SUM(CASE WHEN T2.publisher_name = 'DC Comics' THEN 1 ELSE 0 END) AS _agg1 FROM superhero AS T1 INNER JOIN publisher AS T2 ON T1.publisher_id = T2.id
SQL (Oracle): SELECT SUM(CASE WHEN T2.publisher_name = 'Marvel Comics' THEN CAST(1 AS BINARY_DOUBLE) ELSE 0.0 END) - SUM(CASE WHEN T2.publisher_name = 'DC Comics' THEN CAST(1 AS BINARY_DOUBLE) ELSE 0.0 END) FROM superhero T1 INNER JOIN publisher T2 ON T1.publisher_id = T2.id

Question: Give the publisher ID of Star Trek.
Evidence: Star Trek is the publisher_name;
SQL (SQLite): SELECT id FROM publisher WHERE publisher_name = 'Star Trek'
SQL (SQL Server): SELECT id FROM publisher WHERE publisher_name = 'Star Trek'
SQL (Oracle): SELECT id FROM publisher WHERE publisher_name = 'Star Trek'
Database: superhero

Question: What is the total number of superheroes without full name?
Evidence: superheroes without full name refers to full_name IS NULL
SQL (SQLite): SELECT COUNT(id) FROM superhero WHERE full_name IS NULL
SQL (SQL Server): SELECT COUNT(id) AS _agg1 FROM superhero WHERE full_name IS NULL
SQL (Oracle): SELECT COUNT(id) FROM superhero WHERE full_name IS NULL

Question: What is the average weight of all female superheroes?
Evidence: female refers to gender = 'Female'; average weight refers to AVG(weight_kg)
SQL (SQLite): SELECT AVG(T1.weight_kg) FROM superhero AS T1 INNER JOIN gender AS T2 ON T1.gender_id = T2.id WHERE T2.gender = 'Female'
SQL (SQL Server): SELECT AVG(TRY_CAST(T1.weight_kg AS FLOAT)) AS _agg1 FROM superhero AS T1 INNER JOIN gender AS T2 ON T1.gender_id = T2.id WHERE T2.gender = 'Female'
SQL (Oracle): SELECT AVG(T1.weight_kg) FROM superhero T1 INNER JOIN gender T2 ON T1.gender_id = T2.id WHERE T2.gender = 'Female'

Question: List down at least five superpowers of male superheroes.
Evidence: male refers to gender = 'Male'; superpowers refers to power_name;
SQL (SQLite): SELECT T3.power_name FROM superhero AS T1 INNER JOIN hero_power AS T2 ON T1.id = T2.hero_id INNER JOIN superpower AS T3 ON T3.id = T2.power_id INNER JOIN gender AS T4 ON T4.id = T1.gender_id WHERE T4.gender = 'Male' LIMIT 5
SQL (SQL Server): SELECT TOP 5 T3.power_name FROM superhero AS T1 INNER JOIN hero_power AS T2 ON T1.id = T2.hero_id INNER JOIN superpower AS T3 ON T3.id = T2.power_id INNER JOIN gender AS T4 ON T4.id = T1.gender_id WHERE T4.gender = 'Male'
SQL (Oracle): SELECT T3.power_name FROM superhero T1 INNER JOIN hero_power T2 ON T1.id = T2.hero_id INNER JOIN superpower T3 ON T3.id = T2.power_id INNER JOIN gender T4 ON T4.id = T1.gender_id WHERE T4.gender = 'Male' FETCH FIRST 5 ROWS ONLY

Question: Among the superheroes with height from 170 to 190, list the names of the superheroes with no eye color.
Evidence: height from 170 to 190 refers to height_cm BETWEEN 170 AND 190; no eye color refers to colour = 'No Colour'
SQL (SQLite): SELECT DISTINCT T1.superhero_name FROM superhero AS T1 INNER JOIN colour AS T2 ON T1.eye_colour_id = T2.id WHERE T1.height_cm BETWEEN 170 AND 190 AND T2.colour = 'No Colour'
SQL (SQL Server): SELECT DISTINCT T1.superhero_name FROM superhero AS T1 INNER JOIN colour AS T2 ON T1.eye_colour_id = T2.id WHERE T1.height_cm BETWEEN 170 AND 190 AND T2.colour = 'No Colour'
SQL (Oracle): SELECT DISTINCT T1.superhero_name FROM superhero T1 INNER JOIN colour T2 ON T1.eye_colour_id = T2.id WHERE T1.height_cm BETWEEN 170 AND 190 AND T2.colour = 'No Colour'
Database: superhero

Question: Provide the hair colour of the human superhero who is 185 cm tall.
Evidence: 185 cm tall refers to height_cm = 185; human superhero refers to race = 'human'; hair colour refers to colour where hair_colour_id = colour.id;
SQL (SQLite): SELECT DISTINCT T3.colour FROM superhero AS T1 INNER JOIN race AS T2 ON T1.race_id = T2.id INNER JOIN colour AS T3 ON T1.hair_colour_id = T3.id WHERE T1.height_cm = 185 AND T2.race = 'Human'
SQL (SQL Server): SELECT DISTINCT T3.colour FROM superhero AS T1 INNER JOIN race AS T2 ON T1.race_id = T2.id INNER JOIN colour AS T3 ON T1.hair_colour_id = T3.id WHERE T1.height_cm = 185 AND T2.race = 'Human'
SQL (Oracle): SELECT DISTINCT T3.colour FROM superhero T1 INNER JOIN race T2 ON T1.race_id = T2.id INNER JOIN colour T3 ON T1.hair_colour_id = T3.id WHERE T1.height_cm = 185 AND T2.race = 'Human'

Question: In superheroes with height between 150 to 180, what is the percentage of heroes published by Marvel Comics?
Evidence: height between 150 to 180 refers to height_cm BETWEEN 150 AND 180; heroes published by Marvel Comics refers to publisher_name = 'Marvel Comics'; calculation = MULTIPLY(DIVIDE(SUM(publisher.id = 13)), COUNT(publisher.id), 100)
SQL (SQLite): SELECT CAST(COUNT(CASE WHEN T2.publisher_name = 'Marvel Comics' THEN 1 ELSE NULL END) AS REAL) * 100 / COUNT(T1.id) FROM superhero AS T1 INNER JOIN publisher AS T2 ON T1.publisher_id = T2.id WHERE T1.height_cm BETWEEN 150 AND 180
SQL (SQL Server): SELECT TRY_CAST(COUNT(CASE WHEN T2.publisher_name = 'Marvel Comics' THEN 1 ELSE NULL END) AS FLOAT) * 100 / COUNT(T1.id) AS _agg1 FROM superhero AS T1 INNER JOIN publisher AS T2 ON T1.publisher_id = T2.id WHERE T1.height_cm BETWEEN 150 AND 180
SQL (Oracle): SELECT CAST(COUNT(CASE WHEN T2.publisher_name = 'Marvel Comics' THEN 1 ELSE NULL END) AS BINARY_DOUBLE) * 100 / COUNT(T1.id) FROM superhero T1 INNER JOIN publisher T2 ON T1.publisher_id = T2.id WHERE T1.height_cm BETWEEN 150 AND 180

Question: Among the male superheroes, list the super hero names of superheroes with weight greater than the 79% average weight of all superheroes.
Evidence: super hero names refers to superhero_name;male superheros refers to gender = 'Male';Calculation = weight_kg > MULTIPLY(AVG(weight_kg), 0.79)
SQL (SQLite): SELECT T1.superhero_name FROM superhero AS T1 INNER JOIN gender AS T2 ON T1.gender_id = T2.id WHERE T2.gender = 'Male' AND T1.weight_kg * 100 > ( SELECT AVG(weight_kg) FROM superhero ) * 79
SQL (SQL Server): SELECT T1.superhero_name FROM superhero AS T1 INNER JOIN gender AS T2 ON T1.gender_id = T2.id WHERE T2.gender = 'Male' AND T1.weight_kg * 100 > ( SELECT AVG(TRY_CAST(weight_kg AS FLOAT)) AS _agg1 FROM superhero ) * 79
SQL (Oracle): SELECT T1.superhero_name FROM superhero T1 INNER JOIN gender T2 ON T1.gender_id = T2.id WHERE T2.gender = 'Male' AND T1.weight_kg * 100 > ( SELECT AVG(weight_kg) FROM superhero ) * 79
Database: superhero

Question: What are the superpowers of heroes with ID 1?
Evidence: superpowers refers to power_name; heroes with ID 1 refers to hero_id = 1;
SQL (SQLite): SELECT DISTINCT T2.power_name FROM hero_power AS T1 INNER JOIN superpower AS T2 ON T1.power_id = T2.id WHERE T1.hero_id = 1
SQL (SQL Server): SELECT DISTINCT T2.power_name FROM hero_power AS T1 INNER JOIN superpower AS T2 ON T1.power_id = T2.id WHERE T1.hero_id = 1
SQL (Oracle): SELECT DISTINCT T2.power_name FROM hero_power T1 INNER JOIN superpower T2 ON T1.power_id = T2.id WHERE T1.hero_id = 1

Question: How many heroes have stealth power?
Evidence: stealth power refers to power_name = 'Stealth';
SQL (SQLite): SELECT COUNT(T1.hero_id) FROM hero_power AS T1 INNER JOIN superpower AS T2 ON T1.power_id = T2.id WHERE T2.power_name = 'Stealth'
SQL (SQL Server): SELECT COUNT(T1.hero_id) AS _agg1 FROM hero_power AS T1 INNER JOIN superpower AS T2 ON T1.power_id = T2.id WHERE T2.power_name = 'Stealth'
SQL (Oracle): SELECT COUNT(T1.hero_id) FROM hero_power T1 INNER JOIN superpower T2 ON T1.power_id = T2.id WHERE T2.power_name = 'Stealth'

Question: What is the hero's full name with the highest attribute in strength?
Evidence: highest attribute in strength refers to MAX(attribute_value) WHERE attribute_name = 'strength';
SQL (SQLite): SELECT T1.full_name FROM superhero AS T1 INNER JOIN hero_attribute AS T2 ON T1.id = T2.hero_id INNER JOIN attribute AS T3 ON T2.attribute_id = T3.id WHERE T3.attribute_name = 'Strength' ORDER BY T2.attribute_value DESC LIMIT 1
SQL (SQL Server): SELECT T1.full_name FROM superhero AS T1 INNER JOIN hero_attribute AS T2 ON T1.id = T2.hero_id INNER JOIN attribute AS T3 ON T2.attribute_id = T3.id WHERE T3.attribute_name = 'Strength' ORDER BY T2.attribute_value DESC OFFSET 0 ROWS FETCH NEXT 1 ROWS ONLY
SQL (Oracle): SELECT T1.full_name FROM superhero T1 INNER JOIN hero_attribute T2 ON T1.id = T2.hero_id INNER JOIN attribute T3 ON T2.attribute_id = T3.id WHERE T3.attribute_name = 'Strength' ORDER BY T2.attribute_value DESC NULLS LAST FETCH FIRST 1 ROWS ONLY
Database: superhero

Question: Which superhero has the most durability published by Dark Horse Comics?
Evidence: which superhero refers to superhero_name; most durability refers to MAX(attribute_value) WHERE attribute_name = 'durability'; published by Dark Horse Comics refers to publisher_name = 'Dark Horse Comics';
SQL (SQLite): SELECT T1.superhero_name FROM superhero AS T1 INNER JOIN hero_attribute AS T2 ON T1.id = T2.hero_id INNER JOIN attribute AS T3 ON T3.id = T2.attribute_id INNER JOIN publisher AS T4 ON T4.id = T1.publisher_id WHERE T4.publisher_name = 'Dark Horse Comics' AND T3.attribute_name = 'Durability' ORDER BY T2.attribute_value DESC LIMIT 1
SQL (SQL Server): SELECT T1.superhero_name FROM superhero AS T1 INNER JOIN hero_attribute AS T2 ON T1.id = T2.hero_id INNER JOIN attribute AS T3 ON T3.id = T2.attribute_id INNER JOIN publisher AS T4 ON T4.id = T1.publisher_id WHERE T4.publisher_name = 'Dark Horse Comics' AND T3.attribute_name = 'Durability' ORDER BY T2.attribute_value DESC OFFSET 0 ROWS FETCH NEXT 1 ROWS ONLY
SQL (Oracle): SELECT T1.superhero_name FROM superhero T1 INNER JOIN hero_attribute T2 ON T1.id = T2.hero_id INNER JOIN attribute T3 ON T3.id = T2.attribute_id INNER JOIN publisher T4 ON T4.id = T1.publisher_id WHERE T4.publisher_name = 'Dark Horse Comics' AND T3.attribute_name = 'Durability' ORDER BY T2.attribute_value DESC NULLS LAST FETCH FIRST 1 ROWS ONLY

Question: List the eyes, hair and skin colour of all female superheroes published by Dark Horse Comics.
Evidence: eyes refers to eye_colour_id; hair refers to hair_colour_id; skin colour refers to skin_colour_id; female superheroes refers to gender = 'Female'; published by Dark Horse Comics refers to publisher_name = 'Dark Horse Comics';
SQL (SQLite): SELECT T1.eye_colour_id, T1.hair_colour_id, T1.skin_colour_id FROM superhero AS T1 INNER JOIN publisher AS T2 ON T2.id = T1.publisher_id INNER JOIN gender AS T3 ON T3.id = T1.gender_id WHERE T2.publisher_name = 'Dark Horse Comics' AND T3.gender = 'Female'
SQL (SQL Server): SELECT T1.eye_colour_id, T1.hair_colour_id, T1.skin_colour_id FROM superhero AS T1 INNER JOIN publisher AS T2 ON T2.id = T1.publisher_id INNER JOIN gender AS T3 ON T3.id = T1.gender_id WHERE T2.publisher_name = 'Dark Horse Comics' AND T3.gender = 'Female'
SQL (Oracle): SELECT T1.eye_colour_id, T1.hair_colour_id, T1.skin_colour_id FROM superhero T1 INNER JOIN publisher T2 ON T2.id = T1.publisher_id INNER JOIN gender T3 ON T3.id = T1.gender_id WHERE T2.publisher_name = 'Dark Horse Comics' AND T3.gender = 'Female'
Database: superhero

Question: Which superhero has the same eyes, hair and skin colour? Indicate the publisher of the superhero.
Evidence: which superhero refers to superhero_name; the same eyes, hair and skin colour refers to hair_colour_id = skin_colour_id AND hair_colour_id = eye_colour_id; publisher refers to publisher_name;
SQL (SQLite): SELECT T1.superhero_name, T2.publisher_name FROM superhero AS T1 INNER JOIN publisher AS T2 ON T1.publisher_id = T2.id WHERE T1.eye_colour_id = T1.hair_colour_id AND T1.eye_colour_id = T1.skin_colour_id
SQL (SQL Server): SELECT T1.superhero_name, T2.publisher_name FROM superhero AS T1 INNER JOIN publisher AS T2 ON T1.publisher_id = T2.id WHERE T1.eye_colour_id = T1.hair_colour_id AND T1.eye_colour_id = T1.skin_colour_id
SQL (Oracle): SELECT T1.superhero_name, T2.publisher_name FROM superhero T1 INNER JOIN publisher T2 ON T1.publisher_id = T2.id WHERE T1.eye_colour_id = T1.hair_colour_id AND T1.eye_colour_id = T1.skin_colour_id

Question: What is the percentage of blue female superheroes among all female superheroes?
Evidence: percentage = MULTIPLY(DIVIDE(SUM(colour = 'Blue' WHERE gender = 'Female'), COUNT(gender = 'Female')), 100); blue refers to the color = 'Blue' WHERE skin_colour_id = colour.id; female refers to gender = 'Female';
SQL (SQLite): SELECT CAST(COUNT(CASE WHEN T3.colour = 'Blue' THEN T1.id ELSE NULL END) AS REAL) * 100 / COUNT(T1.id) FROM superhero AS T1 INNER JOIN gender AS T2 ON T1.gender_id = T2.id INNER JOIN colour AS T3 ON T1.skin_colour_id = T3.id WHERE T2.gender = 'Female'
SQL (SQL Server): SELECT TRY_CAST(COUNT(CASE WHEN T3.colour = 'Blue' THEN T1.id ELSE NULL END) AS FLOAT) * 100 / COUNT(T1.id) AS _agg1 FROM superhero AS T1 INNER JOIN gender AS T2 ON T1.gender_id = T2.id INNER JOIN colour AS T3 ON T1.skin_colour_id = T3.id WHERE T2.gender = 'Female'
SQL (Oracle): SELECT CAST(COUNT(CASE WHEN T3.colour = 'Blue' THEN T1.id ELSE NULL END) AS BINARY_DOUBLE) * 100 / COUNT(T1.id) FROM superhero T1 INNER JOIN gender T2 ON T1.gender_id = T2.id INNER JOIN colour T3 ON T1.skin_colour_id = T3.id WHERE T2.gender = 'Female'

Question: How many powers does Amazo hero have?
Evidence: Amazo hero refers to superhero_name = 'Amazo';
SQL (SQLite): SELECT COUNT(T1.power_id) FROM hero_power AS T1 INNER JOIN superhero AS T2 ON T1.hero_id = T2.id WHERE T2.superhero_name = 'Amazo'
SQL (SQL Server): SELECT COUNT(T1.power_id) AS _agg1 FROM hero_power AS T1 INNER JOIN superhero AS T2 ON T1.hero_id = T2.id WHERE T2.superhero_name = 'Amazo'
SQL (Oracle): SELECT COUNT(T1.power_id) FROM hero_power T1 INNER JOIN superhero T2 ON T1.hero_id = T2.id WHERE T2.superhero_name = 'Amazo'
Database: superhero

Question: Provide the heights of the heroes whose eye colours are amber.
Evidence: heights of the heroes refers to height_cm; eye colours are amber refers to colour.colour = 'Amber' WHERE eye_colour_id = colour.id;
SQL (SQLite): SELECT T1.height_cm FROM superhero AS T1 INNER JOIN colour AS T2 ON T1.eye_colour_id = T2.id WHERE T2.colour = 'Amber'
SQL (SQL Server): SELECT T1.height_cm FROM superhero AS T1 INNER JOIN colour AS T2 ON T1.eye_colour_id = T2.id WHERE T2.colour = 'Amber'
SQL (Oracle): SELECT T1.height_cm FROM superhero T1 INNER JOIN colour T2 ON T1.eye_colour_id = T2.id WHERE T2.colour = 'Amber'

Question: List the heroes' names whose eyes and hair colours are both black.
Evidence: heroes' names refers to superhero_name; eyes and hair colours are both black refers to eye_colour_id AND hair_colour_id WHERE colour.colour = 'Black';
SQL (SQLite): SELECT T1.superhero_name FROM superhero AS T1 INNER JOIN colour AS T2 ON T1.eye_colour_id = T2.id AND T1.hair_colour_id = T2.id WHERE T2.colour = 'Black'
SQL (SQL Server): SELECT T1.superhero_name FROM superhero AS T1 INNER JOIN colour AS T2 ON T1.eye_colour_id = T2.id AND T1.hair_colour_id = T2.id WHERE T2.colour = 'Black'
SQL (Oracle): SELECT T1.superhero_name FROM superhero T1 INNER JOIN colour T2 ON T1.eye_colour_id = T2.id AND T1.hair_colour_id = T2.id WHERE T2.colour = 'Black'

Question: Describe the names of neutral alignment superheroes.
Evidence: names of superheroes refers to superhero_name; neutral alignment refers to alignment = 'Neutral';
SQL (SQLite): SELECT T1.superhero_name FROM superhero AS T1 INNER JOIN alignment AS T2 ON T1.alignment_id = T2.id WHERE T2.alignment = 'Neutral'
SQL (SQL Server): SELECT T1.superhero_name FROM superhero AS T1 INNER JOIN alignment AS T2 ON T1.alignment_id = T2.id WHERE T2.alignment = 'Neutral'
SQL (Oracle): SELECT T1.superhero_name FROM superhero T1 INNER JOIN alignment T2 ON T1.alignment_id = T2.id WHERE T2.alignment = 'Neutral'

Question: How many heroes have the highest attribute value in strength?
Evidence: highest attribute value in strength refers to MAX(attribute_value) WHERE attribute_name = 'Strength';
SQL (SQLite): SELECT COUNT(T1.hero_id) FROM hero_attribute AS T1 INNER JOIN attribute AS T2 ON T1.attribute_id = T2.id WHERE T2.attribute_name = 'Strength' AND T1.attribute_value = ( SELECT MAX(attribute_value) FROM hero_attribute )
SQL (SQL Server): SELECT COUNT(T1.hero_id) AS _agg1 FROM hero_attribute AS T1 INNER JOIN attribute AS T2 ON T1.attribute_id = T2.id WHERE T2.attribute_name = 'Strength' AND T1.attribute_value = ( SELECT MAX(attribute_value) AS _agg2 FROM hero_attribute )
SQL (Oracle): SELECT COUNT(T1.hero_id) FROM hero_attribute T1 INNER JOIN attribute T2 ON T1.attribute_id = T2.id WHERE T2.attribute_name = 'Strength' AND T1.attribute_value = ( SELECT MAX(attribute_value) FROM hero_attribute )
Database: superhero

Question: How many percent of female heroes were published by Marvel Comics?
Evidence: percent = MULTIPLY(DIVIDE(SUM(gender = 'Female' WHERE publisher_name = 'Marvel Comics'), COUNT(publisher_name = 'Marvel Comics')), 100); female heroes refers to gender = 'Female'; Marvel Comics refers to publisher_name = 'Marvel Comics';
SQL (SQLite): SELECT CAST(COUNT(CASE WHEN T3.gender = 'Female' AND T2.publisher_name = 'Marvel Comics' THEN 1 ELSE NULL END) AS REAL) / COUNT(CASE WHEN T2.publisher_name = 'Marvel Comics' THEN 1 ELSE NULL END) * 100 FROM superhero AS T1 INNER JOIN publisher AS T2 ON T1.publisher_id = T2.id INNER JOIN gender AS T3 ON T1.gender_id = T3.id
SQL (SQL Server): SELECT TRY_CAST(COUNT(CASE WHEN T3.gender = 'Female' AND T2.publisher_name = 'Marvel Comics' THEN 1 ELSE NULL END) AS FLOAT) / COUNT(CASE WHEN T2.publisher_name = 'Marvel Comics' THEN 1 ELSE NULL END) * 100 FROM superhero AS T1 INNER JOIN publisher AS T2 ON T1.publisher_id = T2.id INNER JOIN gender AS T3 ON T1.gender_id = T3.id
SQL (Oracle): SELECT CAST(COUNT(CASE WHEN T3.gender = 'Female' AND T2.publisher_name = 'Marvel Comics' THEN 1 ELSE NULL END) AS BINARY_DOUBLE) / COUNT(CASE WHEN T2.publisher_name = 'Marvel Comics' THEN 1 ELSE NULL END) * 100 FROM superhero T1 INNER JOIN publisher T2 ON T1.publisher_id = T2.id INNER JOIN gender T3 ON T1.gender_id = T3.id

Question: Calculate the difference between Emil Blonsky's weight and Charles Chandler's weight.
Evidence: difference = SUBTRACT(SUM(weight_kg WHERE full_name = 'Emil Blonsky'), SUM(weight_kg WHERE full_name = 'Charles Chandler')); Emil Blonsky is the full name of superhero; Charles Chandler is the full name of superhero;
SQL (SQLite): SELECT ( SELECT weight_kg FROM superhero WHERE full_name LIKE 'Emil Blonsky' ) - ( SELECT weight_kg FROM superhero WHERE full_name LIKE 'Charles Chandler' ) AS CALCULATE
SQL (SQL Server): SELECT ( SELECT weight_kg FROM superhero WHERE full_name LIKE 'Emil Blonsky' ) - ( SELECT weight_kg FROM superhero WHERE full_name LIKE 'Charles Chandler' ) AS CALCULATE
SQL (Oracle): SELECT ( SELECT weight_kg FROM superhero WHERE full_name LIKE 'Emil Blonsky' ) - ( SELECT weight_kg FROM superhero WHERE full_name LIKE 'Charles Chandler' ) CALCULATE

Question: Calculate the average height for all superhero.
Evidence: average = DIVIDE(SUM(height_cm), COUNT(all heros));
SQL (SQLite): SELECT CAST(SUM(height_cm) AS REAL) / COUNT(id) FROM superhero
SQL (SQL Server): SELECT TRY_CAST(SUM(height_cm) AS FLOAT) / COUNT(id) AS _agg1 FROM superhero
SQL (Oracle): SELECT CAST(SUM(height_cm) AS BINARY_DOUBLE) / COUNT(id) FROM superhero
Database: superhero

Question: What is Abomination's superpower?
Evidence: Abomination refers to superhero_name = 'Abomination'; superpower refers to power_name;
SQL (SQLite): SELECT T3.power_name FROM superhero AS T1 INNER JOIN hero_power AS T2 ON T1.id = T2.hero_id INNER JOIN superpower AS T3 ON T2.power_id = T3.id WHERE T1.superhero_name = 'Abomination'
SQL (SQL Server): SELECT T3.power_name FROM superhero AS T1 INNER JOIN hero_power AS T2 ON T1.id = T2.hero_id INNER JOIN superpower AS T3 ON T2.power_id = T3.id WHERE T1.superhero_name = 'Abomination'
SQL (Oracle): SELECT T3.power_name FROM superhero T1 INNER JOIN hero_power T2 ON T1.id = T2.hero_id INNER JOIN superpower T3 ON T2.power_id = T3.id WHERE T1.superhero_name = 'Abomination'

Question: Which hero was the fastest?
Evidence: which hero refers to superhero_name; fastest refers to MAX(attribute_value) WHERE attribute_name = 'Speed';
SQL (SQLite): SELECT T1.superhero_name FROM superhero AS T1 INNER JOIN hero_attribute AS T2 ON T1.id = T2.hero_id INNER JOIN attribute AS T3 ON T2.attribute_id = T3.id WHERE T3.attribute_name = 'Speed' ORDER BY T2.attribute_value DESC LIMIT 1
SQL (SQL Server): SELECT T1.superhero_name FROM superhero AS T1 INNER JOIN hero_attribute AS T2 ON T1.id = T2.hero_id INNER JOIN attribute AS T3 ON T2.attribute_id = T3.id WHERE T3.attribute_name = 'Speed' ORDER BY T2.attribute_value DESC OFFSET 0 ROWS FETCH NEXT 1 ROWS ONLY
SQL (Oracle): SELECT T1.superhero_name FROM superhero T1 INNER JOIN hero_attribute T2 ON T1.id = T2.hero_id INNER JOIN attribute T3 ON T2.attribute_id = T3.id WHERE T3.attribute_name = 'Speed' ORDER BY T2.attribute_value DESC NULLS LAST FETCH FIRST 1 ROWS ONLY

Question: State all of 3-D Man's attributes along with their values.
Evidence: 3-D Man is the superhero_name. attributes refers to attribute_name; values refers to attribute_value;
SQL (SQLite): SELECT T3.attribute_name, T2.attribute_value FROM superhero AS T1 INNER JOIN hero_attribute AS T2 ON T1.id = T2.hero_id INNER JOIN attribute AS T3 ON T2.attribute_id = T3.id WHERE T1.superhero_name = '3-D Man'
SQL (SQL Server): SELECT T3.attribute_name, T2.attribute_value FROM superhero AS T1 INNER JOIN hero_attribute AS T2 ON T1.id = T2.hero_id INNER JOIN attribute AS T3 ON T2.attribute_id = T3.id WHERE T1.superhero_name = '3-D Man'
SQL (Oracle): SELECT T3.attribute_name, T2.attribute_value FROM superhero T1 INNER JOIN hero_attribute T2 ON T1.id = T2.hero_id INNER JOIN attribute T3 ON T2.attribute_id = T3.id WHERE T1.superhero_name = '3-D Man'