data_yep

[문제]  [MySQL]SELECT MAX(salary * months) AS earnings , COUNT(*)FROM EmployeeGROUP BY salary*monthsORDER BY earnings DESCLIMIT 1

[문제]  [MySQL]SELECT CEIL(AVG(Salary) - AVG(REPLACE(Salary, 0, ''))) AS OutputFROM EMPLOYEES

[문제]  [MySQL]SELECT MAX(POPULATION) - MIN(POPULATION) AS OutputFROM CITY

[문제]  [MySQL]SELECT SUM(POPULATION) AS OutputFROM CITYWHERE COUNTRYCODE = 'JPN'

[문제]  [MySQL]SELECT FLOOR(AVG(POPULATION))FROM CITY

[문제]  [MySQL]SELECT AVG(POPULATION) AS OutputFROM CITYWHERE DISTRICT = 'California'

[문제]  [MySQL]SELECT SUM(POPULATION) AS OutputFROM CITYWHERE DISTRICT = 'California'

[문제]  [MySQL]SELECT COUNT(DISTINCT NAME) AS OutputFROM CITYWHERE POPULATION > 100000

[문제]  [MySQL]SELECT c.* , COUNT(DISTINCT l.lead_manager_code) AS LM , COUNT(DISTINCT s.senior_manager_code) AS CM , COUNT(DISTINCT m.manager_code) AS M , COUNT(DISTINCT e.employee_code) AS EFROM Company AS c LEFT JOIN Lead_Manager AS l ON c.company_code = l.company_code LEFT JOIN Senior_Manager AS s ON l.lead_manager_code = s.lead_manager_code LEFT JOIN Manager AS m ON s..

[문제]  [MySQL]SELECT DISTINCT b1.N , CASE WHEN b1.P IS NULL THEN 'Root' WHEN b2.P IS NULL THEN 'Leaf' ELSE 'Inner' END AS typeFROM BST AS b1 LEFT JOIN BST AS b2 ON b1.N = b2.PORDER BY b1.N ASC