行业网站建设方案,百度云图片转wordpress,做网站公司在丹麦,做背景图 网站文章目录 组合两个表第二高的薪水第N高的薪水分数排名连续出现的数字超过经理收入的员工查找重复的电子邮件从不订购的客户部门工资最高的员工部门工资前三高的所有员工删除重复的电子邮箱上升的温度游戏玩法分析Ⅰ游戏玩法Ⅳ 组合两个表
SELECT firstName,lastName,city,stat… 文章目录 组合两个表第二高的薪水第N高的薪水分数排名连续出现的数字超过经理收入的员工查找重复的电子邮件从不订购的客户部门工资最高的员工部门工资前三高的所有员工删除重复的电子邮箱上升的温度游戏玩法分析Ⅰ游戏玩法Ⅳ 组合两个表
SELECT firstName,lastName,city,state FROM Person LEFT JOIN Address ON Person.PersonIdAddress.PersonId第二高的薪水
SELECT
IFNULL(
(SELECT DISTINCT salary FROM Employee Order by salary desc LIMIT 1,1),null
) AS SecondHighestSalary第N高的薪水
CREATE FUNCTION getNthHighestSalary(N INT) RETURNS INT
BEGIN
SET N:N-1;RETURN (# Write your MySQL query statement below.SELECT IFNULL((SELECT DISTINCT salary FROM Employee ORDER BY salary DESC LIMIT N,1),NULL) );
END分数排名
SELECT score, (SELECT count(DISTINCT score)1 FROM Scores S2 Where S2.scoreScores.score) AS rank FROM Scores ORDER BY score DESC连续出现的数字
SELECT DISTINCT l1.num AS ConsecutiveNums FROM Logs l1,Logs l2,Logs l3 WHERE l1.idl2.id-1 AND l2.idl3.id-1 AND l1.numl2.num AND l2.numl3.num超过经理收入的员工
SELECT e.name AS Employee FROM Employee e,Employee e2 WHERE e2.ide.managerId AND e.salary e2.salary查找重复的电子邮件
SELECT DISTINCT p1.email AS Email FROM Person p1,Person p2 WHERE p1.emailp2.email AND p1.id !p2.id从不订购的客户
SELECT name AS Customers FROM Customers WHERE id NOT IN (SELECT DISTINCT customerId FROM Orders)部门工资最高的员工
SELECT Department.name AS Department,e.name AS Employee,e.salary AS Salary FROM Department,Employee e WHERE Department.ide.departmentId AND e.salary ALL(SELECT salary FROM Employee e2 WHERE e.departmentId e2.departmentId)部门工资前三高的所有员工
SELECT Department.name AS Department, e.name AS Employee,e.salary AS Salary FROM Employee e,Department WHERE e.departmentIdDepartment.id AND
(SELECT COUNT(Distinct e2.salary) FROM Employee e2 WHERE e2.departmentIde.departmentId AND e2.salarye.salary)3 ORDER BY e.departmentId ASC删除重复的电子邮箱
DELETE p2 FROM Person p1,Person p2 WHERE p1.emailp2.email AND p1.idp2.id上升的温度
SELECT w2.id AS Id FROM Weather w1,Weather w2 WHERE DATEDIFF(w2.recordDate,w1.recordDate)1 AND w1.Temperature w2.TemperatureDATEDIFF函数用于计算两个日期的天数差
游戏玩法分析Ⅰ
SELECT player_id,MIN(event_date) AS first_login FROM Activity GROUP BY player_id游戏玩法Ⅳ
SELECT ROUND(
(SELECT COUNT(DISTINCT b.player_id) FROM Activity b WHERE (b.player_id,b.event_date) IN
(
SELECT a.player_id,DATE(MIN(a.event_date)INTERVAL 1 DAY) FROM Activity a GROUP BY a.player_id
))
/
(SELECT COUNT(DISTINCT player_id) FROM Activity),2
) AS fraction
