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Combine Two Tables
解题思路：
合并table Person 和Address，其中提示说不管Address里有没有personId，Person里的记录都要显示全部。这是典型的 left join 功能。
select P.FirstName, P.LastName, A.City, A.State
from Person P
left outer join Address A
on P.PersonId = A.PersonId
知识点：left join (left outer join) 关键字从左表返回所有的行，即使在右表中没有匹配的行
Second Highest Salary
解题思路：
找到第二大的数字。第一反应就是先找到 max，然后从比max小的数里面找个最大的。
select max(Salary) from Employee
where Salary <
(select max(Salary) from Employee)
知识点：max 聚合函数，不能跟where一起，如果要把聚合函数作为条件，用having
Nth Highest Salary
解题思路：
写个找第N大数字的函数，先排序再取数字。
CREATE FUNCTION getNthHighestSalary(N INT) RETURNS INT
BEGIN
set N = N-1;
RETURN (
select distinct Salary from Employee order by Salary desc limit N,1
);
END
知识点： limit N，相当于打印前N条； limit N,m， 打印从第N+1开始的m笔记录。计数从0开始；
limit N-1 是语法错误的，这里只能用 numeric
Rank Scores
解题思路：
对每一个Score，找出>=它的score的数量，这个数量就是rank
select Score ,
(select count(*) from (select distinct Score s from Scores) tmp where s >= Score)
from Scores
order by Score desc
知识点：在内层的select中 s >= Score 表示，对一个Score，每一个s都跟它比较一次。
相当于外面的select是外层for，内嵌的select是内层for的感觉
Employees Earning More Than Their Managers
解题思路：
看成两个表e1和e2，找到满足e1.ManagerId = e2.Id 并且 e1.Salary > e2.Salary的记录
select e1.Name as Employee from Employee e1, Employee e2
where e1.ManagerId = e2.Id
and e1.Salary > e2.Salary
Duplicate Emails
解题思路：
找到group以后count > 1的记录
select Email from Person group by Email having count(Email) > 1
知识点：count作为合计函数，无法与where一起使用，所以出现了having
Customers Who Never Order
解题思路：
先 left join，再找到 CustomerId为空的记录
select C.Name as Customers
from Customers C left join Orders O on C.Id = O.CustomerId
where O.CustomerId is NULL
Department Highest Salary
解题思路：
合并两个table，然后看salary是否是 max的那个
select D.Name as Department, E.Name as Employee, E.Salary
from
Department D,
Employee E
where
D.Id = E.DepartmentId
and E.Salary in
(Select max(Salary) from Employee where DepartmentId = E.DepartmentId);
知识点： max
Department Top Three Salaries
解题思路：
先join，然后再来个select count，计算比当前salary大的数量，0表是当前salary最大，以此类推，找到 count ＝ 0，1，2的就是了
select D.Name as Department, E1.Name as Employee, E1.Salary
from
Employee E1
join Department D
on E1.DepartmentId = D.Id
where 3 > (select count(*) from (select distinct Salary, DepartmentId from Employee) E2
where E2.Salary > E1.Salary
and E1.DepartmentId = E2.DepartmentId)
知识点：count(*) , distinct , 3 > (table_numeric_column)
Delete Duplicate Emails
解题思路：
找到以email Group的最小id，然后删除所有不在这个id 集合里面的所有记录
delete from Person where Id not in (select A.Id from (select min(Id) as Id from Person Group by Email) A)
知识点：not in (table column )
Rising Temperature
解题思路：
当成2个表来进行比较。条件满足 DATEDIFF(a.Data, b.Data) = 1 并且 a.温度 > b.温度
select a.Id from Weather as a, Weather as b where DATEDIFF(a.Date, b.Date) = 1 and a.Temperature > b.Temperature
知识点：DATEDIFF(a, b) = 1 当a-b=1时；
Trips and Users
解题思路：
说不清楚，直接看指令吧
select t.Request_at Day, round(sum(t.Status like "cancelled_%") / count(*), 2) Rate from Trips t join Users u
on t.Client_Id = u.Users_Id
where u.Banned = "No" and t.Request_at >="2013-10-01" and t.Request_at <= "2013-10-03"
group by t.Request_at
知识点：
round( , 2), sum(), like, count, group, where