作为本小册的一个练习，找了一些个通用的sql解法
第N高
这种就是排序，然后分页取就行。
select distinct salary
from Employee
order by salary desc
limit (N-1),1
或者是可以子查询，一个数据，只要找出 >= 的个数，这个个数=N就行：
select distinct t1.Salary
from Employee t1
where N = (
select count(distinct t2.Salary)
from Employee t2
where t2.Salary >= t1.Salary
)
里面的子查询就是找到比当前数据大的数，然后这个 count() = N 就行
树形结构中有N个子级的数目
表自连接，关联父级和子级的字段，然后分组，此时得到的就是子级，然后过滤就行：
select E1.Name
from Employee as E1 join Employee as E2 on(E1.id = E2.managerid)
group by E1.id,E1.Name
having count(E1.id) >= 5
select E1.Name,E2.Name
from Employee as E1 join Employee as E2 on(E1.id = E2.managerid)
上面查询到就是所以匹配的上下级关系，然后对 E1 这个父级进行分组，就是以这个父级为单位的全部结构。
连续出现的数字
看需要几次吧，如果是3次：
SELECT distinct L1.num AS ConsecutiveNums
FROM `Logs` AS L1
JOIN `Logs` AS L2 ON (L1.num = L2.num AND L1.id + 1 = L2.id)
JOIN `Logs` AS L3 ON (L2.num = L3.num AND L2.id + 1 = L3.id)
第一行，然后接到第二行，看第二行的id是否 = 第一行id+1
某个分组下的最高
举一个 leetcode 中的例子，一般这种会在多个表中取数据的，所以不可能太简单就取出来：
部门工资最高的员工
有员工表，部门表
- 先找到每一个部门的最高薪水
select D.id, D.name, max(E1.salary) as m
from Department as D
join Employee as E1
on (D.id = E1.departmentid)
group by D.id,D.name
- 连接员工表
select F.name as `Department`,E.name as `Employee`,E.Salary
from Employee as E
join (
select D.id,D.name,max(E1.salary) as m
from Department as D
join Employee as E1
on (D.id = E1.departmentid)
group by D.id,D.name
) as F
on (E.departmentid = F.id and E.salary = F.m);
