LeetCode_MySQL

以下内容多数参照 https://blog.csdn.net/WhereIsHeroFrom/article/details/85088151 

select 函数的完整语法：

1 select [ALL|DISTINCT|DISTINCTROW|TOP]
2 {*|talbe.*|[table.]field1[AS alias1][,[table.]field2[AS alias2][,…]]}
3 FROM tableexpression[,…][IN externaldatabase]
4 [WHERE…]
5 [GROUP BY…]
6 [HAVING…]
7 [ORDER BY…]

175. 组合两个表

1 select FirstName, LastName, City, State
2 from Person left join Address
3 on Person.PersonId = Address.PersonId
4 ;

 

176. 第二高的薪水

 1 SELECT 
 2     CASE c.cnt
 3     WHEN 0 THEN
 4         NULL
 5     ELSE
 6         c.Salary
 7     END AS SecondHighestSalary
 8 FROM
 9     (SELECT COUNT(Salary) AS cnt, Salary 
10      FROM 
11         (SELECT DISTINCT Salary 
12          FROM Employee 
13          ORDER BY Salary DESC LIMIT 1, 1
14         ) AS b
15     ) AS c

1 SELECT
2     (SELECT DISTINCT
3             Salary
4         FROM
5             Employee
6         ORDER BY Salary DESC
7         LIMIT 1 OFFSET 1) AS SecondHighestSalary
8 ;

1 SELECT 
2     IFNULL(
3         (SELECT DISTINCT Salary 
4              FROM Employee 
5              ORDER BY Salary DESC LIMIT 1, 1
6             ),
7         NULL) SecondHighestSalary

1 SELECT
2     IFNULL(
3       (SELECT DISTINCT Salary
4        FROM Employee
5        ORDER BY Salary DESC
6         LIMIT 1 OFFSET 1),
7     NULL) AS SecondHighestSalary

IFNULL(v1,v2)

如果 v1 的值不为 NULL，则返回 v1，否则返回 v2。

SELECT IFNULL(null,'Hello Word')->Hello Word

LIMIT语法：SELECT * FROM table  LIMIT [offset,] rows | rows OFFSET offset

1 mysql> SELECT * FROM table LIMIT 2,4;  // 检索 3-6 行
2 mysql> SELECT * FROM table LIMIT 6,-1; // 检索 7 之后所有行
3 mysql> SELECT * FROM table LIMIT 5;     //检索前 5 行，相当于 SELECT * FROM table LIMIT 0,5;
4 mysql> SELECT * FROM table LIMIT 2 OFFSET 3;//查询4-5两条记录，想当于 SELECT * FROM orange  LIMIT 3,2;

 

1 CASE expression
2     WHEN condition1 THEN result1
3     WHEN condition2 THEN result2
4    ...
5     WHEN conditionN THEN resultN
6     ELSE result
7 END

CASE 表示函数开始，END 表示函数结束。如果 condition1 成立，则返回 result1, 如果 condition2 成立，则返回 result2，当全部不成立则返回 result，而当有一个成立之后，后面的就不执行了。

177. 第N高的薪水

 1 CREATE FUNCTION getNthHighestSalary(N INT) RETURNS INT
 2 BEGIN
 3   RETURN (
 4       select (IF(
 5           (select count(*) from (select distinct e.Salary from Employee e) e)>=N,
 6           (select min(e.Salary) from (select distinct e.Salary from Employee e order by e.Salary desc limit N) e)
 7           , NULL
 8       ))
 9       
10   );
11 END

 1 CREATE FUNCTION getNthHighestSalary(N INT) RETURNS INT
 2 BEGIN
 3   set N = N - 1;
 4   RETURN (
 5         select 
 6             case c.cnt
 7             when 0 then
 8                 null
 9             else
10                 c.Salary
11             end as xx
12         from
13             (select count(Salary) as cnt, Salary 
14              from 
15                 (select distinct Salary 
16                  from Employee 
17                  order by Salary desc limit N, 1
18                 ) as b
19             ) as c
20   );
21 END

1 CREATE FUNCTION getNthHighestSalary(N INT) RETURNS INT
2 BEGIN
3     set N = N-1;
4   RETURN (
5     select distinct ifnull(salary,null) from  employee order by salary desc limit N,1 
6   );
7 END

MySQL自定义函数-参考

CREATE FUNCTION function_name ([func_parameter[,…]]) 
RETURNS {STRING|INTEGER|REAL|DECIMAL|…} 
routine_body - 函数体

默认情况下自定义函数与当前数据库关联。如果想要把该函数与一个给定数据库关联起来，可以在创建子程序的时候指定其名字为db_name.function_name。

自定义用户变量:
可以先在用户变量中保存值然后在以后引用它；这样可以将值从一个语句传递到另一个语句。用户变量与连接有关。也就是说，一个客户端定义的变量不能被其它客户端看到或使用。当客户端退出时，该客户端连接的所有变量将自动释放。

用户变量的形式为@var_name，其中变量名var_name可以由当前字符集的文字数字字符、‘.’、‘_’和‘$’组成。 默认字符集是cp1252 (Latin1)。可以用mysqld的–default-character-set选项更改字符集。用户变量名对大小写不敏感。
设置用户变量的一个途径是执行SET语句：

1 SET @var_name = expr [, @var_name = expr] ...

对于SET，可以使用=或:=作为分配符。分配给每个变量的expr可以为整数、实数、字符串或者NULL值。

也可以用语句select代替SET来为用户变量分配一个值。在这种情况下，分配符必须为:=而不能用=，因为在非SET语句中=被视为一个比较 操作符：

1 mysql> SET @t1=0, @t2=0, @t3=0;
2 mysql> SELECT @t1:=(@t2:=1)+@t3:=4,@t1,@t2,@t3;

函数体

• 函数体由合法的SQL语法构成;
• 函数体可以是简单的SELECT或INSERT语句;
• 函数体如果为复合结构则使用BEGIN…END语句；
• 复合结构可以包括声明，循环，控制结构。

178. 分数排名

1 select Score, 
2     (select count(distinct Score) 
3      from Scores as b 
4      where a.score < b.score
5     ) + 1  as Rank
6 from Scores as a order by Rank;

1 select distinct a.num ConsecutiveNums from Logs a,(select @val:=0,@count:=0) x 
3 where if( (@val=a.num),  
4           if( (@count:=@count+1) and (@count>=2) ,1, 0 ),  
5             (( @count:=@count-@count) or (@val:=a.num)) and 0
6         )

 

181. 超过经理收入的员工

1 select e1.Name as Employee
2 from Employee e1, 
3 Employee e2 
4 where e1.ManagerId = e2.Id
5 AND e1.Salary > e2.Salary;

182. 查找重复的电子邮箱 

1 select Email
2 from
3     (select Id, Email, count(Email) as e 
4      from Person group by Email
5     ) as x
6 where x.e > 1;

1 select Email 
2 from Person 
3 group by Email 
4 having count(Email) > 1

183. 从不订购的客户

1 select Name Customers 
2 from Customers 
3 where Id in(
4             select Id
5             from Customers 
6             where Id not in(
7                             select CustomerId 
8                             from Orders)
9             )

 1 select c.Name as Customers 
 2 from Customers as c
 3 where c.Id not in 
 4     (select n.CustomerId
 5      from 
 6         (select CustomerId, c.Id, Name from 
 7             Orders o,
 8             Customers c
 9          where c.Id = CustomerId
10         ) as n
11     )

1 select Name as Customers 
2 from Customers
3 where Id not in 
4     (
5         select CustomerId
6         from Orders
7     )

184. 部门工资最高的员工

select d.Name as Department, c.Employee, c.Salary 
from
    (select a.DepartmentId, a.Name as Employee, a.Salary as Salary
     from
        Employee as a,
        (select DepartmentId, max(Salary) as Salary 
         from Employee 
         group by DepartmentId
        ) as b
     where 
             a.Salary = b.Salary
        and  a.DepartmentId = b.DepartmentId
    ) as c,
    Department as d
where d.Id = c.DepartmentId

 1 select 
 2     d.Name as Department,
 3     e.Name as Employee,
 4     e.Salary 
 5 from 
 6     Employee e,Department d 
 7 where
 8     e.DepartmentId=d.id 
 9     and
10         (e.Salary,e.DepartmentId) 
11         in (
12                 select 
13                     max(Salary),DepartmentId 
14                 from 
15                     Employee 
16                 group by 
17                     DepartmentId
18         )

 1 select 
 2     d.Name as Department, 
 3     e.Name as Employee, 
 4     Salary AS Salary 
 5 from 
 6     Employee e inner join Department d 
 7     on e.DepartmentId=d.Id 
 8 where 
 9     Salary in (
10         select 
11             max(Salary) as Salary 
12         from 
13             Employee e1 
14         where 
15             e1.DepartmentId=e.DepartmentId
16     )

185. 部门工资前三高的员工

 2 select 
 3     d.Name as Department, h.Employee, h.Salary
 4 from
 5    (select 
 6         g.DepartmentId, g.Name as Employee, g.Salary
 7     from
 8     (
 9         select a.Id
10         from
11            (select e1.Id, count(distinct e2.Salary) as cnt
12             from
13                 Employee e1,
14                 Employee e2
15             where e1.DepartmentId = e2.DepartmentId
16                 and (
17                     e1.Salary <= e2.Salary
18                 )
19             group by e1.Id
20            ) as a
21         where cnt <= 3
22     ) as b
23     left join Employee g
24     on b.Id = g.Id
25    ) as h
26 left join Department d
27 on h.DepartmentId = d.Id
28 where not IsNull(d.Name)
29 order by h.DepartmentId, h.Salary desc
30 ;

 1 SELECT 
 2     P2.Name AS Department,P3.Name AS Employee,P3.Salary AS Salary
 3 FROM 
 4     Employee AS P3
 5 INNER JOIN 
 6     Department AS P2
 7 ON 
 8     P2.Id = P3.DepartmentId 
 9 WHERE (
10     SELECT 
11         COUNT(DISTINCT Salary)
12     FROM 
13         Employee AS P4
14     WHERE 
15         P3.DepartmentId = P4.DepartmentId
16     AND 
17         P4.Salary >= P3.Salary
18 ) <= 3
19 ORDER BY 
20     DepartmentId,Salary DESC

 1 select 
 2     B.name as Department , A.Name as Employee, A.Salary
 3 from (
 4     select 
 5         Name, DepartmentId, Salary,  DENSE_RANK() over(partition by DepartmentId order by Salary desc) as pm 
 6     from 
 7         Employee
 8 ) A,  Department B  
 9 where 
10     A.pm<=3 and A.DepartmentId = B.Id
11 order by 
12     B.id asc, A.Salary desc

SQL中的窗口函数 OVER窗口函数

Sql 四大排名函数（ROW_NUMBER、RANK、DENSE_RANK、NTILE）简介

187. 重复的DNA序列

1 DELETE p1 FROM Person p1,
2     Person p2
3 WHERE
4     p1.Email = p2.Email AND p1.Id > p2.Id

 1 delete 
 2 from 
 3 　　Person
 4 where Id not in 
 5 (
 6     select 
 7 　　　　*
 8     from
 9     (select min(Id) as Id
10      from Person 
11      group by Email
12     ) as x
13 );

197. 上升的温度

 2 select 
 3     a.Id
 4 from 
 5     Weather a,
 6     Weather b
 7 where 
 8     a.RecordDate in (select interval 1 day + b.RecordDate)
 9     and 
10         a.Temperature > b.Temperature;

1 SELECT
2     weather.Id AS Id
3 FROM
4     weather
5         JOIN
6     weather w ON DATEDIFF(weather.RecordDate, w.RecordDate) = 1
7         AND weather.Temperature > w.Temperature

DATEDIFF(date1,date2) 函数返回两个日期之间的天数(date1-date2)。

262. 行程和用户

 1 select 
 2     Day, round( if(IsNull(up),0,up) / if(IsNull(dwn),1,dwn), 2 ) as 'Cancellation Rate'
 3 from
 4 (
 5     select 
 6         allTrips.Request_at as Day, allTrips.cnt as dwn,  allCancellTrips.cnt as up
 7     from
 8         (
 9          select 
10                 Request_at, count(*) as cnt
11          from 
12             (
13                 select 
14                     ta.Id, ta.Client_Id, ta.Request_at
15                 from 
16                     Trips ta
17                 left join 
18                     Users ua
19                 on 
20                     ta.Client_Id = ua.Users_Id
21                 where 
22                     ua.Banned = "No"
23             ) as noBannedTrips
24          group by 
25                 Request_at
26         ) as allTrips
27     left join
28         (
29          select 
30                 Request_at, count(*) as cnt
31          from 
32             (
33                 select 
34                     ta.Id, ta.Client_Id, ta.Request_at
35                 from 
36                     Trips ta
37                 left join 
38                     Users ua
39                 on 
40                     ta.Client_Id = ua.Users_Id
41                 where 
42                     ua.Banned = "No"
43                 and 
44                     ta.Status != "completed"
45             ) as noBannedAndNotComletedTrips
46          group by 
47                 Request_at
48         ) as allCancellTrips
49     on 
50         allTrips.Request_at = allCancellTrips.Request_at
51 ) as result
52 where 
53     result.Day >= '2013-10-01'
54     and 
55         result.Day <= '2013-10-03';

 1 select
 2     t.Request_at as  Day,
 3     round(
 4         count( if(t.Status != 'completed',1,null)) / count(t.Status),2 )
 5         as  'Cancellation Rate'
 6 from 
 7     Trips as t
 8 left join 
 9     Users as u1
10 On 
11     t.Client_Id = u1.Users_Id
12 left join 
13     Users as u2
14 on 
15     t.Driver_Id = u2.Users_Id
16 where
17     u1.Banned = 'No'
18     and
19         u2.Banned = 'No'
20 and
21     t.Request_at between '2013-10-01' and '2013-10-03'
22 
23 group by Day
24 order by Day

 1 SELECT
 2     t1.Request_at AS DAY,
 3     ROUND(((count( CASE WHEN STATUS = 'cancelled_by_driver' THEN 1 ELSE NULL END ) + count( CASE WHEN STATUS = 'cancelled_by_client' THEN 1 ELSE NULL END )) 
 4     / count( STATUS )),2) AS 'Cancellation Rate' 
 5 FROM
 6     Trips t1
 7     INNER JOIN Users t2 ON t1.Client_Id = t2.Users_Id
 8     INNER JOIN Users t3 ON t1.Driver_Id = t3.Users_Id 
 9 WHERE
10     t2.Banned = 'No' 
11     AND t3.Banned = 'No'
12     AND t1.Request_at >= '2013-10-01'
13     AND t1.Request_at <= '2013-10-03'
14 GROUP BY
15     t1.Request_at 
16 ORDER BY DAY

 1 SELECT 
 2     request_at as Day, 
 3     round( 
 4     count(if(t.status<>'completed',t.status,NULL)) -- 当天非禁止用户的取消数量
 5     /sum(if(t.status,1,0)) -- 当天订单总数
 6     ,
 7     2)as 'Cancellation Rate' 
 8 FROM 
 9     users u 
10     INNER JOIN trips t 
11         on 
12             u.users_id = t.client_id 
13     and t.request_at BETWEEN '2013-10-01' and '2013-10-03' 
14     and u.banned = 'NO' 
15 group by t.request_at

595. 大的国家

1 select 
2     name,population,area
3 from
4     World
5 where
6     area>3000000 or population >25000000

596. 超过5名学生的课

1 select 
2     class
3 from
4     courses 
5 group by
6     class
7 having
8     count(distinct student) >= 5

sql中where和having的区别

Where中不能使用聚合函数，Having中可以使用聚合函数。

601. 体育馆的人流量

1 MySQL自定义排序+本题部分思路
2 select 
3     id,visit_date+@curRank as date,people,@curRank := @curRank + 1 AS rank
4     from stadium,(
5     SELECT @curRank := 0
6     ) q
7     where 
8         people >100
9     ORDER BY visit_date desc

 1 select t.id, t.date, t.people
 2 from
 3    (
 4     select distinct 
 5         s1.id 
 6     from 
 7         stadium s1,
 8         stadium s2,
 9         stadium s3
10     where   (s1.id + 1 = s2.id
11         and s2.id + 1 = s3.id
12         and s1.people >= 100
13         and s2.people >= 100
14         and s3.people >= 100)
15        or (s1.id - 1 = s2.id
16         and s2.id - 1 = s3.id
17         and s1.people >= 100
18         and s2.people >= 100
19         and s3.people >= 100)
20        or (s1.id + 1 = s3.id
21         and s1.id - 1 = s2.id
22         and s1.people >= 100
23         and s2.people >= 100
24         and s3.people >= 100)
25 ) as a
26 left join stadium t
27 on a.id = t.id;

620. 有趣的电影

 1 select 
 2     id,movie,description,rating
 3 from
 4     cinema
 5 where
 6     id % 2 = 1
 7     and
 8         description <> 'boring'
 9 order by
10     rating desc　　

626. 换座位

 1 select 
 2     a.id, if( a.id in (select count(*) from seat) and a.id%2=1, a.student, b.student ) as student
 3 from
 4     (
 5      select id, student, (FLOOR((id+1)/2)*2-(1-id%2)) as next
 6      from seat
 7     ) as a
 8 left join seat b
 9 on a.next = b.id
10 order by a.id;

1 SELECT * FROM(
2     SELECT id-1 AS id,student FROM seat WHERE id%2=0
3     UNION
4     SELECT id+1 AS id,student FROM seat WHERE id%2=1 AND (id+1) <= (SELECT COUNT(*) FROM seat)
5     UNION
6     SELECT id AS id,student FROM seat WHERE id%2=1 AND (id+1) > (SELECT COUNT(*) FROM seat)
7 ) AS T1
8 ORDER BY id ASC

FLOOR(x):返回小于或等于 x 的最大整数(SELECT FLOOR(1.5) -- 小于或等于 1.5 的整数：返回1)

627. 交换工资

2     salary
3 set 
4     sex=if(sex='m','f','m') 

转载于:https://www.cnblogs.com/cjxyu/articles/10323523.html