leetcode数据库刷题一

一

题目
Employee 表包含所有员工信息，每个员工有其对应的 Id, salary 和 department Id。

        +----+-------+--------+--------------+
        | Id | Name  | Salary | DepartmentId |
        +----+-------+--------+--------------+
        | 1  | Joe   | 70000  | 1            |
        | 2  | Henry | 80000  | 2            |
        | 3  | Sam   | 60000  | 2            |
        | 4  | Max   | 90000  | 1            |
        +----+-------+--------+--------------+

Department 表包含公司所有部门的信息。

       +----+----------+
       | Id | Name     |
       +----+----------+
       | 1  | IT       |
       | 2  | Sales    |
       +----+----------+

编写一个 SQL 查询，找出每个部门工资最高的员工。例如，根据上述给定的表格，Max 在 IT 部门有最高工资，Henry 在 Sales 部门有最高工资。

     +------------+----------+--------+
     | Department | Employee | Salary |
     +------------+----------+--------+
     | IT         | Max      | 90000  |
     | Sales      | Henry    | 80000  |
     +------------+----------+--------+

来源：力扣（LeetCode）
链接：https://leetcode-cn.com/problems/department-highest-salary
著作权归领扣网络所有。商业转载请联系官方授权，非商业转载请注明出处。
我的答案

select Department.name Department,Employee.name Employee,max(Salary) as Salary
from Employee join  Department on Employee.DepartmentId=Department.id
group by Department.Id

存在问题
当一个部门，有两个人工资相同高时，不适用。

参考答案

SELECT
    Department.name AS 'Department',
    Employee.name AS 'Employee',
    Salary
FROM
    Employee
        JOIN
    Department ON Employee.DepartmentId = Department.Id
WHERE
    (Employee.DepartmentId , Salary) IN
    (   SELECT
            DepartmentId, MAX(Salary)
        FROM
            Employee
        GROUP BY DepartmentId
    )
;

二

题目
Employee表包含所有员工信息，每个员工有其对应的 Id, salary 和 department Id 。

           +----+-------+--------+--------------+
           | Id | Name  | Salary | DepartmentId |
           +----+-------+--------+--------------+
           | 1  | Joe   | 70000  | 1            |
           | 2  | Henry | 80000  | 2            |
           | 3  | Sam   | 60000  | 2            |
           | 4  | Max   | 90000  | 1            |
           | 5  | Janet | 69000  | 1            |
           | 6  | Randy | 85000  | 1            |
           +----+-------+--------+--------------+

Department 表包含公司所有部门的信息。

          +----+----------+
          | Id | Name     |
          +----+----------+
          | 1  | IT       |
          | 2  | Sales    |
          +----+----------+

编写一个 SQL 查询，找出每个部门工资前三高的员工。例如，根据上述给定的表格，查询结果应返回：

         +------------+----------+--------+
         | Department | Employee | Salary |
         +------------+----------+--------+
         | IT         | Max      | 90000  |
         | IT         | Randy    | 85000  |
         | IT         | Joe      | 70000  |
         | Sales      | Henry    | 80000  |
         | Sales      | Sam      | 60000  |
         +------------+----------+--------+

来源：力扣（LeetCode）
链接：https://leetcode-cn.com/problems/department-top-three-salaries
著作权归领扣网络所有。商业转载请联系官方授权，非商业转载请注明出处。
参考答案：

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(DISTINCT e2.Salary)
        FROM
            Employee e2
        WHERE
            e2.Salary > e1.Salary
                AND e1.DepartmentId = e2.DepartmentId
        )
;

反思
不要妄想一口吃成胖子，对于多表操作在脑子不够用的情况下可以将表分开思考，分别考虑每个表中需要的信息，然后通过嵌套查询或是连接查询的方法连接起来。

三.日期比较

题目

给定一个 Weather 表，编写一个 SQL 查询，来查找与之前（昨天的）日期相比温度更高的所有日期的 Id。

        +---------+------------------+------------------+
        | Id(INT) | RecordDate(DATE) | Temperature(INT) |
        +---------+------------------+------------------+
        |       1 |       2015-01-01 |               10 |
        |       2 |       2015-01-02 |               25 |
        |       3 |       2015-01-03 |               20 |
        |       4 |       2015-01-04 |               30 |
       +---------+------------------+------------------+

例如，根据上述给定的 Weather 表格，返回如下 Id:

我的解答

错误

正确答案

SELECT
    weather.id AS Id
FROM
    weather
        JOIN
    weather w ON DATEDIFF(weather.RecordDate, w.RecordDate) = 1
        AND weather.Temperature > w.Temperature
;

反思

DATEDIFF日期比较
DATEDIFF('2007-12-31','2007-12-30');   # 1
DATEDIFF('2010-12-30','2010-12-31');   # -1

四.sum()函数，round()函数

题目

Trips 表中存所有出租车的行程信息。每段行程有唯一键 Id，Client_Id 和 Driver_Id 是 Users 表中 Users_Id 的外键。Status 是枚举类型，枚举成员为 (‘completed’, ‘cancelled_by_driver’, ‘cancelled_by_client’)。

    +----+-----------+-----------+---------+--------------------+----------+
    | Id | Client_Id | Driver_Id | City_Id |        Status      |Request_at|
    +----+-----------+-----------+---------+--------------------+----------+
    | 1  |     1     |    10     |    1    |     completed      |2013-10-01|
    | 2  |     2     |    11     |    1    | cancelled_by_driver|2013-10-01|
    | 3  |     3     |    12     |    6    |     completed      |2013-10-01|
    | 4  |     4     |    13     |    6    | cancelled_by_client|2013-10-01|
    | 5  |     1     |    10     |    1    |     completed      |2013-10-02|
    | 6  |     2     |    11     |    6    |     completed      |2013-10-02|
    | 7  |     3     |    12     |    6    |     completed      |2013-10-02|
    | 8  |     2     |    12     |    12   |     completed      |2013-10-03|
    | 9  |     3     |    10     |    12   |     completed      |2013-10-03| 
    | 10 |     4     |    13     |    12   | cancelled_by_driver|2013-10-03|
   +----+-----------+-----------+---------+--------------------+----------+

Users 表存所有用户。每个用户有唯一键 Users_Id。Banned 表示这个用户是否被禁止，Role 则是一个表示（‘client’, ‘driver’, ‘partner’）的枚举类型。

       +----------+--------+--------+
       | Users_Id | Banned |  Role  |
      +----------+--------+--------+
      |    1     |   No   | client |
      |    2     |   Yes  | client |
      |    3     |   No   | client |
      |    4     |   No   | client |
      |    10    |   No   | driver |
      |    11    |   No   | driver |
      |    12    |   No   | driver |
      |    13    |   No   | driver |
      +----------+--------+--------+

写一段 SQL 语句查出 2013年10月1日至 2013年10月3日期间非禁止用户的取消率。基于上表，你的 SQL 语句应返回如下结果，取消率（Cancellation Rate）保留两位小数。

    +------------+-------------------+
    |     Day    | Cancellation Rate |
    +------------+-------------------+
    | 2013-10-01 |       0.33        |
    | 2013-10-02 |       0.00        |
    | 2013-10-03 |       0.50        |
    +------------+-------------------+

来源：力扣（LeetCode）
链接：https://leetcode-cn.com/problems/trips-and-users
著作权归领扣网络所有。商业转载请联系官方授权，非商业转载请注明出处。

代码

select tp.Request_at as Day , round(sum(tp.Status!='completed')/count(*),2) as 'Cancellation Rate' 
from Trips as tp left join Users as us on tp.Client_Id=us.Users_Id 
where (tp.Request_at between '2013-10-01' and '2013-10-03') and (us.Banned='No') 
group by tp.Request_at

反思

count(),统计符合条件的行，除了count(*)外会跳过空值的行
sum

1.sum() 函数用于计算某一字段中所有行的数值之和（ sum 求和时会对 null 进行过滤，不计算）
2.使用 sum(条件) 进行对符合条件的结果行数进行求和;

    select sum(user_id is null),sum(aid = 1) from user_auth;
    #如果没有符合条件的元组，返回null,不是0

五

题目

X 市建了一个新的体育馆，每日人流量信息被记录在这三列信息中：序号 (id)、日期 (visit_date)、人流量 (people)。
请编写一个查询语句，找出人流量的高峰期。高峰期时，至少连续三行记录中的人流量不少于100。

例如，表 stadium：

    +------+------------+-----------+
    | id   | visit_date | people    |
    +------+------------+-----------+
    | 1    | 2017-01-01 | 10        |
    | 2    | 2017-01-02 | 109       |
    | 3    | 2017-01-03 | 150       |
    | 4    | 2017-01-04 | 99        |
    | 5    | 2017-01-05 | 145       |
    | 6    | 2017-01-06 | 1455      |
    | 7    | 2017-01-07 | 199       |
    | 8    | 2017-01-08 | 188       |
    +------+------------+-----------+

对于上面的示例数据，输出为：

   +------+------------+-----------+
   | id   | visit_date | people    |
   +------+------------+-----------+
   | 5    | 2017-01-05 | 145       |
   | 6    | 2017-01-06 | 1455      |
   | 7    | 2017-01-07 | 199       |
   | 8    | 2017-01-08 | 188       |
   +------+------------+-----------+

提示：
每天只有一行记录，日期随着 id 的增加而增加。
来源：力扣（LeetCode）
链接：https://leetcode-cn.com/problems/human-traffic-of-stadium
著作权归领扣网络所有。商业转载请联系官方授权，非商业转载请注明出处。

我的答案（微笑）

(select s1.id id ,s1.visit_date visit_date,s1.people people
from stadium s1 join stadium s2 on s1.id+1=s2.id join  stadium s3 on s1.id+2=s3.id
where s1.people>=100 and s2.people>=100 and s3.people>=100)

union

(select s3.id id ,s3.visit_date visit_date,s3.people people
from stadium s1 join stadium s2 on s1.id+1=s2.id join  stadium s3 on s1.id+2=s3.id
where s1.people>=100 and s2.people>=100 and s3.people>=100
)
union
(select s2.id id ,s2.visit_date visit_date,s2.people people
from stadium s1 join stadium s2 on s1.id+1=s2.id join  stadium s3 on s1.id+2=s3.id
where s1.people>=100 and s2.people>=100 and s3.people>=100)

order by id

其他人的答案

SELECT distinct a.*
FROM stadium as a,stadium as b,stadium as c
where ((a.id = b.id-1 and b.id+1 = c.id) or
       (a.id-1 = b.id and a.id+1 = c.id) or
       (a.id-1 = c.id and c.id-1 = b.id))
  and (a.people>=100 and b.people>=100 and c.people>=100)
order by a.id;

六

题目

小美是一所中学的信息科技老师，她有一张 seat 座位表，平时用来储存学生名字和与他们相对应的座位 id。其中纵列的 id 是连续递增的小美想改变相邻俩学生的座位。你能不能帮她写一个 SQL query 来输出小美想要的结果呢？
示例：

      +---------+---------+
      |    id   | student |
     +---------+---------+
     |    1    | Abbot   |
     |    2    | Doris   |
     |    3    | Emerson |
     |    4    | Green   |
     |    5    | Jeames  |
    +---------+---------+

假如数据输入的是上表，则输出结果如下：

    +---------+---------+
    |    id   | student |
    +---------+---------+
    |    1    | Doris   |
    |    2    | Abbot   |
    |    3    | Green   |
    |    4    | Emerson |
    |    5    | Jeames  |
    +---------+---------+

注意：
如果学生人数是奇数，则不需要改变最后一个同学的座位。
来源：力扣（LeetCode）
链接：https://leetcode-cn.com/problems/exchange-seats
著作权归领扣网络所有。商业转载请联系官方授权，非商业转载请注明出处。

答案

FROM seat a LEFT JOIN seat b ON if(a.id&1,a.id=b.id-1,a.id=b.id+1) 
ORDER BY id;

反思

七

题目

好评率是会员对平台评价的重要指标。现在需要统计2018年1月1日到2018年1月31日，用户'小明'提交的母婴类目"花王"品牌的好评率（好评率=“好评”评价量/总评价量）:
用户评价详情表：a
字段：id（评价id，主键），
create_time（评价创建时间，格式'2017-01-01'），
user_name(用户名称)，
goods_id(商品id，外键) ，
sub_time（评价提交时间，格式'2017-01-01 23:10:32'），
sat_name（好评率类型，包含：“好评”、“中评”、“差评”）
商品详情表：b
字段：good_id（商品id，主键），
bu_name（商品类目）,
brand_name(品牌名称)

答案

slelect sum(CASE  WHEN sat_time = '好评' THEN 1 ELSE 0 END)/count(sat_time) as good_rate
from a join b on a.goods_id=b.good_id
where a.user_name="小明" and brand_name="花王" and bu_name="母婴" and sub_time between 2018-01-01 and 2018-01-31

反思：

SELECT SUM(if(category=1,size,0)) ,COUNT(if(category=1,true,null)) FORM t_file;

解析：

sum（if(category=1,size,0))中 sum函数返回一个值类型的数值，如果category=1,则返回size,如果category不等于1就返回0。
count(if(category=1,true,null))中count函数返回一个布尔值类型的数值，如果category=1,返回true,如果category不等于1返回null，如果写成count(If(category=1,1,0) 则返回的全是true,也就是说全都会计数，而count()间断内容是true还是null,如果不是null就计数，如果是null就不计数。

所以count(if())的写法应该是count(if(表达式表达式，true,null));

八

题目

Table: Books

 +----------------+---------+
 | Column Name    | Type    |
+----------------+---------+
| book_id        | int     |
| name           | varchar |
| available_from | date    |
+----------------+---------+
book_id is the primary key of this table.

Table: Orders

 +----------------+---------+
 | Column Name    | Type    |
+----------------+---------+
| order_id       | int     |
| book_id        | int     |
| quantity       | int     |
| dispatch_date  | date    |
+----------------+---------+

order_id is the primary key of this table.
book_id is a foreign key to the Books table.

Write an SQL query that reports the books that have sold less than 10 copies in the last year, excluding books that have been available for less than 1 month from today. Assume today is 2019-06-23.

The query result format is in the following example:

Books table:

   +---------+--------------------+----------------+
   | book_id | name               | available_from |
   +---------+--------------------+----------------+
   | 1       | "Kalila And Demna" | 2010-01-01     |
   | 2       | "28 Letters"       | 2012-05-12     |
   | 3       | "The Hobbit"       | 2019-06-10     |
   | 4       | "13 Reasons Why"   | 2019-06-01     |
   | 5       | "The Hunger Games" | 2008-09-21     |
  +---------+--------------------+----------------+

Orders table:

 +----------+---------+----------+---------------+
 | order_id | book_id | quantity | dispatch_date |
 +----------+---------+----------+---------------+
 | 1        | 1       | 2        | 2018-07-26    |
 | 2        | 1       | 1        | 2018-11-05    |
 | 3        | 3       | 8        | 2019-06-11    |
 | 4        | 4       | 6        | 2019-06-05    |
 | 5        | 4       | 5        | 2019-06-20    |
 | 6        | 5       | 9        | 2009-02-02    |
 | 7        | 5       | 8        | 2010-04-13    |
 +----------+---------+----------+---------------+

Result table:

+-----------+--------------------+
| book_id   | name               |
+-----------+--------------------+
| 1         | "Kalila And Demna" |
| 2         | "28 Letters"       |
| 5         | "The Hunger Games" |
+-----------+--------------------+

来源：力扣（LeetCode）
链接：https://leetcode-cn.com/problems/unpopular-books
著作权归领扣网络所有。商业转载请联系官方授权，非商业转载请注明出处。

答案

select b.book_id,b.name
from Books b left  join Orders o on b.book_id=o.book_id
where available_from<"2019-05-23"
group by b.book_id
having ifnull(sum(if(o.dispatch_date<"2018-06-23",quantity=0,quantity)),0)<10

反思

IFNULL() 函数用于判断第一个表达式是否为 NULL，如果为 NULL 则返回第二个参数的值，如果不为 NULL 则返回第一个参数的值。
IFNULL() 函数语法格式为：
IFNULL(expression, alt_value)