【LeetCode】精选数据库70题（2022-10-14完结啦~）
文章目录
- 题目链接
- 511 游戏玩法分析
- 512 游戏玩法分析II
- 534 游戏玩法分析III
- 550 游戏玩法分析IV
- 569 员工薪水中位数（困难...还没解决...窗口函数）
- 570 至少有5名直接下属的经理
- 571 给定数字的频率查询中位数
- 574 当选者
- 577 员工奖金
- 578 查询回答率最高的问题
- 579 查询员工的累计薪水
- 580 统计各专业学生人数
- 584 寻找用户推荐人
- 585 2016年的投资
- 586 订单最多的客户
- 597 好友申请
- 602 好友申请II
- 603 连续空余座位
- 607 销售员
- 608 树节点
- 610 判断三角形
- 612 平面上的最近距离
- 613 直线上的最近距离
- 614 二级关注者
- 615 平均工资：部门与公司比较
- 618 学生地理信息报告
- 619 只出现一次的最大数字
- 1045 买下所有产品的客户
- 1050 合作过至少三次的演员和导演
- 1068 产品销售分析
- 1069 产品销售分析II
- 1070 产品销售分析III
- 1075 项目员工 I
- 1076 项目员工II
- 1077 项目员工III
- 1082 销售分析
- 1083 销售分析II
- 1084 销售分析III
- 1097 游戏玩法分析 V
- 1098 小众书籍
- 1107 每日新用户统计
- 1112 每位学生的最高成绩
- 1113 报告的记录
- 1126 查询活跃业务
- 1127 用户购买平台
- 1132 报告的记录III
- 1141 查询近30天活跃用户
- 1142 过去30天的用户活动II
- 1148 文章浏览I
- 1149 文章浏览
- 1158 市场分析I
- 1159 市场分析II
- 1164 指定日期的产品价格
- 1173 即时食物配送
- 1174 即时食物配送II
- 1193 每月交易I
- 1194 锦标赛优胜者
- 1204 最后一个进入电梯的人
- 1205 每月交易II
- 1211 查询结果的质量和占比
- 1212 查询球队积分
- 1225 报告系统状态的连续日期
- 1241 每个帖子的评论数
- 1251 平均售价
- 1264 页面推荐
- 1270 向公司CEO汇报工作的所有人
- 1280 学生们参加各科测试的次数
- 1285 找到连续区间的开始和结束数字
- 1294 不同国家的天气类型
- 1303 求团队人数
- 1308 不同性别每日分数总计
- 1321 餐馆营业额变化增长
题目链接
LeetCode精选数据库70题
511 游戏玩法分析
按照player_id进行分组，找出其中最小的日期
SELECT player_id, MIN(event_date) first_login
FROM Activity
GROUP BY player_id;
512 游戏玩法分析II
SELECT player_id, device_id
FROM Activity
WHERE (player_id, event_date) IN (
SELECT player_id, MIN(event_date)
FROM Activity
GROUP BY player_id
);
534 游戏玩法分析III
自连接
SELECT t1.player_id, t1.event_date, SUM(t2.games_played) games_played_so_far
FROM Activity t1
CROSS JOIN Activity t2
WHERE t1.player_id = t2.player_id AND t1.event_date >= t2.event_date
GROUP BY t1.player_id,t1.event_date;
窗口函数
SELECT
player_id,
event_date,
SUM(games_played) OVER(PARTITION BY player_id ORDER BY event_date) games_played_so_far
FROM Activity;
550 游戏玩法分析IV
题目要求：计算在首次登陆的第二天再次登录的玩家的比率
也就是计算从首次登陆日起开始至少连续两天登陆的玩家的数量，然后除以玩家总数。
从首次登陆日期开始至少连续两天，等价于event_date必须含有首次登陆日期+1，至于event_date是不是包含别的日期，不是这道题关心的，满足了这一条件，即可说明在首次登陆的第二天再次登陆了。
首先找出每个player_id首次登陆的第二天的日期
SELECT player_id,DATE(MIN(event_date)+1)
FROM Activity
GROUP BY player_id;
在Activity表中把event_date和上述SQL的结果进行匹配。找出哪些player_id的event_date里含有首次登陆后的第二天的日期
SELECT ROUND(
COUNT(player_id)/(SELECT COUNT(DISTINCT player_id) FROM Activity) ,2
) fraction
FROM Activity
WHERE (player_id,event_date) IN (
SELECT player_id,DATE(MIN(event_date)+1)
FROM Activity
GROUP BY player_id
);
569 员工薪水中位数（困难…还没解决…窗口函数）
表的自连接
SELECT t1.id,t1.company,t1.salary
FROM Employee t1
JOIN EMployee t2
ON t1.company = t2.company
GROUP BY t1.id
HAVING SUM(t1.salary>=t2.salary) >= COUNT(t1.id)/2
AND SUM(t1.salary<=t2.salary) >= COUNT(t1.id)/2
输入
{"headers": {"Employee": ["id", "company", "salary"]}, "rows": {"Employee": [[1, "A", 2341],[2, "A", 341],[3, "A", 15],[4, "A", 15314],[5, "A", 451],[6, "A", 513],[7, "B", 15],[8, "B", 13],[9, "B", 1154],[10, "B", 1345],[11, "B", 1221],[12, "B", 234],[13, "C", 2345],[14, "C", 2645],[15, "C", 2645],[16, "C", 2652],[17, "C", 65]]}}
方法一：窗口函数+表的自连接
在原始表数据后面添加了ROW_NUMBER()进行排序，排序以后再进行表连接判断
比如，如果一个company下有6个（偶数个）员工，那么在HAVING子句中，只有需要为3、4的满足条件，如果一个company下有5个（奇数个）员工，那么在HAVING子句中，只有3满足条件，这样就可以解决偶数个员工输出两个，奇数个员工输出一个的问题了
WITH temp AS (
SELECT
id,
company,
salary,
ROW_NUMBER() OVER(PARTITION BY company ORDER BY salary) rk
FROM Employee
)
SELECT t1.id,any_value(t1.company) company,any_value(t1.salary) salary
FROM temp t1
JOIN temp t2
ON t1.company = t2.company
GROUP BY t1.id
HAVING SUM(t1.rk>=t2.rk) >= COUNT(t1.id)/2 AND SUM(t1.rk<=t2.rk)>=COUNT(t1.id)/2
方法二：正反序
正反序，就是按照salary排序，如果salary相同，再比较id
SELECT
id,
company,
salary,
ROW_NUMBER() OVER(PARTITION BY company ORDER BY salary,id) rk1,
ROW_number() OVER(PARTITION BY company ORDER BY salary DESC,id DESC) rk2,
COUNT(id) OVER(PARTITION BY company) cnt
FROM Employee
必须满足rk1>=cnt/2，且rk2>=cnt/2
SELECT
id,
company,
salary
FROM (
SELECT
id,
company,
salary,
ROW_NUMBER() OVER(PARTITION BY company ORDER BY salary,id) rk1,
ROW_number() OVER(PARTITION BY company ORDER BY salary DESC,id DESC) rk2,
COUNT(id) OVER(PARTITION BY company) cnt
FROM Employee
) temp
WHERE rk1 >= cnt/2 AND rk2 >= cnt/2;
方法三：奇偶数讨论
和方法一有点类似，与方法二的区别在于，where子句的条件不同，但是，异曲同工
SELECT
id,
company,
salary
FROM (
SELECT
id,
company,
salary,
ROW_NUMBER() OVER(PARTITION BY company ORDER BY salary,id) rk,
COUNT(id) OVER(PARTITION BY company) cnt
FROM Employee
) temp
WHERE rk IN (cnt/2,cnt/2+1,(cnt+1)/2)
570 至少有5名直接下属的经理
表的自连接
SELECT t2.name
FROM Employee t1
JOIN Employee t2 ON t1.managerId = t2.id
GROUP BY t1.managerId
HAVING COUNt(t1.managerId) >= 5;
571 给定数字的频率查询中位数
中位数就是将所有数字按照升序或者降序排列，然后取最中间的数字
- 数字个数是奇数的话，那么中位数会在这个序列中
- 数字个数是偶数的话，那么中位数是最中间的两个数的平均值
步骤：
-
使用
sum(frequency) over() total_frequency计算出所有数字的个数，计算总数还可以使用SELECT SUM(frequency) total_frequency FROM Numbers -
sum(frequency) over(order by num desc) desc_frequency，使用窗口函数，按照num降序计算出当前数字和之前数字出现的次数 -
sum(frequency) over(order by num asc) asc_frequency，使用窗口函数，按照num升序计算出当前数字和之前数字出现的次数
将查询出来的num，desc_frequency，asc_frequency，total_frequency作为临时表，筛选条件是desc_frequency >= total_frequency/2，asc_frequency >= total_frequency/2
SELECT
num,
SUM(frequency) OVER(ORDER BY num ASC) desc_frequency,
SUM(frequency) OVER(ORDER BY num DESC) asc_frequency,
SUM(frequency) OVER() total_frequency
FROM Numbers
{"headers":
["num", "desc_frequency", "asc_frequency", "total_frequency"],
"values":
[[3, 12, 1, 12],
[2, 11, 4, 12],
[1, 8, 5, 12],
[0, 7, 12, 12]]}
SELECT ROUND(AVG(num),1) median
FROM (
SELECT
num,
SUM(frequency) OVER(ORDER BY num ASC) desc_frequency,
SUM(frequency) OVER(ORDER BY num DESC) asc_frequency,
SUM(frequency) OVER() total_frequency
FROM Numbers
)temp
WHERE desc_frequency>=total_frequency/2 AND asc_frequency>=total_frequency/2;
574 当选者
SELECT name
FROM candidate t1
JOIN (
SELECT candidateId, COUNT(*) votes
FROM Vote
GROUP BY candidateId
) t2 ON t1.id = t2.candidateId
ORDER BY votes DESC
LIMIT 1;
SELECT name
FROM candidate t1
JOIN (
SELECT candidateId, COUNT(*) votes
FROM Vote
GROUP BY candidateId
ORDER BY votes DESC
LIMIT 1
) t2 ON t1.id = t2.candidateId;
577 员工奖金
题目要求选出所有bonus<1000的员工的name及其bonus，如果该员工没有bonus，也要输出，不过bonus那一栏为null，所以在将两个表连接时，使用外连接
SELECT name,bonus
FROM Employee t1
LEFT JOIN Bonus t2 ON t1.empId = t2.empId
WHERE t1.empId NOT IN (
SELECT empId
FROM Bonus
WHERE bonus >= 1000
);
SELECT name,bonus
FROM Employee t1
LEFT JOIN Bonus t2 ON t1.empId = t2.empId
WHERE bonus < 1000 OR bonus IS NULL;
578 查询回答率最高的问题
SELECT question_id survey_log
FROM (
SELECT question_id, SUM(IF(action='answer',1,0))/SUM(IF(action='show',1,0)) rate
FROM SurveyLog
GROUP BY question_id
ORDER BY rate DESC,question_id ASC
LIMIT 1
) t;
SELECT question_id survey_log
FROM SurveyLog
GROUP BY question_id
ORDER BY SUM(IF(action='answer',1,0))/SUM(IF(action='show',1,0)) DESC, question_id ASC
LIMIT 1;
579 查询员工的累计薪水
自连接
首先把每个员工最近一个月的薪水信息排除，生成一个临时表temp
然后将temp表进行自连接，连接的条件是Id相等，Month的差的范围在0，2，也就是最近三个月，最后使用SUM进行累加即可
WITH temp AS(
SELECT Id,`Month`,Salary
FROM Employee
WHERE (Id,`Month`) NOT IN (
SELECT Id, MAX(`Month`) max_month
FROM Employee
GROUP BY Id
)
)
SELECT
t1.Id,
t1.`Month`,
SUM(t2.Salary) Salary
FROM temp t1
JOIN temp t2 ON t1.Id = t2.Id AND t1.`Month`-t2.`Month`>=0 AND t1.`Month`-t2.`Month`<=2
GROUP BY t1.Id,t1.`Month`
ORDER BY Id ASC, `Month` DESC;
窗口函数
SELECT
id,
`Month`,
SUM(Salary) OVER(PARTITION BY Id ORDER BY `Month` RANGE 2 PRECEDING) Salary
FROM Employee
WHERE (Id,`Month`) NOT IN (
SELECT Id,MAX(`Month`)
FROM Employee
GROUP BY Id
)
ORDER BY Id ASC,`Month` DESC;
580 统计各专业学生人数
SELECT dept_name, IFNULL(COUNT(t2.dept_id),0) student_number
FROM Department t1
LEFT JOIN Student t2 ON t1.dept_id = t2.dept_id
GROUP BY t1.dept_id
ORDER BY student_number DESC,dept_name ASC;
584 寻找用户推荐人
只加入referee_id != 2的条件，无法得到referee_id 为 NULL的记录，所以需要再加上referee_id IS NULL的条件
SELECT name
FROM customer
WHERE referee_id IS NULL OR referee_id != 2
585 2016年的投资
检查每一个TIV_2015是否是唯一的，如果不是唯一的且同时坐标是唯一的，那么这条记录就符合题目要求。应该被统计到答案中。
SELECT ROUND(SUM(TIV_2016),2) TIV_2016
FROM insurance
WHERE TIV_2015 IN (
SELECT TIV_2015
FROM insurance
GROUP BY TIV_2015
HAVING COUNT(*) > 1
) AND (LAT,LON) IN (
SELECT LAT,LON
FROM insurance
GROUP BY LAT,LON
HAVING COUNT(*)=1
);
上述SQL执行结果
SELECT ROUND(SUM(TIV_2016),2) TIV_2016
FROM (
SELECT TIV_2016,
COUNT(*) OVER(PARTITION BY TIV_2015) count_tiv_2015,
COUNT(*) OVER(PARTITION BY LAT,LON) count_lat_lon
FROM insurance
) t
WHERE count_tiv_2015 > 1 AND count_lat_lon = 1;
上述SQL执行结果
586 订单最多的客户
按照customer_number对每条记录进行分组，统计每个客户的订单数量，并按照订单数量从大到小进行排序，取出第一个，也就是订单记录最多的用户
SELECT customer_number
FROM (
SELECT customer_number,COUNT(*) order_count
FROM Orders
GROUP BY customer_number
ORDER BY order_count DESC
LIMIT 1
) t;
597 好友申请
通过这题了解到，SELECT里面可以再写SELECT，本来一直想的是分开查询，最后怎么把两个查询结果合在一起…
COUNT(expr)函数返回不为空的记录总数，不过当RequestAccepted表中的记录数为0时，两个COUNT函数的结果相除，会是NULL值，所以外面要加上一层IFNULL函数
SELECT
ROUND(
IFNULL(
(SELECT COUNT(DISTINCT requester_id,accepter_id) FROM RequestAccepted) /
(SELECT COUNT(DISTINCT sender_id,send_to_id) FROM FriendRequest),0),2
) accept_rate;
602 好友申请II
先分别用requester_id 和 accepter_id 分组，统计出每个人拥有的好友数量
- 因为成为好友是双向的，比如，1向3提出好友申请，3同意了，那么，站在1的角度，1有一个朋友是3，站在3的角度，3有一个朋友是1
最后将用requester_id 和 accepter_id 分组的两个表进行合并，再按照id再次合并，题目保证生成的测试用例保证拥有最多好友数目的只有一个人，那么我们就可以使用ORDER BY + LIMIT的组合，取出好友数最多的那个人的信息
SELECT id,SUM(num) num
FROM (
SELECT requester_id id,COUNT(requester_id) num
FROM RequestAccepted
GROUP BY requester_id
UNION ALL
SELECT accepter_id id,COUNT(accepter_id) num
FROM RequestAccepted
GROUP BY accepter_id
) t
GROUP BY id
ORDER BY num DESC
Limit 1;
603 连续空余座位
用表进行自连接，不过From Cinema c1, Cinema c2不加任何过滤条件的话，就是笛卡尔积，所以需要加上两个seat_id值相减作为条件，最后在使用DISTINCT去重。
不使用DISTINCT，则会出现，比如3 4 5 连续，3 4，4 3，4 5，5 4，根据过滤条件，4会出现两次。
SELECT DISTINCT c1.seat_id
FROM Cinema c1,Cinema c2
WHERE ABS(c1.seat_id - c2.seat_id) = 1 AND c1.free = 1 AND c2.free = 1
ORDER BY c1.seat_id;
窗口函数 - MySQL-ROW_NUMBER()
with temp as (
select
seat_id,
seat_id - row_number() over() as k
from cinema where free = 1
)
select seat_id from temp where k in (
select k from temp group by k having count(*) >= 2
);
607 销售员
SELECT name
FROM SalesPerson
WHERE sales_id NOT IN (
SELECT DISTINCT sales_id
FROM Orders
WHERE com_id = (SELECT com_id
FROM Company
WHERE name = 'RED')
);
608 树节点
使用union
SELECT id,'Root' Type
FROM tree
WHERE p_id IS NULL
union
SELECT id,'Inner' Type
FROM tree
WHERE id IN (
SELECT DISTINCT p_id
FROM tree
WHERE p_id IS NOT NULL
) AND p_id IS NOT NULL
union
SELECT id,'Leaf' Type
FROM tree
WHERE id NOT IN (
SELECT DISTINCT p_id
FROM tree
WHERE p_id IS NOT NULL
) AND p_id IS NOT NULL
ORDER BY id ASC;
CASE … WHEN …
当父节点不为空，并且本身也是父节点的节点，就是内部节点
SELECT id,
CASE
WHEN p_id IS NULL
THEN 'Root'
WHEN id IN (SELECT DISTINCT p_id FROM tree ) # AND p_id IS NOT NULL
THEN 'Inner'
ELSE 'Leaf'
END AS Type
FROM
tree
ORDER BY id ASC;
610 判断三角形
这题点击“执行代码”以后，总是出错，但是点击“提交”，却能正常通过…
SELECT x,y,z,(
CASE WHEN x+y>z AND y+z>x AND x+z>y THEN 'Yes'
ELSE 'No'
END
) triangle
FROM Triangle;
612 平面上的最近距离
SELECT ROUND(
MIN(SQRT(
POWER(t1.x-t2.x,2)+power(t1.y-t2.y,2)
))
,2) shortest
FROM Point2D t1 CROSS JOIN Point2D t2
WHERE (t1.x,t1.y)!=(t2.x,t2.y);
其实没有必要计算每一个点与其他所有点的距离，因为这些距离可能已经被计算过了。那么该如何避免重复计算呢？
当把一个表与它自己连接以后，可以只计算每个点与X坐标比它大的点坐标的距离。这样可以避免很多重复的计算
SELECT ROUND(
MIN(SQRT(POWER(t1.x-t2.x,2)+POWER(t1.y-t2.y,2)))
,2) shortest
FROM Point2D t1 CROSS JOIN Point2D t2
WHERE (t1.x <= t2.x AND t1.y > t2.y) OR (t1.x <= t2.x AND t1.y < t2.y) OR (t1.x < t2.x AND t1.y = t2.y);
613 直线上的最近距离
point表进行自连接，不能自己和自己连，即可
SELECT MIN(ABS(t1.x - t2.x)) shortest
FROM `point` t1, `point` t2
WHERE t1.x != t2.x
ORDER BY t1.x ASC
614 二级关注者
SELECT DISTINCT follower,num
FROM follow t1
JOIN (
SELECT followee,COUNT(followee) num
FROM follow
GROUP BY followee
) t2 ON t1.follower = t2.followee
ORDER BY follower ASC;
615 平均工资：部门与公司比较
先按照年-月进行分组，求出每一个月公司的平均工资
再按照年月+部门分组，求出每一个月，每一个部门的平均工资，最后，将两个表按照日期进行连接，比较公司平均工资和部门平均工资
SELECT
t1.pay_month,
department_id,
CASE WHEN t1.avg_amount > department_avg_amount THEN 'lower'
WHEN t1.avg_amount = department_avg_amount THEN 'same'
WHEN t1.avg_amount < department_avg_amount THEN 'higher' END comparison
FROM (
SELECT
DATE_FORMAT(pay_date,'%Y-%m') pay_month,
SUM(amount)/COUNT(t1.employee_id) avg_amount
FROM Salary t1
JOIN employee t2
ON t1.employee_id = t2.employee_id
GROUP BY DATE_FORMAT(pay_date,'%Y-%m')
) t1
JOIN (
SELECT
DATE_FORMAT(pay_date,'%Y-%m') pay_month,
t2.department_id,
SUM(amount)/COUNT(t1.employee_id) department_avg_amount
FROM Salary t1
JOIN employee t2
ON t1.employee_id = t2.employee_id
GROUP BY DATE_FORMAT(pay_date,'%Y-%m'),t2.department_id
) t2
ON t1.pay_month = t2.pay_month;
窗口函数
SELECT DISTINCT *
FROM (
SELECT
DATE_FORMAT(t1.pay_date,'%Y-%m') pay_month,
t2.department_id,
CASE
WHEN AVG(amount) OVER(PARTITION BY t2.department_id,t1.pay_date) < AVG(amount) OVER(PARTITION BY t1.pay_date) THEN 'lower'
WHEN AVG(amount) OVER(PARTITION BY t2.department_id,t1.pay_date) = AVG(amount) OVER(PARTITION BY t1.pay_date) THEN 'same'
WHEN AVG(amount) OVER(PARTITION BY t2.department_id,t1.pay_date) > AVG(amount) OVER(PARTITION BY t1.pay_date) THEN 'higher'
END comparison
FROM salary t1
JOIN employee t2
ON t1.employee_id = t2.employee_id
) temp;
618 学生地理信息报告
这种类型的题之前没有见过，输出的结果就像是把表转了个方向
首先，FROM后面的子表按照continent分组，按照name排序
SELECT *,ROW_NUMBER() OVER(PARTITION BY continent ORDER BY `name`) rn
FROM student
题目中指明最后的结果：每个学生按照姓名的字母顺序依次排列在对应的大洲的下面，所以，ROW_NUMBER()结果相同的，都在同一行
最后使用CASE WHEN 取出对应的name
SELECT
MAX(CASE continent WHEN 'America' THEN `name` ELSE NULL END )America,
MAX(CASE continent WHEN 'Asia' THEN `name` ELSE NULL END) Asia,
MAX(CASE continent WHEN 'Europe' THEN `name` ELSE NULL END) Europe
FROM (
SELECT *,ROW_NUMBER() OVER(PARTITION BY continent ORDER BY `name`) rn
FROM student
) temp
GROUP BY rn;
或者把MAX换成MIN都可以，根据rn分组以后的，每一组肯定只有一个数，但是直接输出会报错，只要把每一个组中的值取出来就行，MIN或者MAX都可以
SELECT
MIN(CASE continent WHEN 'America' THEN `name` ELSE NULL END )America,
MIN(CASE continent WHEN 'Asia' THEN `name` ELSE NULL END) Asia,
MIN(CASE continent WHEN 'Europe' THEN `name` ELSE NULL END) Europe
FROM (
SELECT *,ROW_NUMBER() OVER(PARTITION BY continent ORDER BY `name`) rn
FROM student
) temp
GROUP BY rn;
619 只出现一次的最大数字
# Write your MySQL query statement below
SELECT IFNULL(
(SELECT num FROM MyNumbers
GROUP BY num
HAVING COUNT(num) = 1
ORDER BY num DESC
LIMIT 1),
NULL) num;
1045 买下所有产品的客户
SELECT customer_id
FROM Customer
GROUP BY customer_id
HAVING COUNT(DISTINCT product_key) = (
SELECT COUNT(product_key) prduct_cnt
FROM Product
);
1050 合作过至少三次的演员和导演
SELECT actor_id, director_id
FROM ActorDirector
GROUP BY actor_id,director_id
HAVING COUNT(*) >= 3;
1068 产品销售分析
表连接操作
SELECT product_name, year, price
FROM Sales t1
LEFT JOIN Product t2 ON t1.product_id = t2.product_id;
1069 产品销售分析II
SELECT product_id, SUM(quantity) total_quantity
FROM Sales
GROUP BY product_id;
1070 产品销售分析III
SELECT product_id,year first_year,quantity,price
FROM Sales
WHERE (product_id,year) IN (
SELECT product_id,MIN(year) first_year
FROM Sales
GROUP BY product_id
);
1075 项目员工 I
SELECT project_id, ROUND(AVG(experience_years),2) average_years
FROM Project t1
JOIN Employee t2 ON t1.employee_id = t2.employee_id
GROUP BY project_id;
1076 项目员工II
SELECT project_id
FROM Project
GROUP BY project_id
HAVING COUNT(*) >= ALL(
SELECT COUNT(*)
FROM Project
GROUP BY project_id
);
1077 项目员工III
SELECT project_id,t1.employee_id
FROM Project t1
JOIN Employee t2 ON t1.employee_id = t2.employee_id
WHERE (project_id,experience_years) IN (
SELECT project_id,MAX(experience_years)
FROM Project t1
JOIN Employee t2 ON t1.employee_id = t2.employee_id
GROUP BY project_id
);
1082 销售分析
SELECT seller_id
FROM Sales
GROUP BY seller_id
HAVING SUM(price) >= ALL(
SELECT SUM(price)
FROM Sales
GROUP BY seller_id
);
1083 销售分析II
查找购买了S8但是没有购买iPhone的买家
FROM子句里面的子查询，查找的是购买了S8的买家，WHERE子句里查找的是，没有购买iPhone的买家
SELECT DISTINCT buyer_id
FROM (
SELECT buyer_id
FROM Sales t1
JOIN Product t2 ON t1.product_id = t2.product_id
WHERE product_name = 'S8'
) t
WHERE buyer_id NOT IN (
SELECT buyer_id
FROM Sales t1
JOIN Product t2 ON t1.product_id = t2.product_id
WHERE product_name = 'iPhone'
);
SELECT DISTINCT buyer_id
FROM Sales t1
JOIN Product t2 ON t1.product_id = t2.product_id
WHERE product_name = 'S8' AND buyer_id NOT IN (
SELECT buyer_id
FROM Sales t1
JOIN Product t2 ON t1.product_id = t2.product_id
WHERE product_name = 'iPhone'
);
SELECT buyer_id
FROM Sales t1
JOIN Product t2 ON t1.product_id = t2.product_id
GROUP BY buyer_id
HAVING SUM(IF(product_name='S8',1,0))>=1
AND SUM(IF(product_name='iPhone',1,0))=0
1084 销售分析III
SELECT t1.product_id, product_name
FROM Sales t1
JOIN Product t2 ON t1.product_id = t2.product_id
GROUP BY t1.product_id
HAVING (MAX(sale_date) BETWEEN '2019-01-01' AND '2019-03-31') AND (MIN(sale_date) BETWEEN '2019-01-01' AND '2019-03-31');
SELECT t1.product_id, product_name
FROM Sales t1
JOIN Product t2 ON t1.product_id = t2.product_id
GROUP BY t1.product_id
HAVING MAX(sale_date) <= '2019-03-31' AND MIN(sale_date) >= '2019-01-01';
1097 游戏玩法分析 V
先按照player_id进行分组，找到每一个player的登陆日期，然后计算出登陆日期后一天的日期next_date，将结果作为一张临时表
将该临时表与Activity表进行左外连接，连接的条件是两张表的player_id的值要是一样的，并且Activity中得有next_date的日期，按照login_date进行分组，最后统计相应的数据即可
WITH temp AS (
SELECT
player_id,
MIN(event_date) login_date,
DATE_SUB(MIN(event_date),INTERVAL -1 DAY) next_day
FROM Activity
GROUP BY player_id
)
SELECT
t1.login_date install_dt,
COUNT(t1.login_date) installs,
ROUND(COUNT(t2.event_date)/COUNT(t1.login_date),2) Day1_retention
FROM temp t1
LEFT JOIN Activity t2
ON t1.player_id = t2.player_id AND t1.next_day = t2.event_date
GROUP BY login_date
1098 小众书籍
SELECT t1.book_id,t1.name
FROM (
SELECT book_id,name
FROM Books
WHERE DATEDIFF('2019-06-23',available_from) > 31
) t1
LEFT JOIN (
SELECT book_id,quantity,dispatch_date
FROM Orders
WHERE DATEDIFF('2019-06-23',dispatch_date) <= 365
) t2 ON t1.book_id = t2.book_id
GROUP BY t1.book_id
HAVING SUM(IFNULL(t2.quantity,0))<10 ;
SELECT t1.book_id,t1.name
FROM books t1
LEFT JOIN Orders t2
ON t1.book_id = t2.book_id AND DATEDIFF('2019-06-23',dispatch_date)<=365
WHERE DATEDIFF('2019-06-23',available_from) > 31
GROUP BY t1.book_id
HAVING SUM(IFNULL(quantity,0))<10;
1107 每日新用户统计
SELECT min_activity_date login_date, COUNT(min_activity_date) user_count
FROM (
SELECT user_id, MIN(activity_date) min_activity_date
FROM Traffic
WHERE activity = 'login'
GROUP BY user_id
HAVING MIN(activity_date) >= '2019-04-01'
) t1
GROUP BY min_activity_date;
SELECT min_activity_date login_date, COUNT(min_activity_date) user_count
FROM (
SELECT user_id, MIN(activity_date) min_activity_date
FROM Traffic
WHERE activity = 'login'
GROUP BY user_id
) t1
WHERE min_activity_date >= '2019-04-01'
GROUP BY min_activity_date;
1112 每位学生的最高成绩
SELECT student_id, MIN(course_id) course_id, grade
FROM Enrollments
WHERE (student_id,grade) IN (
SELECT student_id, MAX(grade)
FROM Enrollments
GROUP BY student_id
)
GROUP BY student_id
ORDER BY student_id ASC;
1113 报告的记录
SELECT extra report_reason, COUNT(DISTINCT post_id) report_count
FROM Actions
WHERE action_date = '2019-07-04' AND action = 'report' AND extra IS NOT NULL
GROUP BY extra;
1126 查询活跃业务
SELECT t1.business_id
FROM Events t1
JOIN (
SELECT event_type, AVG(occurences) avg_occurences
FROM Events
GROUP BY event_type
) t2 ON t1.event_type = t2.event_type
GROUP BY t1.business_id
HAVING SUM(t1.occurences > t2.avg_occurences) >= 2
SELECT t1.business_id
FROM Events t1
JOIN (
SELECT event_type, AVG(occurences) avg_occurences
FROM Events
GROUP BY event_type
) t2 ON t1.event_type = t2.event_type
WHERE t1.occurences > t2.avg_occurences
GROUP BY t1.business_id
HAVING COUNT(*) >= 2;
1127 用户购买平台
首先按照spend_date和user_id分组，根据购买记录确定用户的platform是both、mobile还是desktop，生成子表t1
其次，将both、mobile、desktop作为单独一个表t2
t2与t1进行笛卡尔积连接
最后按照t1.platform和t2.spend_date进行分组，得到结果
SELECT
t2.spend_date,
t1.platform,
SUM(IF(t1.platform=t2.platform,amount,0)) total_amount,
COUNT(IF(t1.platform=t2.platform,1,NULL)) total_users
FROM (
SELECT 'both' platform
UNION
SELECT 'mobile' platform
UNION
SELECT 'desktop' platform
) t1,
(
SELECT
spend_date,
IF(COUNT(DISTINCT platform)=2,'both',any_value(platform)) platform,
user_id,
SUM(amount) amount
FROM Spending
GROUP BY spend_date,user_id
) t2
GROUP BY t1.platform,t2.spend_date
1132 报告的记录III
注意：相同的post_id只能算一个
SELECT ROUND(AVG(avg_daily_percent)*100,2) average_daily_percent
FROM (
SELECT COUNT(DISTINCT t2.post_id)/COUNT(DISTINCT t1.post_id) avg_daily_percent
FROM Actions t1
LEFT JOIN Removals t2 ON t1.post_id = t2.post_id
WHERE extra = 'spam'
GROUP BY t1.action_date
) temp;
1141 查询近30天活跃用户
DATEDIFF的结果除了要小于29之外，还要大于等于0，如果没有大于等于0的条件，会查找到2019-07-27之后的日子
SELECT activity_date day,COUNT(DISTINCT user_id) active_users
FROM Activity
WHERE DATEDIFF('2019-07-27',activity_date)<=29
AND DATEDIFF('2019-07-27',activity_date)>=0
GROUP BY activity_date;
1142 过去30天的用户活动II
SELECT ROUND(IFNULL(COUNT(DISTINCT session_id)/COUNT(DISTINCT user_id),0),2)average_sessions_per_user
FROM Activity
WHERE DATEDIFF('2019-07-27',activity_date)<=29;
1148 文章浏览I
SELECT DISTINCT viewer_id id
FROM Views t1
WHERE viewer_id = author_id
ORDER BY id ASC;
1149 文章浏览
注意：需要加上DISTINCT，因为，可能存在同一个用户，在不同的天都读了两篇及以上的文章，这样，在最后输出的时候，同一个id会出现多次
SELECT DISTINCT viewer_id id
FROM Views
GROUP BY viewer_id, view_date
HAVING COUNT(DISTINCT article_id) >= 2
ORDER BY id ASC;
1158 市场分析I
SELECT user_id buyer_id, join_date, COUNT(IF(YEAR(order_date)=2019,1,NULL)) orders_in_2019
FROM Users t1
LEFT JOIN Orders t2
ON t1.user_id = t2.buyer_id
GROUP BY t1.user_id
1159 市场分析II
使用ROW_NUMBER()
WITH temp AS (
SELECT
seller_id,
item_brand,
favorite_brand,
ROW_NUMBER() OVER(PARTITION BY t1.seller_id ORDER BY order_date) ranking
FROM Orders t1,Items t2, Users t3
WHERE t1.seller_id = t3.user_id AND t1.item_id = t2.item_id
)
SELECT t1.user_id,IF(t2.seller_id IS NULL,'no','yes') 2nd_item_fav_brand
FROM (
SELECT user_id
FROM Users
)t1
LEFT JOIN (
SELECT seller_id
FROM temp
WHERE ranking = 2 AND item_brand = favorite_brand
) t2 ON t1.user_id = t2.seller_id
使用NTH_VALUE()
注意，在使用NTH_VALUE时，有可能某一个seller_id卖的物品数大于两个，所以，在进行LEFT JOIN的时候，会出现重复的值，所以要在t2这个子查询中使用DISTINCT
WITH temp AS (
SELECT
seller_id,
favorite_brand,
NTH_VALUE(item_brand,2) OVER(PARTITION BY t1.seller_id ORDER BY order_date) item_brand
FROM Orders t1,Items t2, Users t3
WHERE t1.seller_id = t3.user_id AND t1.item_id = t2.item_id
)
SELECT t1.user_id seller_id ,IF(t2.seller_id IS NULL,'no','yes') 2nd_item_fav_brand
FROM (
SELECT user_id
FROM Users
)t1
LEFT JOIN (
SELECT DISTINCT seller_id
FROM temp
WHERE item_brand IS NOT NULL AND item_brand = favorite_brand
) t2 ON t1.user_id = t2.seller_id
其他思路：使用自连接，having+sum判断
第一步，先将表自连接，两个Orders表，一个Items表
SELECT *
FROM Orders t1,Orders t2,Items t3
WHERE t1.seller_id = t2.seller_id AND t1.item_id = t3.item_id
ORDER BY t1.order_date,t1.seller_id;
按照order_id进行分组，统计比当前订单更早的订单数，若为1，说明当前订单是对应用户卖出的第二单
SELECT t1.user_id seller_id,IF(t2.item_brand=t1.favorite_brand,'yes','no') 2nd_item_fav_brand
FROM Users t1
LEFT JOIN (
SELECT
any_value(t1.seller_id) seller_id,
any_value(item_brand) item_brand
FROM Orders t1,Orders t2,Items t3
WHERE t1.seller_id = t2.seller_id AND t1.item_id = t3.item_id
GROUP BY t1.order_id
HAVING SUM(t1.order_date > t2.order_date) = 1
) t2
ON t1.user_id = t2.seller_id
1164 指定日期的产品价格
本题关键点在于找到2019-08-16前所有有改动的产品及其最新价格和没有修改过价格的产品
可以先找到所有的产品，再找到所有2019-08-16前有修改的产品和他们最新的价格，使用left join将两个查询联合。如果产品没有价格，说明没有修改过，设置为10，如果有价格，设置为最新的价格。
找出所有的产品
SELECT DISTINCT product_id
FROM Products;
找出2019-08-16前所有有改动的产品的最新价格
- 使用max函数找到产品最新修改的时间，使用where查询限制时间小于等于
2019-08-16
SELECT product_id, MAX(change_date)
FROM Products
WHERE change_date <= '2019-08-16'
GROUP BY product_id;
- 使用where子查询，根据prodct_id和change_date找到对应的价格
SELECT product_id, new_price
FROM Products
WHERE (product_id,change_date) IN (
SELECT product_id, MAX(change_date)
FROM Products
WHERE change_date <= '2019-08-16'
GROUP BY product_id
);
经过上面两步已经找到了所有的产品和已经修改过价格的产品。使用left join得到所有产品的最新价格，如果没有，设置为10
SELECT t1.product_id,IFNULL(t2.price,10) price
FROM (
SELECT DISTINCT product_id
FROM Products
) t1
LEFT JOIN (
SELECT product_id, new_price price
FROM Products
WHERE (product_id,change_date) IN (
SELECT product_id, MAX(change_date)
FROM Products
WHERE change_date <= '2019-08-16'
GROUP BY product_id
)
) t2 ON t1.product_id = t2.product_id;
1173 即时食物配送
SELECT ROUND((
SELECT COUNT(*)*100 FROM Delivery WHERE order_date = customer_pref_delivery_date
) / COUNT(*),2) immediate_percentage
FROM Delivery;
1174 即时食物配送II
SELECT ROUND(
SUM(order_date = customer_pref_delivery_date) * 100 / COUNT(DISTINCT customer_id)
,2) immediate_percentage
FROM Delivery
WHERE (customer_id,order_date) IN (
SELECT customer_id,MIN(order_date) min_order_date
FROM Delivery
GROUP BY customer_id
);
1193 每月交易I
SELECT
DATE_FORMAT(trans_date,'%Y-%m') `month`,
country,
COUNT(id) trans_count,
SUM(state='approved') approved_count,
SUM(amount) trans_total_amount,
SUM(IF(state='approved',amount,0)) approved_total_amount
FROM Transactions
GROUP BY `month`,country;
1194 锦标赛优胜者
首先计算出每位选手的总分，然后用ROW_NUMBER()给选手排序，按照group_id分组，按照score降序排列，如果score一样，则player_id小的优先，最后返回每一个组需要唯一的选手
注意在使用UNION时，只能使用UNION ALL，不能使用UNION，因为UNION有去重的作用
WITH temp AS (
SELECT
player,
SUM(score) score
FROM(
(
SELECT first_player player,SUM(first_score) score
FROM Matches
GROUP BY first_player
)
UNION ALL
(
SELECT second_player player,SUM(second_score) score
FROM Matches
GROUP BY second_player
)
)t1
GROUP BY player
)
SELECT group_id,player_id
FROM(
SELECT
player_id,
group_id,
ROW_NUMBER() OVER(PARTITION BY group_id ORDER BY score DESC ,player_id ASC) rk
FROM temp
JOIN Players t2
ON temp.player = t2.player_id
) t
WHERE rk = 1
1204 最后一个进入电梯的人
需要根据turn排序，并累加weight，找到最后一个使得总和小于等于1000的person_name。
只需要计算出每个人进去后的总和，找到总和小于等于1000的最后一个人即可。
可以使用自连接，将b表中的每一条数据和a表中的每一条数据连接。假设Queue有3条记录，那么自连接后将会有9条数据，分别为(a1 b1) (a1 b2) (a1 b3) (a2 b1) (a2 b2) (a2 b3) (a3 b1) (a3 b2) (a3 b3)
接下来对数据进行处理，使用a表的person_id表示自身，b表中的数据表示为包括自己在内的所有人。使用GROUP BY a.person_id处理每个人的数据。因为要计算每个人的weght加上之前所有人的weight，使用查询条件a.turn>=b.turn找到所有在他之间以及他自己的重量。再使用SUM计算总和并过滤掉大于1000的数据
SELECT q1.person_name
FROM Queue q1
CROSS JOIN Queue q2
WHERE q1.turn >= q2.turn
GROUP BY q1.person_id
HAVING SUM(q2.weight) <= 1000
ORDER BY SUM(q2.weight) DESC
LIMIT 1;
1205 每月交易II
本题难点在于如何处理Chargebacks表，每项退单都对应于之前进行的交易，即使未经批准，说明我们应该使用full outer join构造chargeback这个state，然后再对年月GROUP BY计算各类指标。
- union all
SELECT * FROM Transactions
UNION ALL
SELECT
t1.trans_id id,
t2.country,
'chargeback' state,
t2.amount,
t1.trans_date
FROM Chargebacks t1
LEFT JOIN Transactions t2
ON t1.trans_id = t2.id;
- 计算指标
按照（month，country）分组
还要忽略所有为零的行，也就是当approved_count ，chargeback_count 都为0时，不输出
WITH cet AS(
SELECT * FROM Transactions
UNION ALL
SELECT
t1.trans_id id,
t2.country,
'chargeback' state,
t2.amount,
t1.trans_date
FROM Chargebacks t1
LEFT JOIN Transactions t2
ON t1.trans_id = t2.id
)
SELECT
DATE_FORMAT(trans_date,'%Y-%m') `month`,
country,
SUM(state='approved') approved_count,
SUM(IF(state='approved',amount,0)) approved_amount,
SUM(state='chargeback') chargeback_count,
SUM(IF(state='chargeback',amount,0)) chargeback_amount
FROM cet
GROUP BY `month`,country
HAVING approved_count OR chargeback_count;
# HAVING approved_amount OR chargeback_amount;
1211 查询结果的质量和占比
SELECT
query_name,
ROUND(SUM(rating/position)/COUNT(*),2) quality,
ROUND(SUM(IF(rating<3,1,0))*100/COUNT(*),2) poor_query_percentage
FROM Queries
GROUP BY query_name;
1212 查询球队积分
SELECT team_id,team_name,SUM(num_points) num_points
FROM (
# 作为主队
SELECT
t1.team_id,
t1.team_name,
SUM(IFNULL(
CASE WHEN host_goals > guest_goals THEN 3
WHEN host_goals = guest_goals THEN 1 END
,0)) num_points
FROM Teams t1
LEFT JOIN Matches t2
ON t1.team_id = t2.host_team
GROUP BY t1.team_id
UNION ALL
#作为客队
SELECT
t1.team_id,
t1.team_name,
SUM(IFNULL(
CASE WHEN host_goals < guest_goals THEN 3
WHEN host_goals = guest_goals THEN 1 END
,0)) num_points
FROM Teams t1
LEFT JOIN Matches t2
ON t1.team_id = t2.guest_team
GROUP BY t1.team_id
) t
GROUP BY team_id
ORDER BY num_points DESC,team_id ASC;
1225 报告系统状态的连续日期
首先将Succeeded表和Failed表中的数据打标签，标明该条数据来源于哪一张表，最后用UNION将两张表连接起来
SELECT
fail_date AS `date`,
'failed' AS period_state
FROM Failed
UNION
SELECT
success_date,
'succeeded'
FROM Succeeded
ORDER BY `date`
求解某些记录是否连续就会想到使用ROW_NUMBER，当某几条记录和ROW_NUMBER的差是一样的时候，就说明它们是连续的，但是这也是针对与ROW_NUMBER相减的是数字才行，现在的类型是Date，该怎么办呢？
首先，按照date进行排序，对它们进行标号，然后按照period_state进行分组，再按照date进行排序，再对它们进行一次标号，两次标号进行相减，值一样的，就说明日期是连续的
WITH temp AS(
SELECT
fail_date AS `date`,
'failed' AS period_state
FROM Failed
UNION
SELECT
success_date,
'succeeded'
FROM Succeeded
ORDER BY `date`
)
SELECT
period_state,
`date`,
row_number() over(ORDER BY `date`) date_num,
row_number() over(PARTITION BY period_state ORDER BY `date`) period_num,
row_number() over(ORDER BY `date`) -
row_number() over(PARTITION BY period_state ORDER BY `date`) AS rnk_diff
FROM temp
WHERE YEAR(`date`) = 2019
WITH temp AS(
SELECT
fail_date AS `date`,
'failed' AS period_state
FROM Failed
UNION
SELECT
success_date,
'succeeded'
FROM Succeeded
ORDER BY `date`
)
SELECT
period_state,
MIN(`date`) start_date,
MAX(`date`) end_date
FROM (
SELECT
period_state,
`date`,
row_number() over(ORDER BY `date`) -
row_number() over(PARTITION BY period_state ORDER BY `date`) AS rnk_diff
FROM temp
WHERE YEAR(`date`) = 2019
) t
GROUP BY period_state,rnk_diff
ORDER BY start_date ASC;
1241 每个帖子的评论数
SELECT t1.sub_id post_id, COUNT(DISTINCT t2.sub_id) number_of_comments
FROM (
SELECT DISTINCT sub_id
FROM Submissions
WHERE parent_id IS NULL
) t1
LEFT JOIN Submissions t2
ON t1.sub_id = t2.parent_id
GROUP BY t1.sub_id
ORDER BY post_id ASC;
1251 平均售价
除了使用product_id进行表连接时，还需要保证purchase_date在start_date和end_date之间
SELECT t1.product_id, ROUND(SUM(units*price)/SUM(units),2) average_price
FROM Prices t1
JOIN UnitsSold t2 ON t1.product_id = t2.product_id AND purchase_date BETWEEN start_date AND end_date
GROUP BY product_id;
1264 页面推荐
SELECT DISTINCT page_id recommended_page
FROM Likes
WHERE page_id NOT IN (
# id为1的用户喜欢的页面
SELECT page_id
FROM Likes
WHERE user_id = 1
) AND user_id IN (
# 朋友id
SELECT user2_id
FROM Friendship
WHERE user1_id = 1
UNION
SELECT user1_id
FROM Friendship
WHERE user2_id = 1
);
1270 向公司CEO汇报工作的所有人
方案一 UNION ALL + 子查询
本题规定经理之间的间接关系不超过三个经理，那么我们可以分别求出3层的人数，最后汇总到一起。
首先第一层是CEO的直接汇报人，需要把CEO排除掉：
SELECT employee_id
FROM Employees
WHERE manager_id = 1 AND employee_id != 1
第二层汇报人的maager_id就是第一层需要向CEO汇报的人，可以使用子查询，求出第二层的人
SELECT employee_id
FROM Employees
WHERE manager_id IN (
SELECT employee_id
FROM Employees
WHERE manager_id = 1 AND employee_id != 1
)
同样，第三层的汇报人，就是第二步求出来的人
SELECT employee_id
FROM Employees
WHERE manager_id IN (
SELECT employee_id
FROM Employees
WHERE manager_id IN (
SELECT employee_id
FROM Employees
WHERE manager_id = 1 AND employee_id != 1
)
);
至此，就求出来了所有汇报的人，然后只需要用UNION ALL将他们汇总到一起，输出即可。
答案如下
(
SELECT employee_id
FROM Employees
WHERE manager_id = 1 AND employee_id != 1
UNION ALL
SELECT employee_id
FROM Employees
WHERE manager_id IN (
SELECT employee_id
FROM Employees
WHERE manager_id = 1 AND employee_id != 1
)
)
UNION ALL
SELECT employee_id
FROM Employees
WHERE manager_id IN (
SELECT employee_id
FROM Employees
WHERE manager_id IN (
SELECT employee_id
FROM Employees
WHERE manager_id = 1 AND employee_id != 1
)
);
1280 学生们参加各科测试的次数
COUNT()里面写 t2.subject_name 和 t3.subject_name的效果还是不一样的
SELECT t1.student_id, t1.student_name, t2.subject_name, COUNT(t3.subject_name) attended_exams
FROM Students t1 CROSS JOIN Subjects t2
LEFT JOIN Examinations t3 ON t1.student_id = t3.student_id AND t2.subject_name = t3.subject_name
GROUP BY t1.student_id,t2.subject_name
ORDER BY t1.student_id,t2.subject_name;
1285 找到连续区间的开始和结束数字
方案一 窗口函数
如何知道连续区间的开始和结束数字，首先需要知道这个区间
一个连续区间的数减去某个个规律的数结果应该是一样的
基于这个思路会想到，将这些数按照从小到大的顺序进行排列，那么这个数字减去它在这张表中的排列顺序，如果是一个区间内的数，那么结果应该是一样的。
SELECT
log_id,
log_id - row_number() over() diff
FROM Logs
上面SQL的运行结果，可以看到，如果是连续的数字，log_id - row_number()的值是一样的
SELECT
MIN(log_id) start_id,
MAX(log_id) end_id
FROM (
SELECT
log_id,
log_id - row_number() over() diff
FROM Logs
) t
GROUP BY diff;
方案二
第一步，一个连续的区间，start_id -1 和 end_id + 1肯定不在表中
将log_id-1的值肯定不在Logs中，将符合条件的这些值的log_id查询出来就是start_id
SELECT log_id start_id, row_number() over(ORDER BY log_id) rank_id
FROM Logs
WHERE log_id -1 NOT IN (SELECT log_id FROM Logs);
第二步，同理，end_id+1的值肯定不在Logs中
SELECT log_id end_id, row_number() over(ORDER BY log_id) rank_id
FROM Logs
WHERE log_id + 1 NOT IN (SELECT log_id FROM Logs);
将第一步的结果作为临时表t1，将第二步的结果作为临时表t2
-
一个区间有start_id，就一定会有end_id，所以，start_id和end_id的数量应该是相等的
-
因为start_id和end_id的数量相同，所以可以使用排序函数分别对t1和t2表进行排序 row_number() over(order by log_id) rank_id
-
两表的连接条件就是rank_id相等
SELECT start_id, end_id
FROM (
SELECT log_id start_id, row_number() over(ORDER BY log_id) rank_id
FROM Logs
WHERE log_id - 1 NOT IN (SELECT log_id FROM Logs)
) t1 JOIN (
SELECT log_id end_id, row_number() over(ORDER BY log_id) rank_id
FROM Logs
WHERE log_id + 1 NOT IN (SELECT log_id FROM Logs)
) t2 ON t1.rank_id = t2.rank_id;
1294 不同国家的天气类型
SELECT country_name, CASE WHEN AVG(weather_state) <= 15 THEN 'Cold'
WHEN AVG(weather_state) >= 25 THEN 'Hot'
ELSE 'Warm' END weather_type
FROM Weather t1
JOIN Countries t2 ON t1.country_id = t2.country_id
WHERE DATE_FORMAT(day,'%Y%m') = '201911'
GROUP BY country_name;
1303 求团队人数
子查询
SELECT employee_id,team_size
FROM Employee t1
JOIN (
SELECT team_id,COUNT(team_id) team_size
FROM Employee
GROUP BY team_id
) t2 ON t1.team_id = t2.team_id;
窗口函数
SELECT employee_id,COUNT(*) OVER(PARTITION BY team_id) team_size
FROM Employee;
1308 不同性别每日分数总计
方案一 表连接
SELECT t1.gender, t1.`day`,SUM(t2.score_points) total
FROM Scores t1
JOIN Scores t2 ON t1.gender = t2.gender AND t1.day >= t2.day
GROUP BY t1.gender,t1.`day`
ORDER BY t1.gender ASC,t1.`day` ASC;
方案二 窗口函数（不太理解）
在窗口的每条记录动态应用聚合函数SUM，可以动态计算在指定的窗口内的累计分数
# Write your MySQL query statement below
SELECT
gender, `day`,
SUM(score_points) OVER (PARTITION BY gender ORDER BY `day`) total
FROM scores;
1321 餐馆营业额变化增长
为什么要在SELECT中加入DISTINCT？
就所给示例来说，2019-01-10这一天有两个顾客，在进行表连接时，这两条数据连接的最近7天的内容是一样的，外加另一条2019-01-10那条数据，所以在展示时，只要展示一个就可以
为什么HAVING COUNT中加入DISTINCT？
就所给示例来说，2019-01-10这一天有两个顾客，再计算它所连接的另一个表中的记录的时候，会算上另一个2019-01-10那个条数据，COUNT的结果是8，但是，我们想要得到的是最近七天，也就是把同一天的看成一个整体，所以，需要去重
SELECT DISTINCT t1.visited_on, SUM(t2.amount) amount,ROUND(SUM(t2.amount)/7,2)average_amount
FROM Customer t1
JOIN Customer t2
ON DATEDIFF(t1.visited_on,t2.visited_on)<=6 AND DATEDIFF(t1.visited_on,t2.visited_on)>=0
GROUP BY t1.visited_on,t1.customer_id
HAVING COUNT(DISTINCT t2.visited_on) = 7
ORDER BY t1.visited_on;