「LeetCode SQL入门学习计划」

Leetcode MySQL 入门学习计划

选择

595. 大的国家
思路：

1. 简单过滤，无顺序要求
2. 显示字段 name, population, area
3. area >= 3 000 000 or population >=25 000 000

select name, population, area from World where area >= 3000000 
or population >=25000000

1757. 可回收且低脂的产品
思路：

1. 简单过滤， 无顺序要求
2. 显示字段product_id
3. 低脂 and 可回收

select product_id from Products where low_fats='Y' and recyclable='Y'

584. 寻找用户推荐人
思路：

1. 简单过滤，无顺序要求
2. 显示字段name
3. 推荐人编号不是2有两种情况，为空或者非空非2

select name from customer where referee_id<>2 or referee_id is null

183. 从不订购的客户
思路：

1. 涉及两表，所以需要连接两张表然后进行条件过滤
   1. 自查询作为筛选范围
   2. join连接
2. 两张表靠Customers表的id连接可用left join

select Name as Customers from Customers as c 
where c.id not in (select CustomerId from Orders)

select Name as Customers from Customers c
left join Orders o on c.id = o. CustomerId 
where o.CustomerId is null

排序 & 修改

1873. 计算特殊奖金
思路：

1. 需要构造bonus字段在查询结果
2. 结果需要按照employee_id升序
3. case when then else end
4. 取余函数mod(, ), 字符串中的charAt - left(, )

select employee_id,
case when mod(employee_id, 2)=1 and left(name,1)!='M' then salary
else 0
end bonus from Employees
order by employee_id

627. 变更性别
思路：

1. 用 case when then else end 控制赋值

update Salary set sex = case 
when sex = 'f' then 'm'
else 'f'
end

196. 删除重复的电子邮箱
思路：

1. 删除的语句时delete from where
2. 这里需要构建一个相同的表进行比较删除

字符串处理函数/正则

1667. 修复表中的名字

思路：

1. 用upper函数全变为大写再使用left()函数

select user_id,
concat(left(upper(name),1),lower(substring(name,2))) as name
from Users order by user_id; 

1484. 按日期分组销售产品
思路：

1. 分组计数，连接名

select sell_date, 
count(distinct product) num_sold, 
group_concat(distinct product) products 
from Activities
group by sell_date

1527. 患某种疾病的患者
思路：

select patient_id, patient_name, conditions
from Patients where conditions like 'DIAB1%'
or conditions like "% DIAB1%"

组合查询 & 指定选取

1965. 丢失信息的雇员
思路：

1. 将两个表的employee_id合并
2. 找到合并后个数为1的employee_id即为缺信息的id

select 
    employee_id 
from 
    (
    select employee_id from employees
    union all 
    select employee_id from salaries
) as t
group by 
    employee_id
having 
    count(employee_id) = 1
order by 
    employee_id;

1795. 每个产品在不同商店的价格
思路：

1. 一列一列处理：把“列名”做为新列的value（如本题的store），把原来的value也作为新列（如本题的price），这是一个查询，其他列不要
2. 用union all拼接每一列的结果
3. 注意本题如果这一产品在商店里没有出售，则不输出这一行，所以要原列 is not null的筛选条件

select product_id, 'store1' store, store1 price
from products
where store1 is not null
union all 
select product_id, 'store2' store, store2 price
from products 
where store2 is not null
union all 
select product_id, 'store3' store, store3 price
from products
where store3 is not null;

608. 树节点
思路一：

1. 通过union将符合条件的结果分别查出来拼接
   union会自动压缩多个结果集合中的重复结果，而union all则将所有的结果全部显示出来，不管是不是重复

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) 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

A not in B的原理是拿A表值与B表值做是否不等的比较, 也就是a != b. 在sql中, null是缺失未知值而不是空值(详情请见MySQL reference).

当你判断任意值a != null时, 官方说, “You cannot use arithmetic comparison operators such as =, <, or <> to test for NULL”, 任何与null值的对比都将返回null. 因此返回结果为否,这点可以用代码 select if(1 = null, ‘true’, ‘false’)证实.

从上述原理可见, 当询问 id not in (select p_id from tree)时, 因为p_id有null值, 返回结果全为false, 于是跳到else的结果, 返回值为inner. 所以在答案中,leaf结果从未彰显,全被inner取代.

select
    id,
    case when p_id is null then "Root"
         when id not in (select p_id from tree where p_id is not null) then "Leaf"
         else "Inner"
    end as Type
from
    tree

176. 第二高的薪水
思路：

1. 要想获取第二高，需要排序，使用 order by（默认是升序 asc，即从小到大），若想降序则使用关键字 desc
2. 去重，如果有多个相同的数据，使用关键字 distinct 去重
3. 判断临界输出，如果不存在第二高的薪水，查询应返回 null，使用 ifNull（查询，null）方法
4. 起别名，使用关键字 as …
5. 因为去了重，又按顺序排序，使用 limit（）方法，查询第二大的数据，即第二高的薪水，即 limit(1,1) （因为默认从0开始，所以第一个1是查询第二大的数，第二个1是表示往后显示多少条数据，这里只需要一条）

select ifNull((
    select distinct salary from employee
    order by salary desc limit 1,1),null) 
    as SecondHighestSalary

合并

175. 组合两个表

select p.firstName, p.lastName, a.city , a.state 
from person p left join address a on p.personid = a.personid

1581. 进店却未进行过交易的顾客
思路一：
查询进店未消费的人然后分组统计

select v.customer_id, count(v.customer_id) count_no_trans from visits v
where v.visit_id not in (select visit_id from transactions)
group by v.customer_id

思路二：
链接两张表，用transaction_id来筛选微交易的customer_id, 然后分组统计

select v.customer_id as customer_id, count(v.customer_id) as count_no_trans 
from Visits v left join Transactions t on v.visit_id = t.visit_id 
where transaction_id is null group by v.customer_id;

1148. 文章浏览 I

select distinct author_id id from views
where author_id = viewer_id
order by viewer_id

197. 上升的温度
思路：
MySQL 使用 DATEDIFF 来比较两个日期类型的值。
因此，我们可以通过将 weather 与自身相结合，并使用 DATEDIFF() 函数。

SELECT w2.Id
FROM Weather w1, Weather w2
WHERE DATEDIFF(w2.RecordDate, w1.RecordDate) = 1
AND w1.Temperature < w2.Temperature

607. 销售员

select name from salesperson
where sales_id not in
(select o.sales_id from orders o
left join company c on o.com_id = c.com_id where c.name = "RED")

select name from salesperson
where sales_id not in
(select o.sales_id from orders o, company c
where o.com_id = c.com_id and c.name = "RED")

计算函数

1141. 查询近30天活跃用户数

select activity_date day, count(distinct user_id) active_users from activity
where datediff('2019-07-27', activity_date) >= 0 AND datediff('2019-07-27', activity_date) <30
group by activity_date

1693. 每天的领导和合伙人

select date_id, make_name, count(distinct lead_id) unique_leads, count(distinct partner_id) unique_partners from dailysales group by date_id, make_name

1729. 求关注者的数量

select user_id, count(follower_id) followers_count from followers
group by user_id
order by user_id

586. 订单最多的客户

SELECT
    customer_number
FROM
    orders
GROUP BY customer_number
ORDER BY COUNT(*) DESC
LIMIT 1
;

511. 游戏玩法分析 I

select player_id, min(event_date) as first_login from activity
group by player_id;

1890. 2020年最后一次登录

select user_id, max(time_stamp) last_stamp 
    from logins t
        where year(time_stamp) = 2020
            group by t.user_id;

1741. 查找每个员工花费的总时间

select event_day as day, emp_id, sum(out_time - in_time) as total_time 
from Employees 
group by emp_id, event_day 
order by event_day, emp_id;

控制流

1393. 股票的资本损益

select stock_name, sum(if(operation='Buy', -price, price)) capital_gain_loss
from stocks
group by stock_name

1407. 排名靠前的旅行者

select u.name, sum(if(r.distance is not null, r.distance, 0)) travelled_distance from users u left join rides r
on u.id = r.user_id
group by r.user_id
order by travelled_distance desc, u.name

1158. 市场分析 I
思路：
外连接时要注意where和on的区别，on是在连接构造临时表时执行的，不管on中条件是否成立都会返回主表（也就是left join左边的表）的内容，where是在临时表形成后执行筛选作用的，不满足条件的整行都会被过滤掉。如果这里用的是 where year(order_date)=‘2019’ 那么得到的结果将会把不满足条件的user_id为3，4的行给删掉。用on的话会保留user_id为3，4的行。

select user_id as buyer_id, join_date, count(order_id) as orders_in_2019
from Users as u left join Orders as o on u.user_id = o.buyer_id and year(order_date)='2019'
group by user_id

过滤

182. 查找重复的电子邮箱
思路：
where>group by>having>order by

select Email
from Person
group by Email
having count(Email) > 1;

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

1587. 银行账户概要 II

select u.name NAME, sum(t.amount) BALANCE from users u left join transactions t
on u.account = t.account
group by t.account
having sum(t.amount) >= 10000

1084. 销售分析III

SELECT 
    s.product_id, p.product_name
FROM 
    sales s
LEFT JOIN 
    product p 
ON 
    s.product_id = p.product_id
GROUP BY 
    s.product_id
HAVING MIN(sale_date) >= '2019-01-01' 
AND MAX(sale_date) <= '2019-03-31';