力扣SQL刷题6

1076. 项目员工II-----重要

题型：分组计数，并列第一的多个类别都输出
解答：order by后limit 1 只能输出一个类别
用count(1) >= all(select count(1) from 表 group by 列A)

下面的代码，对于多个并列第一的会报错

select project_id
from Project
group by project_id
order by count(project_id) desc
limit 1

any()和all()函数

select project_id
from Project
group by project_id
having count(project_id) >= all(select count(1) from project group by project_id)

1077. 项目员工 III-----重要

题型：对列A分组，每一个类型中，列B值最大的记录输出（可能有并列）
解答1：先max筛选出最大的，然后where筛选
where （列A，列B）in (select 列A，max(列B) from 表 group by 列A)
解法2：窗口函数dense_rank()排序，输出排序为1的

有两个表关联，可以先关联出来名为表t,后面用起来： with t as (…)

解法1：

# 链接2表
with t as
(select
a.project_id,
a.employee_id,
b.experience_years 
from project a
join employee b
using(employee_id))

# 子查询匹配
select project_id,employee_id
from t
where (project_id, experience_years)
in
(select project_id, max(experience_years)
from t 
group by 1)

解法2：

with t as
(select
a.project_id,
a.employee_id,
b.experience_years 
from project a
join employee b
using(employee_id))

# 窗口函数
select project_id,employee_id
from 
(select 
    project_id,
    employee_id,
    dense_rank() over(partition by project_id order by experience_years desc) as rnk
from t) tmp
where rnk =1

1098. 小众书籍

题型：表A与B连接，A中没有出现在B中的类别也要显示出来
解答：ifnull(语句,0)

易错点：
（1）容易忽略销量为0的书籍
（2）使用左链接，但是限制"近一年“使用了where字段，导致销量为null依然被忽略 --条件写在on里而不在where里
（3）having条件忘记is null，导致销量为null被忽略 ----ifnull()函数

select b.book_id,b.name
from books b left join orders o on o.book_id=b.book_id and dispatch_date >= '2018-06-23'
where b.available_from <='2019-05-23' 
group by b.book_id
having ifnull(sum(quantity), 0)<10

1112. 每位学生的最高成绩

题型：根据列A分组，每组类别中列C最大，列B最小的行记录输出
解答：窗口函数在这种排序问题中是一个很好的解法，别忘

select student_id,course_id,grade 
from
(select *,
rank() over(partition by student_id order by grade desc,course_id asc) rk
from Enrollments) a
where rk=1
order by student_id

窗口函数总结：
<窗口函数> over (partition by <分组列名> order by <排序列名>)

rank() : 阶梯排序-前两个是并列的1，接下来就是第3名
dense_rank(): 连续排序-前两个是并列的1，接下来就是第2名
row_number(): 不会出现重复的排序

1126. 查询活跃业务—会

题型：见题目
解答：一步一步解就好，业务的平均次数先求出来—表t，大于平均次数—打标签k，次数至少两次–sum(k)>=2

with t as
(select event_type,avg(occurences ) as avgo
from Events e
group by event_type)


select business_id
from
(select e.business_id,e.event_type,e.occurences,if(e.occurences>t.avgo,,0) as k
from Events e left join t on e.event_type =t.event_type
) a
group by business_id
having sum(k)>=2