1581. 进店却未进行过交易的顾客

https://leetcode.cn/problems/customer-who-visited-but-did-not-make-any-transactions/

本来想偷懒，直接写个一拖三，但是。。。效率有点吓人的低

select customer_id,count(0) count_no_trans 
from visits v
where visit_id not in (select visit_id from transactions t)
group by customer_id

在这里插入图片描述
那就换个一拖三的办法

select customer_id,count(0) count_no_trans 
from visits v
left join transactions t on t.visit_id=v.visit_id
where t.visit_id is null
group by customer_id

在这里插入图片描述

CSDN 文盲老顾的博客，https://blog.csdn.net/superwfei

1587. 银行账户概要 II

https://leetcode.cn/problems/bank-account-summary-ii/

还以为和前边一样，要case一下再 sum ，结果。。。。直接正负数就已经列出来了。。。

Hmmmm。注意，是高于10000，不小心习惯性的写成 >= 运算符，白瞎了一次提交。

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

1633. 各赛事的用户注册率

https://leetcode.cn/problems/percentage-of-users-attended-a-contest/

额。。。。用子查询求出用户总数作为除数就好，嗯，注意 group 跟的字段，mysql 可以省略一些内容。。。

# mysql
select r.contest_id,round(count(r.user_id)/b.total * 100,2) percentage
from register r,(select count(user_id) total from users) b
group by r.contest_id
order by count(0) desc,r.contest_id
# mssql
select r.contest_id,convert(decimal(6,2),count(0) * 1.0/b.total * 100) percentage
from register r,(select count(0) total from users) b
group by r.contest_id,b.total
order by count(0) desc,r.contest_id
# oracle
select r.contest_id,round(count(r.user_id)/b.total * 100,2) percentage
from register r,(select count(user_id) total from users) b
group by r.contest_id,b.total
order by count(0) desc,r.contest_id

1661. 每台机器的进程平均运行时间

https://leetcode.cn/problems/average-time-of-process-per-machine/

嗯。。。。其实算是行转列类的需求，不过只有两个对应数据，所以，就直接自关联一下就可以了。

# oracle && mysql
select a.machine_id,round(avg(ts),3) processing_time
from (
    select a.machine_id,b.timestamp-a.timestamp ts 
    from activity a
    inner join activity b on a.machine_id=b.machine_id and a.process_id=b.process_id
    where a.activity_type='start' and b.activity_type='end'
) a
group by a.machine_id
# mssql
select a.machine_id,convert(decimal(6,3),avg(ts)) processing_time
from (
    select a.machine_id,b.timestamp-a.timestamp ts 
    from activity a
    inner join activity b on a.machine_id=b.machine_id and a.process_id=b.process_id
    where a.activity_type='start' and b.activity_type='end'
) a
group by a.machine_id

说起来，mssql 的保留指定位数的小数，实在有点太恶心了，必须自己指定类型来干掉浮点数多余的小数位

1667. 修复表中的名字

https://leetcode.cn/problems/fix-names-in-a-table/

哦，字符串操作函数也上场了。注意 substr/substring 的写法，以及 concat 的用法，mssql 可以用 + 运算代替 concat，其他的不行

# mssql
select user_id,upper(left(name,1)) + substring(lower(name),2,len(name)) name
from Users
order by user_id
# mysql
select user_id,concat(upper(substring(name,1,1)),substring(lower(name),2,length(name))) name
from users
order by user_id
# oracle
select user_id,concat(upper(substr(name,1,1)),substr(lower(name),2,length(name))) name
from users
order by user_id

1683. 无效的推文

https://leetcode.cn/problems/invalid-tweets/

呵，数据库字符串操作入门题，一个 len/length 足够，还以为会有全角字符，unicode ，结果根本没用上 blen 之类的东西。

# mssql
select tweet_id
from tweets
where len(content)>15
# oracle && mysql
select tweet_id
from tweets
where length(content)>15

1693. 每天的领导和合伙人

https://leetcode.cn/problems/daily-leads-and-partners/

额。。。。简单的 count distinct 即可。。。。oracle 的就不写了，其实指令是一拖三，不过 oracle 需要用 to_char 处理下日期格式，干掉时间部分

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

https://leetcode.cn/problems/find-followers-count/

额，统计入门题，直接 group count 。。。。。

select user_id,count(0) followers_count
from followers
group by user_id
order by user_id

1731. 每位经理的下属员工数量

https://leetcode.cn/problems/the-number-of-employees-which-report-to-each-employee/

。。。。又是 group ，力扣到底是有多喜欢用 group 啊。。。。无语了

# mysql && oracle
select a.employee_id,a.name,count(0) as reports_count,round(avg(b.age),0) average_age 
from employees a
inner join employees b on a.employee_id=b.reports_to
group by a.employee_id,a.name
having(count(0)>0)
order by a.employee_id
# mssql
select a.employee_id,a.name,count(0) as reports_count,convert(int,round(avg(b.age * 1.0),0)) average_age 
from employees a
inner join employees b on a.employee_id=b.reports_to
group by a.employee_id,a.name
having(count(0)>0)
order by a.employee_id

1741. 查找每个员工花费的总时间

https://leetcode.cn/problems/find-total-time-spent-by-each-employee/

写了一大堆的 group 了，恶心，想吐，还想吃点酸的。。。。。。。

# oracle
select to_char(event_day,'YYYY-mm-DD') day,emp_id,sum(out_time-in_time) total_time 
from employees
group by event_day,emp_id
# mysql && mssql
select event_day day,emp_id,sum(out_time-in_time) total_time 
from employees
group by event_day,emp_id