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【SQL篇】面试篇之子查询

article 2024/9/27 12:16:11

1303 求团队人数

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# 写法1
# Write your MySQL query statement below
select employee_id, count(*) over(partition by team_id) as team_size
from Employee

# 写法2
# Write your MySQL query statement below
select employee_id, team_size
from Employee e
join (
    select team_id, count(*) as team_size
    from Employee
    group by team_id
) tmp
on e.team_id = tmp.team_id

总结

考察子查询
学会使用窗口函数

512 游戏玩法分析 II

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# 写法1
# Write your MySQL query statement below
select player_id, device_id
from (
    select player_id, device_id, 
    rank() over(partition by player_id order by event_date) as rk
    from Activity 
) tmp
where tmp.rk = 1

# 写法2
# Write your MySQL query statement below
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
);

184 部门工资最高的员工

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# 写法1
select d.name as Department, e.name as Employee, e.salary as salary
from Employee e
join Department d
on e.departmentId = d.id
where (e.departmentId, salary) in (
    select departmentId, max(salary)
    from Employee
    group by departmentId
);

# 写法2 rank()
select d.name as Department, e.name as  Employee, salary
from (
    select *,
    rank() over(partition by departmentId order by salary desc) as rk
    from Employee
) e, Department d
where e.departmentId = d.id and e.rk = 1;

# 写法3 dense_rank()
select d.name as Department, e.name as  Employee, salary
from (
    select *,
    dense_rank() over(partition by departmentId order by salary desc) as rk
    from Employee
) e, Department d
where e.departmentId = d.id and e.rk = 1;

总结

row_number, rank(), dense_rank()的区别

rank()排序相同时会重复，总数不变，即会出现1、1、3这样的排序结果；
dense_rank()排序相同时会重复，总数会减少，即会出现1、1、2这样的排序结果；
row_number()排序相同时不会重复，会根据顺序排序。

注意点

row_number函数得到的列别名可用于order by 排序，因为order by执行在select之后。
where, group by, having都不可引用该列，因为这些语句执行在select之前，此时函数尚未计算出值。

1549 每件商品的最新订单

1549题

# 写法1
# Write your MySQL query statement below
select p.product_name, p.product_id, order_id, order_date
from Products p
join Orders o
on p.product_id = o.product_id
where (p.product_id, order_date) in (
    select product_id, max(order_date)
    from Orders
    group by product_id
)
order by product_name, product_id, order_id

# 写法2
# Write your MySQL query statement below
select p.product_name, p.product_id, tmp.order_id, tmp.order_date
from (
    select *, dense_rank() over(partition by product_id order by order_date desc) as rk
    from Orders
    ) tmp, Products p
where tmp.product_id = p.product_id and rk = 1
order by product_name, product_id, order_id

总结

该题与顾客表无关，我们只需要订单表和商品表就可。
注意dense_rank()和row_number()的区别，不要用错。

1532 最近的三笔订单

1532题

# 写法1
# Write your MySQL query statement below
select c.name as customer_name, c.customer_id, o.order_id, o.order_date
from (
    select *, rank() over(partition by customer_id order by order_date desc) as rk
    from Orders
) o, Customers c
where o.customer_id = c.customer_id and rk <= 3
order by customer_name, customer_id, order_date desc;

1831 每天的最大交易

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# 写法1
# Write your MySQL query statement below
select transaction_id
from Transactions
where (day(day), amount) in (
    select day(day) as t, max(amount) as mx
    from Transactions
    group by t
)
order by transaction_id;

# 写法2
select transaction_id
from (
    select transaction_id, rank() over(partition by day(day) order by amount desc) as rk
    from Transactions t
) tmp
where rk = 1
order by transaction_id;