Original question: https://leetcode-cn.com/problems/customers-who-never-order/
I made the problem-solving record for the database problem of leetcode:
Problem-solving directory https://blog.csdn.net/weixin_42845682/ article / details / 105196004
Title description:
A website contains two tables, the Customers table and the Orders table. Write a SQL query to find all customers who never order anything.
+----+-------+
| Id | Name |
+----+-------+
| 1 | Joe |
| 2 | Henry |
| 3 | Sam |
| 4 | Max |
+----+-------+
Orders table:
+----+------------+
| Id | CustomerId |
+----+------------+
| 1 | 3 |
| 2 | 1 |
+----+------------+
For example, given the above table, your query should return:
+-----------+
| Customers |
+-----------+
| Henry |
| Max |
+-----------+
answer:
The first answer
I think this question is very simple, nothing to do. . .
select
name customers
from Customers
where id not in
(
select
distinct(c.id)
from Customers c
join orders o on c.id=o.Customerid
)
The second answer
Take a look at the table structure, I feel that it is not so troublesome ...
select
name customers
from Customers
where id not in
(
select
distinct(CustomerId)
from Orders
)
The third answer
The first two methods are to seek union and then do not exist. You can also find the difference set and then exist (however, I remember there is no difference set in mysql, so I used oracle):
SELECT
name customers
from
customers c
where id in
(
SELECT id FROM Customers
minus
SELECT distinct(customerid) FROM Orders
)
But this way of writing seems to be very slow to execute, and leetcode shows 1914 ms. I don't know if it's a problem with leetcode or my writing is wrong.
The fourth answer
Actually, this problem can be done with join. If you do n’t understand, I recommend this:
Use join to achieve the intersection, union, difference, and complement effects
https://blog.csdn.net/weixin_42845682/article/details/105414734
select
distinct(c.name) customers
from customers c
left join orders o on c.id = o.customerid
where o.id is null
But, I can't figure it out, why did I report it wrong when I added distinct. . . But it's not a big mistake, this is based on business needs, maybe people have changed their names.
So the correct answer is:
select
c.name customers
from customers c
left join orders o on c.id = o.customerid
where o.id is null
