1693. Daily Leads and Partners

題目敘述

SQL Schema >

Create table If Not Exists DailySales(date_id date, make_name varchar(20), lead_id int, partner_id int)
Truncate table DailySales
insert into DailySales (date_id, make_name, lead_id, partner_id) values ('2020-12-8', 'toyota', '0', '1')
insert into DailySales (date_id, make_name, lead_id, partner_id) values ('2020-12-8', 'toyota', '1', '0')
insert into DailySales (date_id, make_name, lead_id, partner_id) values ('2020-12-8', 'toyota', '1', '2')
insert into DailySales (date_id, make_name, lead_id, partner_id) values ('2020-12-7', 'toyota', '0', '2')
insert into DailySales (date_id, make_name, lead_id, partner_id) values ('2020-12-7', 'toyota', '0', '1')
insert into DailySales (date_id, make_name, lead_id, partner_id) values ('2020-12-8', 'honda', '1', '2')
insert into DailySales (date_id, make_name, lead_id, partner_id) values ('2020-12-8', 'honda', '2', '1')
insert into DailySales (date_id, make_name, lead_id, partner_id) values ('2020-12-7', 'honda', '0', '1')
insert into DailySales (date_id, make_name, lead_id, partner_id) values ('2020-12-7', 'honda', '1', '2')
insert into DailySales (date_id, make_name, lead_id, partner_id) values ('2020-12-7', 'honda', '2', '1')

Table: DailySales

+-------------+---------+
| Column Name | Type    |
+-------------+---------+
| date_id     | date    |
| make_name   | varchar |
| lead_id     | int     |
| partner_id  | int     |
+-------------+---------+
This table does not have a primary key.
This table contains the date and the name of the product sold and the IDs of the lead and partner it was sold to.
The name consists of only lowercase English letters.

Write an SQL query that will, for each date_id and make_name, return the number of distinct lead_id's and distinct partner_id's.

Return the result table in any order.

The query result format is in the following example.

Example 1:

Input:
DailySales table:
+-----------+-----------+---------+------------+
| date_id   | make_name | lead_id | partner_id |
+-----------+-----------+---------+------------+
| 2020-12-8 | toyota    | 0       | 1          |
| 2020-12-8 | toyota    | 1       | 0          |
| 2020-12-8 | toyota    | 1       | 2          |
| 2020-12-7 | toyota    | 0       | 2          |
| 2020-12-7 | toyota    | 0       | 1          |
| 2020-12-8 | honda     | 1       | 2          |
| 2020-12-8 | honda     | 2       | 1          |
| 2020-12-7 | honda     | 0       | 1          |
| 2020-12-7 | honda     | 1       | 2          |
| 2020-12-7 | honda     | 2       | 1          |
+-----------+-----------+---------+------------+
Output:
+-----------+-----------+--------------+-----------------+
| date_id   | make_name | unique_leads | unique_partners |
+-----------+-----------+--------------+-----------------+
| 2020-12-8 | toyota    | 2            | 3               |
| 2020-12-7 | toyota    | 1            | 2               |
| 2020-12-8 | honda     | 2            | 2               |
| 2020-12-7 | honda     | 3            | 2               |
+-----------+-----------+--------------+-----------------+
Explanation:
For 2020-12-8, toyota gets leads = [0, 1] and partners = [0, 1, 2] while honda gets leads = [1, 2] and partners = [1, 2].
For 2020-12-7, toyota gets leads = [0] and partners = [1, 2] while honda gets leads = [0, 1, 2] and partners = [1, 2].

題目翻譯

要從資料表 DailySales 中取出每個 date_id 和 make_name 的組合中，不重複的 lead_id 和 partner_id 總數。

解法解析

光看到不重複的第一直覺就是要使用到 DISTINCT，然後再來是有組合的，就會需要搭配使用 GROUP BY 了。最後因為要算個數所以再搭配使用 COUNT 來做計算

解法範例

MySQL

# Write your MySQL query statement below
SELECT date_id,
    make_name,
    COUNT(DISTINCT lead_id) AS unique_leads,
    COUNT(DISTINCT partner_id) AS unique_partners
FROM DailySales
GROUP BY date_id,
    make_name