# 1693. 每天的领导和合伙人
# https://leetcode-cn.com/problems/daily-leads-and-partners/

# SQL架构
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');

# 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;

# clean-up
drop table DailySales;