SQL查询返回零行

Question

我必须找到每个用户，加入日期和他们在2019年作为买家发出的订单数量。

Table: Users
+----------------+---------+
| Column Name    | Type    |
+----------------+---------+
| user_id        | int     |
| join_date      | date    |
| favorite_brand | varchar |
+----------------+---------+

user_id是该表的主键。该表包含一个在线购物网站的用户信息，用户可以在这个网站上买卖商品。

表:订单

+---------------+---------+
| Column Name   | Type    |
+---------------+---------+
| order_id      | int     |
| order_date    | date    |
| item_id       | int     |
| buyer_id      | int     |
| seller_id     | int     |
+---------------+---------+

order_id是该表的主键。item_id是Items表的外键。buyer_id和seller_id是Users表的外键。

表:项目

+---------------+---------+
| Column Name   | Type    |
+---------------+---------+
| item_id       | int     |
| item_brand    | varchar |
+---------------+---------+
item_id is the primary key of this table.

输入:

Users表:

+---------+------------+----------------+
| user_id | join_date  | favorite_brand |
+---------+------------+----------------+
| 1       | 2018-01-01 | Lenovo         |
| 2       | 2018-02-09 | Samsung        |
| 3       | 2018-01-19 | LG             |
| 4       | 2018-05-21 | HP             |
+---------+------------+----------------+

订单表:

+----------+------------+---------+----------+-----------+
| order_id | order_date | item_id | buyer_id | seller_id |
+----------+------------+---------+----------+-----------+
| 1        | 2019-08-01 | 4       | 1        | 2         |
| 2        | 2018-08-02 | 2       | 1        | 3         |
| 3        | 2019-08-03 | 3       | 2        | 3         |
| 4        | 2018-08-04 | 1       | 4        | 2         |
| 5        | 2018-08-04 | 1       | 3        | 4         |
| 6        | 2019-08-05 | 2       | 2        | 4         |
+----------+------------+---------+----------+-----------+

项目表:

+---------+------------+
| item_id | item_brand |
+---------+------------+
| 1       | Samsung    |
| 2       | Lenovo     |
| 3       | LG         |
| 4       | HP         |
+---------+------------+

输出:

+-----------+------------+----------------+
| buyer_id  | join_date  | orders_in_2019 |
+-----------+------------+----------------+
| 1         | 2018-01-01 | 1              |
| 2         | 2018-02-09 | 2              |
| 3         | 2018-01-19 | 0              |
| 4         | 2018-05-21 | 0              |
+-----------+------------+----------------+

我写了下面的解决方案，但它不起作用。请说明是什么问题。

select u.user_id as buyer_id, u.join_date as join_date, count(o.order_id) as orders_in_2019
from Users u join Orders o on u.user_id = o.buyer_id
where o.order_date < date('2010-01-01') and o.order_date >= date('2019-01-01')
group by u.user_id;

结果为空。

Answer 1

除了日期打错(0。order_date & lt;'2010-01-01')你似乎想要所有的用户，即使他们在2019年没有订单，所以左连接和条件聚合似乎是合适的

select u.user_id as buyer_id, u.join_date as join_date
,sum(case when o.order_date < '2020-01-01' and o.order_date >= '2019-01-01' then 1 else 0 end) cnt
from users u 
left join Orders o on u.user_id = o.buyer_id
group by u.user_id ;
+----------+------------+------+
| buyer_id | join_date  | cnt  |
+----------+------------+------+
|        1 | 2018-01-01 |    1 |
|        2 | 2018-02-09 |    2 |
|        3 | 2018-01-19 |    0 |
|        4 | 2018-05-21 |    0 |
+----------+------------+------+
4 rows in set (0.001 sec)

Answer 2

不能选择非该组中的直接元素

select g_user.buyer_id, Users.join_date as join_date , g_user.orders_in_2019 From (select u.user_id as buyer_id, count(o.order_id) as orders_in_2019 From Users u 
join Orders o on u.user_id = o.buyer_id
where o.order_date < date('2010-01-01') and o.order_date >= date('2019-01-01')
group by u.user_id) 
as g_user join Users on g_user.buyer_id = Users.user_id;

Answer 3

这是因为您的日期在sql:

中不正确

o.order_date < date('2010-01-01') 

很明显没有您的订单是在2010年之前。你应该改成

o.order_date < date('2020-01-01')