UNION ALL 并用 NULL 填充日期列

Question

CREATE TABLE sales (
    id SERIAL PRIMARY KEY,
    last_order DATE,
    customer VARCHAR(255)
);

INSERT INTO sales
(last_order, customer)
VALUES 
('2020-04-10', 'user_01'),
('2020-06-15', 'user_02'),
('2020-08-26', 'user_03');



CREATE TABLE customers (
    id SERIAL PRIMARY KEY,
    first_order DATE,
    customer VARCHAR(255)
);

INSERT INTO customers
(first_order, customer)
VALUES 
('2020-03-10', 'user_01'),
('2020-05-15', 'user_02'),
('2020-01-17', 'user_03');

预期结果：

last_order   |   first_order     |
-------------|-------------------|----
2020-04-10   |                   |
2020-06-15   |                   |
2020-08-26   |                   |
             |    2020-03-10     |
             |    2020-05-15     |
             |    2020-01-17     |
             |                   |

在 table sales 我有 last_order 的日期，在 table customers 我有 first_order 的日期.
现在我想 UNION ALL 两个 tables，因此需要用 NULL[=42= 填充 last_order 列和 - 分别 - first_order 列] 以保持结构相等。

在 postgresSQL 中，我正在使用这个查询，它没有任何问题：

SELECT s.last_order AS last_order, NULL AS first_order FROM sales s UNION ALL SELECT NULL AS last_order, c.first_order AS first_order FROM customers c

但是，当我将此查询应用于 amazon-redshift 时，出现错误：

Invalid operation: UNION types date and text cannot be matched

我需要如何修改amazon-redshift中的查询才能达到预期的结果？

Answer 1

按照@Serg 的建议，请尝试将 null 转换为日期，如下所示：

查询：

 SELECT
 s.last_order AS last_order,
 NULL::date AS first_order
 FROM sales s
 
 UNION ALL
 
 SELECT
 NULL::date AS last_order,
 c.first_order AS first_order
 FROM customers c

输出：

last_order	first_order
2020-04-10	null
2020-06-15	null
2020-08-26	null
null	2020-03-10
null	2020-05-15
null	2020-01-17

db<>fiddle here

Answer 2

一个有趣的方法是：

select s.last_order, c.first_order
from sales s full join
     customers c
     on 1 = 0;  -- never true!

UNION ALL 并用 NULL 填充日期列

UNION ALL and fill date column with NULL

sql

amazon-redshift