表:
id | date_in | date_out |
1 | 2016-01-01 12:00:00 | 2016-01-03 12:00:00 |
2 | 2016-01-01 12:00:00 | 2016-01-02 12:00:00 |
我需要一个返回结果的查询:
id | alias
1 | 2016-01-01 12:00:00
2 | 2016-01-01 12:00:00
2 | 2016-01-02 12:00:00
1 | 2016-01-03 12:00:00
答案 0 :(得分:5)
我会假设union
的某些内容就足够了,因为没有上下文我只能假设这是你正在寻找的东西?
SELECT t.id, t.date_in AS alias FROM test t
UNION ALL
SELECT t2.id, t2.date_out AS alias FROM test t2
ORDER BY alias ASC;
会给你一个如下结果:
+----+---------------------+
| id | alias |
+----+---------------------+
| 2 | 2016-01-01 12:00:00 |
| 1 | 2016-01-01 12:00:00 |
| 2 | 2016-01-02 12:00:00 |
| 1 | 2016-01-03 12:00:00 |
+----+---------------------+
4 rows in set (0.00 sec)
答案 1 :(得分:0)
请尝试以下
SELECT
CONCAT(date_in,date_out)为Alias FROM table_name;
答案 2 :(得分:0)
使用像Ash建议的UNION ALL查询回答了这个问题。