Question

我的表

| ID |   Date   | Next Date | 
| 1  |2014-12-23| 2014-12-25|
| 2  |2014-12-20| 2014-12-22|
| 3  |2014-12-19| 2014-12-21|
| 4  |2014-12-15| 2014-12-18|

如何在搜索结果查询中的日期范围内搜索时显示重复项：

select * from my_table where date >= '2014-12-14' AND date <='2015-01-22' ORDER BY (date) ASC

我的结果是：

1. ID: 4, Date: 2014-12-15, Next Date: 2014-12-18
2. ID: 3, Date: 2014-12-19, Next Date: 2014-12-21
3. ID: 2, Date: 2014-12-20, Next Date: 2014-12-22
4. ID: 1, Date: 2014-12-23, Next Date: 2014-12-25

我想要做的是显示重复项，包括日期，下一个日期和按日期排序，下一个日期。

1. ID: 4, Date: *2014-12-15*
2. ID: 4, Next Date: *2014-12-18*
3. ID: 3, Date: *2014-12-19*
4. ID: 2, Date: *2014-12-20*
5. ID: 3, Next Date: *2014-12-21*
6. ID: 2, Next Date: *2014-12-22*
7. ID: 1, Date: *2014-12-23*
8. ID: 1, Next Date: *2014-12-25*

对我的情况有什么好处？

由于

Answer 1

我认为这就是你想要的：

select "ID", "Order Date"
from
(
    (select "ID", "Date", "Next Date", "Date" as "Order Date" from Table1)
    union all
    (select "ID", "Date", "Next Date", "Next Date" as "Order Date" from Table1)
) as Table2
order by "Order Date"

注意：将Table1更改为表格的实际名称。

如何获取重复的行但具有不同的日期

1 个答案: