我试图从3个表中获取所有组合的完整结果集,但它不起作用,因为我无法弄清楚如何告诉SQL-Server 2008我想要什么。
我已将其简化为以下类似问题...我有3个表:
table `date_ranges`:
--------------------------------------------------------
| start_day | end_day |
| -------------------------------------------------------- |
| November, 01 2015 14:37:00 | November, 02 2015 00:00:00 |
| November, 02 2015 00:00:00 | November, 03 2015 00:00:00 |
| November, 03 2015 00:00:00 | November, 04 2015 00:00:00 |
| November, 04 2015 00:00:00 | November, 04 2015 02:00:00 |
--------------------------------------------------------
table `sites`:
----
| site |
| ---- |
| 1 |
| 2 |
| 3 |
| 4 |
----
table `all_data`:
--------------------------------------
| data_date | data_site |
| ---------------------------|---------- |
| November, 02 2015 15:35:00 | 1 |
--------------------------------------
我希望得到以下结果:
-------------------------------------------------------------------------------------------------------
| data_date | data_site | start_day | end_day | site |
| ------------------------------------------------------------------------------------------------------- |
| null | null | November, 01 2015 14:37:00 | November, 02 2015 00:00:00 | 1 |
| November, 02 2015 15:35:00 | 1 | November, 02 2015 00:00:00 | November, 03 2015 00:00:00 | 1 |
| null | null | November, 03 2015 00:00:00 | November, 04 2015 00:00:00 | 1 |
| null | null | November, 04 2015 00:00:00 | November, 04 2015 02:00:00 | 1 |
| null | null | November, 01 2015 14:37:00 | November, 02 2015 00:00:00 | 2 |
| null | null | November, 02 2015 00:00:00 | November, 03 2015 00:00:00 | 2 |
| null | null | November, 03 2015 00:00:00 | November, 04 2015 00:00:00 | 2 |
| null | null | November, 04 2015 00:00:00 | November, 04 2015 02:00:00 | 2 |
| null | null | November, 01 2015 14:37:00 | November, 02 2015 00:00:00 | 3 |
| null | null | November, 02 2015 00:00:00 | November, 03 2015 00:00:00 | 3 |
| null | null | November, 03 2015 00:00:00 | November, 04 2015 00:00:00 | 3 |
| null | null | November, 04 2015 00:00:00 | November, 04 2015 02:00:00 | 3 |
| null | null | November, 01 2015 14:37:00 | November, 02 2015 00:00:00 | 4 |
| null | null | November, 02 2015 00:00:00 | November, 03 2015 00:00:00 | 4 |
| null | null | November, 03 2015 00:00:00 | November, 04 2015 00:00:00 | 4 |
| null | null | November, 04 2015 00:00:00 | November, 04 2015 02:00:00 | 4 |
-------------------------------------------------------------------------------------------------------
但我只能弄清楚如何获得以下结果(see fiddle here):
-------------------------------------------------------------------------------------------------------
| data_date | data_site | start_day | end_day | site |
| ------------------------------------------------------------------------------------------------------- |
| null | null | November, 01 2015 14:37:00 | November, 02 2015 00:00:00 | null |
| November, 02 2015 15:35:00 | 1 | November, 02 2015 00:00:00 | November, 03 2015 00:00:00 | 1 |
| null | null | November, 03 2015 00:00:00 | November, 04 2015 00:00:00 | null |
| null | null | November, 04 2015 00:00:00 | November, 04 2015 02:00:00 | null |
| null | null | null | null | 2 |
| null | null | null | null | 3 |
| null | null | null | null | 4 |
-------------------------------------------------------------------------------------------------------
使用以下错误查询:
select * from all_data d
full outer join date_ranges r on (r.start_day <= d.data_date and d.data_date < r.end_day)
full outer join sites s on s.site = d.data_site
答案 0 :(得分:5)
CROSS JOIN
和LEFT JOIN
的组合可以帮到你。
像这样......
SELECT *
FROM DATE_RANGES R
CROSS JOIN SITES S
LEFT OUTER JOIN ALL_DATA D ON (R.START_DAY <= D.DATA_DATE AND D.DATA_DATE < R.END_DAY)
AND S.SITE = D.DATA_SITE