我有一张桌子像:
| Date | Week | Name | No | Count |
|-----------|------|--------|----|-------|
| 2019/4/1 | 14 | John | 1 | 1 |
| 2019/4/1 | 14 | Mary | 2 | 1 |
| 2019/4/9 | 15 | Kevin | 3 | 2 |
| 2019/4/9 | 15 | John | 4 | 1 |
| 2019/4/9 | 15 | Jessie | 5 | 1 |
| 2019/4/18 | 16 | Kevin | 6 | 1 |
| 2019/4/18 | 16 | John | 7 | 1 |
| 2019/4/18 | 16 | Jessie | 8 | 2 |
| 2019/4/18 | 16 | Mary | 9 | 3 |
| 2019/4/18 | 16 | Mary | 10 | 1 |
| 2019/4/18 | 16 | Jessie | 11 | 1 |
| 2019/4/24 | 17 | Mary | 12 | 1 |
| 2019/4/24 | 17 | Jessie | 13 | 1 |
我想要做的是计算每周人们的总数。并按总计数排序。
我知道GROUP BY可以实现这一点,我已经尝试过,但是无法理解。
这就是我的期望:
| Name | 14 | 15 | 16 | 17 | Total |
|--------|----|----|----|----|-------|
| Mary | 1 | 0 | 4 | 1 | 6 |
| Jessie | 0 | 1 | 3 | 1 | 5 |
| John | 1 | 1 | 1 | 0 | 3 |
| Kevin | 0 | 2 | 1 | 0 | 3 |
| Total | 2 | 4 | 9 | 2 | 17 |
我能怎么做?
Select [Name]
,sum(case when [Week] = 14 then [Count] else 0 end) as Week14
,sum(case when [Week] = 15 then [Count] else 0 end) as Week15
,sum(case when [Week] = 16 then [Count] else 0 end) as Week16
,sum(case when [Week] = 17 then [Count] else 0 end) as Week17
,sum([Count]) as Total
from [table]
group by [Name]
order by Total
我不确定您使用的是哪个版本的DB2(LUW / zOS / i),所以这是一般答案。可以使周数更灵活,但需要在数周内进行一定数量的硬编码。