问题描述
+----+-------+
| id | value |
+----+-------+
| 1 | A |
| 2 | B |
| 3 | C |
| 4 | D |
| 5 | D |
| 6 | D |
| 7 | N |
| 8 | P |
| 9 | P |
+----+-------+
所需的输出
+----+-------+---------------------+
| id | value | calc ↓ |
+----+-------+---------------------+
| 1 | A | 1 |
| 2 | B | 2 |
| 3 | C | 3 |
| 4 | D | 6 |
| 5 | D | 6 |
| 6 | D | 6 |
| 7 | N | 7 |
| 8 | P | 9 |
| 9 | P | 9 |
| 10 | D | 11 |
| 11 | D | 11 |
| 12 | Z | 12 |
+----+-------+---------------------+
您能帮我解决这个问题吗? id是身份,输出中必须存在id,输出中必须具有相同的9行。
新注释:我添加了10、11、12行。请注意,具有字母“ D”的ID 10和11与ID 4,5,6处于不同的组
谢谢
解决方法
对于此采样日期,您需要MAX()窗口函数:
SELECT id,value,MAX(id) OVER (PARTITION BY value) calc
FROM tablename
,
如果分组还取决于周围的id,那么这将变成类似间隙和孤岛的问题https://www.red-gate.com/simple-talk/sql/t-sql-programming/the-sql-of-gaps-and-islands-in-sequences/#:~:text=The%20SQL%20of%20Gaps%20and%20Islands%20in%20Sequences,...%204%20Performance%20Comparison%20of%20Gaps%20Solutions.%20
您可以使用Tabibitosan方法https://rwijk.blogspot.com/2014/01/tabibitosan.html
在这里,您还需要按“值”列进行分组,但这并不会使其复杂化:
select id,max(id) over (partition by value,island) calc
from (
select id,id - row_number() over(partition by value order by id) island
from my_table
) as sq
order by id;
id - row_number() over(partition by value order by id)
表达式为您提供一个数字,每当ID值对每个值的值变化大于1时,该数字都会变化。这将包含在max(id) over (partition by value,island)
表达式中。岛号仅对该特定值有效。在您的情况下,值N和D的计算出的孤岛数均为6,但需要区别对待。
Db小提琴https://www.db-fiddle.com/f/jahP7T6xBt3cpbLRhZZdQG/1
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