我想添加一列，该数字表示通过在PERSON_ID上划分分区并按INVITED_DATE排序时递增地计算非null之前的null列的数量，从而在接受之前邀请一个人。
我的表格具有以下格式：
| UNIQUE_ID | PERSON_ID | INVITED_DATE | ACCEPTED_DATE |
| 12345 | 567 | 12-01-18 | NULL |
| 12346 | 567 | 12-02-18 | NULL |
| 12347 | 567 | 12-03-18 | NULL |
| 12348 | 567 | 12-04-18 | 12-04-18 |
| 12349 | 567 | 12-05-18 | NULL |
| 12350 | 568 | 12-01-18 | NULL |
| 12351 | 568 | 12-02-18 | 12-02-18 |
理想情况下，输出应如下所示：
| UNIQUE_ID | PERSON_ID | INVITED_DATE | ACCEPTED_DATE | INVITES_BEFORE_ACCEPT |
| 12345 | 567 | 12-01-18 | NULL | 1 |
| 12346 | 567 | 12-02-18 | NULL | 2 |
| 12347 | 567 | 12-03-18 | NULL | 3 |
| 12348 | 567 | 12-04-18 | 12-04-18 | 0 |
| 12349 | 567 | 12-05-18 | NULL | 1 |
| 12350 | 568 | 12-01-18 | NULL | 1 |
| 12351 | 568 | 12-02-18 | 12-02-18 | 0 |
到目前为止，我已经尝试了使用OVER和PARTITION对ROW NUMBER进行多次迭代，但是我发现它必须是外部应用。以下OUTER APPLY会计算数据，但不会成功接受就重新开始计数。
SELECT t.* , invites.INVITES_BEFORE_ACCEPT
FROM table t
OUTER APPLY (
SELECT COUNT(*) INVITES_BEFORE_ACCEPT
FROM table t2
WHERE t.PERSON_ID = t2.PERSON_ID and t.INVITED_DATE < t2.ACCEPTED_DATE
) invites
答案 0 :(得分：2)
一种方法是
WITH t
AS (SELECT *,
COUNT(ACCEPTED_DATE)
OVER (
PARTITION BY PERSON_ID
ORDER BY INVITED_DATE) AS Grp
FROM [table])
SELECT *,
SUM(CASE
WHEN ACCEPTED_DATE IS NULL
THEN 1
ELSE 0
END)
OVER (
PARTITION BY PERSON_ID, Grp
ORDER BY INVITED_DATE) AS INVITES_BEFORE_ACCEPT
FROM t