将多行合并为一行的SQL

我希望将多行合并为一行，并且只保留值不为NULL的值

以下是我想要实现的目标：

我想从这个

+----+-----------------+-----------------+-----------------+--------------------+
| ID | 1stNofification | 2ndNotification | 3rdNotification | NotificationNumber |
+----+-----------------+-----------------+-----------------+--------------------+
| 1  | 01.01.2019      | NULL            | NULL            | 1                  |
+----+-----------------+-----------------+-----------------+--------------------+
| 1  | NULL            | 02.02.2019      | NULL            | 2                  |
+----+-----------------+-----------------+-----------------+--------------------+
| 1  | NULL            | NULL            | 03.03.2019      | 3                  |
+----+-----------------+-----------------+-----------------+--------------------+
| 2  | 06.01.2019      | NULL            | NULL            | 1                  |
+----+-----------------+-----------------+-----------------+--------------------+
| 2  | NULL            | 09.02.2019      | NULL            | 2                  |
+----+-----------------+-----------------+-----------------+--------------------+
| 2  | NULL            | NULL            | 11.03.2019      | 3                  |
+----+-----------------+-----------------+-----------------+--------------------+

为此：

+----+-----------------+-----------------+-----------------+
| ID | 1stNofification | 2ndNotification | 3rdNotification |
+----+-----------------+-----------------+-----------------+
| 1  | 01.01.2019      | 02.02.2019      | 03.03.2019      |
+----+-----------------+-----------------+-----------------+
| 2  | 06.01.2019      | 09.02.2019      | 11.03.2019      |
+----+-----------------+-----------------+-----------------+

我试过这样的方法：

选择
身份证件
最大值（当a.NotificationNumber=1时，则1通知结束）1通知，
最大值（当a.NotificationNumber=2时，则为第2个通知结束）第2个通知，
最大值（当a.NotificationNumber=3时，则为3通知结束）3通知
来自通知
按ID分组

但不幸的是，这并没有给我预期的结果

---

如果有人能帮助我，我将不胜感激：）

您只需在没有任何案例的情况下使用max即可

SELECT 
ID, 
MAX(1stNotification) AS 1stNotification,
MAX(2ndNotification) AS 2ndNotification, 
MAX(3rdNotification) AS 3rdNotification

FROM Notifications

GROUP BY  ID

---

你只需要在没有任何案例的情况下使用max

SELECT 
ID, 
MAX(1stNotification) AS 1stNotification,
MAX(2ndNotification) AS 2ndNotification, 
MAX(3rdNotification) AS 3rdNotification

FROM Notifications

GROUP BY  ID

---

我想你需要这样的东西

 ; with cte as (
     select 1 as id, 'dd' as not1, null as not2, null as not3 , 1 as notifications
     union all 
     select 1, null, 'df', null  , 2 
     union all 
     select 1, null, null, 'vc', 3  
     union all 
     select 2, 'ws', null, null, 1  
     union all 
     select 2, null, 'xs', null, 2  
     union all 
     select 2, null, null, 'nm', 3  
 )
 , ct as (
    select id, coalesce(not1, not2, not3) as ol, notifications , 
    'notification' + cast(notifications as varchar(5)) as Col 
    from cte
 )
 select * from (
 select id, ol,  col from ct )
 as d 
 pivot (
 max(ol) for col in ( [notification1], [notification2], [notification3] )
 ) as P

---

据我所知，这里的结果中的通知栏实际上是在行中提及的通知编号

我想你需要这样的东西

 ; with cte as (
     select 1 as id, 'dd' as not1, null as not2, null as not3 , 1 as notifications
     union all 
     select 1, null, 'df', null  , 2 
     union all 
     select 1, null, null, 'vc', 3  
     union all 
     select 2, 'ws', null, null, 1  
     union all 
     select 2, null, 'xs', null, 2  
     union all 
     select 2, null, null, 'nm', 3  
 )
 , ct as (
    select id, coalesce(not1, not2, not3) as ol, notifications , 
    'notification' + cast(notifications as varchar(5)) as Col 
    from cte
 )
 select * from (
 select id, ol,  col from ct )
 as d 
 pivot (
 max(ol) for col in ( [notification1], [notification2], [notification3] )
 ) as P

---

据我所知，这里的结果中的通知栏实际上是在行中提及的通知编号

因此，可以确定每个

Id

只有一个

NotificationNumber

@DarkRob是的，每个Id只有一个NotificationNumber，因此可以确定每个

Id

只有一个

NotificationNumber

@DarkRob是的，每个Id只有一个NotificationNumber这对我不起作用，它仍然给我包含空值的列，而不是它应该包含的日期pick@calgara12 . . . 这应该行得通。如果没有，那么你的数据与你的问题不同。@GordonLinoff你是对的，我犯了一个错误，现在它起作用了。谢谢这对我不起作用，它仍然给我包含空值的列，而不是应该包含的日期pick@calgara12 . . . 这应该行得通。如果没有，那么你的数据与你的问题不同。@GordonLinoff你是对的，我犯了一个错误，现在它起作用了。谢谢