Sql 在unpivot中包含空值

很长一段时间以来，我一直在寻找解决这个问题的办法。但是，找不到

我的表格如下：

Month   Col_1   Col_2   Col_3   Col_4   Col_5   
---------------------------------------------
Jan     NULL    NULL    1       1       1   

Month    Name    Value
-----------------------
Jan      Col_3   1
Jan      Col_4   1
Jan      Col_5   1

我想取消此表的IVOT，以便与字段名（Col_1、Col2等）上的另一个表联接

我的查询：

select Month,Name,value from 
TableName
    unpivot
    (
        Value 
        for Name in (Col_1,Col_2,Col_3,Col_4,Col_5)  
    ) u 

select c.column_name,value
from INFORMATION_SCHEMA.COLUMNS c
left join(select * from 'tablename') t1
unpivot(value for column_name in (col_1,col_2,col_3,col_4,col_5)) t2
on t2.column_name=c.COLUMN_NAME where c.TABLE_NAME='tablename'

当前结果：

select Month,Name,value from 
TableName
    unpivot
    (
        Value 
        for Name in (Col_1,Col_2,Col_3,Col_4,Col_5)  
    ) u 

select c.column_name,value
from INFORMATION_SCHEMA.COLUMNS c
left join(select * from 'tablename') t1
unpivot(value for column_name in (col_1,col_2,col_3,col_4,col_5)) t2
on t2.column_name=c.COLUMN_NAME where c.TABLE_NAME='tablename'

这使我不需要

NULL

值，如下所示：

Month   Col_1   Col_2   Col_3   Col_4   Col_5   
---------------------------------------------
Jan     NULL    NULL    1       1       1   

Month    Name    Value
-----------------------
Jan      Col_3   1
Jan      Col_4   1
Jan      Col_5   1

预期结果：

select Month,Name,value from 
TableName
    unpivot
    (
        Value 
        for Name in (Col_1,Col_2,Col_3,Col_4,Col_5)  
    ) u 

select c.column_name,value
from INFORMATION_SCHEMA.COLUMNS c
left join(select * from 'tablename') t1
unpivot(value for column_name in (col_1,col_2,col_3,col_4,col_5)) t2
on t2.column_name=c.COLUMN_NAME where c.TABLE_NAME='tablename'

我希望结果中包含

NULL

s

Month    Name    Value
-----------------------
Jan      Col_1   NULL
Jan      Col_2   NULL
Jan      Col_3   1
Jan      Col_4   1
Jan      Col_5   1

---

任何帮助都将不胜感激。

查询：

select Month,Name,value from 
TableName
    unpivot
    (
        Value 
        for Name in (Col_1,Col_2,Col_3,Col_4,Col_5)  
    ) u 

select c.column_name,value
from INFORMATION_SCHEMA.COLUMNS c
left join(select * from 'tablename') t1
unpivot(value for column_name in (col_1,col_2,col_3,col_4,col_5)) t2
on t2.column_name=c.COLUMN_NAME where c.TABLE_NAME='tablename'

输出

name    value
Col_1   NULL
Col_2   NULL
Col_3   1
Col_4   1
Col_5   1

---

如果保证

列1、列2，…

不采用特定值，例如

-1

，则可以使用以下查询作为解决方法：

select [Month], Name, NULLIF(value, -1) AS value 
from (
   select [Month],
          coalesce(Col_1, -1) AS Col_1,
          coalesce(Col_2, -1) AS Col_2,
          coalesce(Col_3, -1) AS Col_3,
          coalesce(Col_4, -1) AS Col_4,
          coalesce(Col_5, -1) AS Col_5
   from TableName) AS t
unpivot
(
    Value 
    for Name in (Col_1,Col_2,Col_3,Col_4,Col_5)
) AS u 

---

希望能有所帮助，我以前试过这个。不工作。我得到了同样的结果。这应该会给我预期的结果。因此，我想问题在于我在表中的额外列。编辑问题以添加列

Month

。