在 Case When 函数中有非布尔结果

Question

我正在尝试创建一个将结果分成不同列的查询。我能找到的最佳公式是 Case When 函数，但它表示等式的 Then 部分必须是布尔值（或 true/false 结果）。例如，Then 有没有办法计算数字 3-1？

Case
when
    DATEDIFF(day, T0.[DocDueDate], getdate()) > 0 
    AND DATEDIFF(day, T0.[DocDueDate], getdate()) < 30 
then
(T0.[DocTotal] - T0.[PaidToDate]) 
else
    ' ' 
end
as "Greater than 1", 
Case
when
    DATEDIFF(day, T0.[DocDueDate], getdate()) > 30 
    AND DATEDIFF(day, T0.[DocDueDate], getdate()) < 60 
then
(T0.[DocTotal] - T0.[PaidToDate]) 
else
    ' ' 
end
as "Greater than 30"

Answer 1

如果您使用 sql 服务器，那么您的代码片段将如下所示。你必须在 then 和 else 之后保持相同的数据类型，因为你在使用 else 之后使用字符串类型，所以你必须在以后转换它 then

Case
when
    DATEDIFF(day, T0.[DocDueDate], getdate()) > 0 
    AND DATEDIFF(day, T0.[DocDueDate], getdate()) < 30 
then
( convert(varchar(255), T0.[DocTotal] - T0.[PaidToDate]))
else
    ' ' 
end
as Greater_than_1, 
Case
when
    DATEDIFF(day, T0.[DocDueDate], getdate()) > 30 
    AND DATEDIFF(day, T0.[DocDueDate], getdate()) < 60 
then
(convert(varchar(255), T0.[DocTotal] - T0.[PaidToDate])) 
else
    ' ' 
end
as Greater_than_30

Answer 2

您遇到了类型兼容性问题。我建议您只使用 NULL 表示不匹配：

(case when DATEDIFF(day, T0.[DocDueDate], getdate()) > 0 AND DATEDIFF(day, T0.[DocDueDate], getdate()) < 30 
      then (T0.[DocTotal] - T0.[PaidToDate]) 

 end) as Greater_than_1, 
(case when DATEDIFF(day, T0.[DocDueDate], getdate()) > 30  and DATEDIFF(day, T0.[DocDueDate], getdate()) < 60 
      then (T0.[DocTotal] - T0.[PaidToDate]) 
 end) as Greater_than_30

我还猜测您打算 <= 30 作为第一个条件。

Answer 3

发现我需要将等式转换为整数，如下所示。

Case when 
    DATEDIFF(day, T0.[DocDueDate], getdate()) > 0 
    AND DATEDIFF(day, T0.[DocDueDate], getdate()) <30 
then 
    cast( (T0.[DocTotal]-T0.[PaidToDate]) as varchar(12) ) 
else 
    ' ' 
end as "Greater than 1"

Answer 4

以下是您可以使用多种方法执行此操作的方法。 TSET 只是生成从 6 月 1 日到今天的测试日期。

WITH tset(td)
     AS (
     SELECT CAST('2018-08-01' AS DATE) AS td
     UNION ALL
     SELECT DATEADD(d, -1, td)
     FROM   tset
     WHERE  td > CAST('2018-06-01' AS DATE))
     -- Actual examples start here
     SELECT td
          -- Diff is just showing the difference in days so you can see the group assignements
          , DATEDIFF(d, td, GETDATE()) AS diff
          -- This column groups into < 30, 31-60, and > 60
          , CASE
                WHEN DATEDIFF(d, td, GETDATE()) < 30 THEN 1 ELSE CASE
                                                                     WHEN DATEDIFF(d, td, GETDATE()) < 60 THEN 2 ELSE 3
                                                                 END
            END three_sets
          -- this example will group  into any number of 30 day sets.
          , cast ( DATEDIFF(d, td, GETDATE()) as int) / 30 any_number_of_sets
     FROM   tset
     ORDER BY td;

一些示例结果：

td        diff  three_sets  any_number_of_sets
2018-06-01  66  3   2
2018-06-02  65  3   2
2018-06-03  64  3   2
. . .

2018-06-12  55  2   1
2018-06-13  54  2   1
2018-06-14  53  2   1
. . .
2018-07-09  28  1   0
2018-07-10  27  1   0
2018-07-11  26  1   0

在 Case When 函数中有非布尔结果

Have non-boolean result in Case When Function

sql

case-when