Question

我正在尝试创建一个查询，该查询会将结果分为不同的列。我可以找到的最佳公式是Case When函数，但它表示方程式的Then部分必须是布尔值（或正确/错误的结果）。那么，Then是否有一种方法可以计算出3-1的数字？

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

您对类型兼容性有疑问。我建议您简单地使用NULL进行匹配：

(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

我还会猜到您打算将<= 30用于第一个条件。

Answer 2

如果您使用sql server，则您的代码片段如下所示。您必须在then和else之后保持与在使用else之后使用字符串类型相同的数据类型，因此您必须在以后的then上进行转换

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 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

这里是您可以通过多种方法执行此操作的方法。 TSET只是生成6月1日到今天的日期进行测试。

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一些示例结果：

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;

在“函数何时起作用”的情况下产生非布尔结果

4 个答案: