获取小数位数的长度

Question

我有一张Excel工作表，其中有几列，其中一列是“ Amount ，在这里我必须验证每个单元格并检查小数点后的长度是否大于2，如果是，则抛出错误。

Public Function CheckLength(value As String) As Integer
    Dim n As Double
    Dim itg As Integer
    Dim dcm As Double
    n = value 'Say Value is 50.01 here 
    itg = Int(n) 'Will be 50
    dcm = Split(n - itg, ".")(1) 'But n-itg will yield something and dcm`s value will be 
    '1000001 some strange value where as it has to be `01` or say whatever comes after decimal part
    CheckLength = Len(dcm)
End Function

Answer 1

您可以这样做：

Public Function CheckLength(value As String) As Integer
    Dim n As Double
    Dim itg As Integer
    Dim dcm As Double
    n = value
    itg = Int(n)
    dcm = Len(CStr(n)) - InStr(CStr(n), ".")

    CheckLength = dcm
End Function

注意：如果没有"." 在n中，它将返回总长度（因为它将是Len(CStr(n)) - 0 ），因此您可以检查字符串是否包含"." 之前，或者您可以检查dcm是否与Len(CStr(n)) ，然后返回0 。

Answer 2

如果您实际上正在检查数字，那么它将起作用：

Function CheckLength(value As Double) As Integer
    If InStr(CStr(value), ".") Then
        CheckLength = Len(Split(CStr(value), ".")(1))
    Else
       CheckLength = 0
    End If
End Function

---

它将数字转换为字符串，并使用"."分割"." 作为定界符，然后返回所返回数组中第二项的长度（它是"."之后的任何值）

Const myNumber As Double = 50.13245
Debug.Print CheckLength(myNumber) '// Returns 5

'// Split 50.13245 on "." returns an array with 2 parts:
'//    (0) = "50"
'//    (1) = "13245"
'//    Length of (1) '13245' = 5