從另一個函數內部調用VBA函數

Question

下面是我編寫的代碼，但是我一直對我添加註釋的行感到不解，並且只是該行。我已經評論了所有其他的路線，並將其隔離爲問題路線，但是對於我的生活以及我已經完成的一小時或更多的研究，我無法弄清楚問題所在。這可能是一個非常明顯的，但我真的被卡住了，這讓我瘋狂。

不管怎樣，代碼是用來取含換擋時間和語言能力數據的範圍，並顯示有多少人有特定的語言如何都可以在給定的時間段（The_Time在下面的代碼）

任何幫助將不勝感激！

Function ReturnAvailability(The_Time As String, The_Info As Range) 

Dim The_Lang As String 
Dim The_Shift_Start As String 
Dim The_Shift_End As String 
Dim stGotIt As String 
Dim stCell As Integer 
Dim Counter As Integer 

Counter = 0 

For Each r In The_Info.Rows 
    For Each c In r.Cells 
     stCell = c.Value 
     If InStr(stCell, "Eng") > 0 Then 
      The_Lang = "Eng" 
     ElseIf InStr(c, ":") > 0 Then 
      stGotIt = StrReverse(c) 
      stGotIt = Left(c, InStr(1, c, " ", vbTextCompare)) 
      The_Shift_End = StrReverse(Trim(stGotIt)) 
      stGotIt = Left(The_Shift, InStr(1, The_Shift, " ", vbTextCompare)) 
      The_Shift_Start = stGotIt 
      stCell = ReturnAvailabilityEnglish(The_Time, The_Shift_Start, The_Shift_End) ' this is the line causing the error 
     End If 
    Next c 
Next r 

ReturnAvailability = Counter 

End Function 


Function ReturnAvailabilityEnglish(The_Time As String, The_Shift_Start As String, The_Shift_End As String) 

Dim Time_Hour As Integer 
Dim Time_Min As Integer 
Dim Start_Hour As Integer 
Dim Start_Min As Integer 
Dim End_Hour As Integer 
Dim End_Min As Integer 
Dim Available As Integer 

Available = 13 

Time_Hour = CInt(Left(The_Time, 2)) 
Time_Min = CInt(Right(The_Time, 2)) 
Start_Hour = CInt(Left(The_Shift_Start, 2)) 
Start_Min = CInt(Right(The_Shift_Start, 2)) 
End_Hour = CInt(Left(The_Shift_End, 2)) 
End_Min = CInt(Right(The_Shift_End, 2)) 

If Start_Hour <= Time_Hour And Start_Min <= Time_Min Then 
    If End_Hour > Time_Hour And End_Min > Time_Min Then 
     Available = 1 
    Else 
     Available = 0 
    End If 
End If 

ReturnAvailabilityEnglish = Available 

End Function 

感謝， 達拉赫Ĵ

Answer 1

您已經聲明

Dim stCell As Integer 

這意味着，這部分不能工作：

stCell = c.Value 
If InStr(stCell, "Eng") > 0 Then 

無論c.Value的分配將失敗，因爲它包含文本，或InStr（stCell，「Eng」）永遠不會是真的，因爲e範圍內的所有單元都是數字。

你缺少文本比較：

If InStr(1, stCell, "Eng", vbTextCompare) > 0 Then 

這也是一個問題，你需要添加一個檢查所示：

If The_Time = vbNullString Or The_Shift_Start = vbNullString _ 
    Or The_Shift_End = vbNullString Then 
    Available = -1 
Else 

    Time_Hour = CInt(Left(The_Time, 2)) 
    Time_Min = CInt(Right(The_Time, 2)) 
    Start_Hour = CInt(Left(The_Shift_Start, 2)) 
    Start_Min = CInt(Right(The_Shift_Start, 2)) 
    End_Hour = CInt(Left(The_Shift_End, 2)) 
    End_Min = CInt(Right(The_Shift_End, 2)) 

    If Start_Hour <= Time_Hour And Start_Min <= Time_Min Then 
     If End_Hour > Time_Hour And End_Min > Time_Min Then 
      Available = 1 
     Else 
      Available = 0 
     End If 
    End If 
End If 
ReturnAvailabilityEnglish = Available 

最後，也是最重要的是，你的函數總是會返回0，因爲您在開始時將計數器設置爲0，並且從不更新它。