拆分函数中的 VBA 类型不匹配
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VBA type mismatch in split function
提问by TMikonos
I have written a code where I need to get the number from a string like this: "3/1". I need to store those 2 numbers as integer in 2 different variables. I have written this code in 2 classes: This function is my split function in the class (BigOne)
我写了一个代码，我需要从这样的字符串中获取数字：“3/1”。我需要将这 2 个数字作为整数存储在 2 个不同的变量中。我已经在 2 个类中编写了这段代码：这个函数是我在类 (BigOne) 中的拆分函数
Public Function SplitValues(pInput As String, pdelim As String) As String()
'Declaration of variables
Dim strSplit() As String
Dim countDelim As Integer
'Initialization of variables
countDelim = countCharacter(pInput, pdelim)
If countDelim > 0 Then
ReDim strSplit(countDelim)
strSplit = Split(pInput, pdelim)
SplitValues = strSplit
End If
End Function
In the main class I have a function calling to this function that splits the number to get the values that I want. However I am getting a "Type Mismatch error" I am not able to detect the reason of this type mismatch.
在主类中，我有一个函数调用此函数，该函数将数字拆分为我想要的值。但是我收到“类型不匹配错误”我无法检测到这种类型不匹配的原因。
Public Function get_MaxChars(pInput As String) As Integer
'declaration of variables
Dim gen As cBigOne
Dim values As String
'Main code
pInput = CStr(pInput)
Debug.Print (pInput)
values = gen.SplitValues(pInput, "/")
get_MaxChars = CInt(values(0))
End Function
So, I am not able to see why it is not working correctly and I am getting the type mismatch error. Because, I believe that everywhere I am passing the same type.
所以，我不明白为什么它不能正常工作，并且我收到了类型不匹配错误。因为，我相信在任何地方我都在传递相同的类型。
回答by bobajob
SplitValuesreturns a String array, and you are trying to assign it to a String. Try dimming valuesas String()instead.
SplitValues返回一个字符串数组，而您正试图将其分配给一个字符串。尝试调光values作为String()代替。
You'll still have an issue when calling SplitValuesas you haven't created an instance of your class, just said that genwill be one. After Dim gen As cBigOne, you should have Set gen As New cBigOne
您在调用时仍然会遇到问题，SplitValues因为您还没有创建类的实例，只是说gen将是一个实例。之后Dim gen As cBigOne，你应该有Set gen As New cBigOne
回答by user2485790
I had the same problem. Then I found that you have to declare the array as VARIANT
我有同样的问题。然后我发现你必须将数组声明为 VARIANT
This is my code, where I save an attachment for every mail that arrives in Outlook to a specified folder
这是我的代码，我将到达 Outlook 的每封邮件的附件保存到指定文件夹
Private Sub Application_NewMailEx(ByVal EntryIDCollection As String)
Dim objOL As Outlook.Application
Dim arr As Variant
Dim i As Integer
Dim ns As Outlook.NameSpace
Dim itm As MailItem
Dim m As Outlook.MailItem
Dim j As Long
Dim lngCount As Long
Dim strFile As String
Dim strFolderpath As String
Dim strDeletedFiles As String
Dim fs As FileSystemObject
Dim mldat As Date
Dim strsndr As String
'On Error Resume Next
Set ns = Application.Session
arr = Split(EntryIDCollection, ",")
'*******************************************************************************************
' Set the Attachment folder.
strFolderpath = "z:\dropbox (AAA-DZ)_Attach\"
' Check each selected item for attachments. If attachments exist,
' save them to the strFolderPath folder and strip them from the item.
For ij = 0 To UBound(arr)
Set itm = ns.GetItemFromID(arr(ij))
If itm.Class = olMail Then
With itm
' This code only strips attachments from mail items.
' If objMsg.class=olMail Then
' Get the Attachments collection of the item.
Set objAttachments = .Attachments
lngCount = objAttachments.Count
strDeletedFiles = ""
If lngCount > 0 Then
' We need to use a count down loop for removing items
' from a collection. Otherwise, the loop counter gets
' confused and only every other item is removed.
Set fs = New FileSystemObject
For i = lngCount To 1 Step -1
' Save attachment before deleting from item.
' Get the file name.
strFile = Right("0000" + Trim(Str$(Year(.SentOn))), 4) + "_" + Right("00" + Trim(Str$(Month(.SentOn))), 2) + "_" + Right("00" + Trim(Str$(Day(.SentOn))), 2) + "_" + Right("00" + Trim(Str$(Hour(.SentOn))), 2) + "_" + Right("00" + Trim(Str$(Minute(.SentOn))), 2) + "_" + Right("00" + Trim(Str$(Second(.SentOn))), 2) + "_" + .SenderEmailAddress + "_" + .Parent + "_" + objAttachments.Item(i).FileName
' Combine with the path to the Temp folder.
strFile = strFolderpath & strFile
' Save the attachment as a file.
If Left(objAttachments.Item(i).FileName, 5) <> "image" Then
objAttachments.Item(i).SaveAsFile strFile
' Delete the attachment.
objAttachments.Item(i).Delete
'write the save as path to a string to add to the message
'check for html and use html tags in link
If .BodyFormat <> olFormatHTML Then
strDeletedFiles = strDeletedFiles & vbCrLf & "<file://" & strFile & ">"
Else
strDeletedFiles = strDeletedFiles & "<br>" & "<a href='file://" & _
strFile & "'>" & strFile & "</a>"
End If
Else
objAttachments.Item(i).Delete
End If
Next i
' Adds the filename string to the message body and save it
' Check for HTML body
If Len(strDeletedFiles) > 5 Then
If .BodyFormat <> olFormatHTML Then
.Body = vbCrLf & "The file(s) were saved to " & strDeletedFiles & vbCrLf & .Body
Else
.HTMLBody = "<p>" & "The file(s) were saved to " & strDeletedFiles & "</p>" & .HTMLBody
End If
.Save
End If
Else
objAttachments.Item(1).Delete
End If
End With
End If
Next ij
ExitSub:
Set objAttachments = Nothing
Set objSelection.Item(ij) = Nothing
Set objSelection = Nothing
Set objOL = Nothing
'********************************************************************************************
Set ns = Nothing
Set itm = Nothing
Set m = Nothing
End Sub
回答by Dhiraj Mishra
"Type Mismatch" Error 13 When we split as per below data type we can get this error,
“类型不匹配”错误 13 当我们按照以下数据类型拆分时，我们会得到这个错误，
**Public Sub Array_Split()
**公共子数组_Split()
This data_Dim X() As Variant_type would be change to string that mean if we are entring wrong data type we will get Type Mismatch Errorthat shit
此 data_ Dim X() As Variant_type 将更改为字符串，这意味着如果我们输入错误的数据类型，我们将得到该狗屎的类型不匹配错误
Dim X() As Variant
Dim X() 作为变体
Dim VALU As Variant
Dim VALU 作为变体
VALU = "Raj,Kumar"__ also we can select cell refrence like range("A2") or cells(1,1)
VALU = "Raj,Kumar"__ 我们也可以选择单元格引用，如 range("A2") 或单元格 (1,1)
X = VBA.split(VALU, ",")
X = VBA.split(VALU, ",")
MsgBox X(0) & vbNewLine & X(1)
MsgBox X(0) & vbNewLine & X(1)
End Sub
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