如何创建ActiveX控件命令按钮并将其设置为变量Excel VBA

时间:2018-12-19 17:08:04

标签: excel vba excel-vba variables

我正在尝试使用Excel VBA创建新的ActiveX控件命令按钮。我有一个过去工作过的循环VBA,theFile1.1.xlsm具有工作簿的主列表。我需要在〜3200个工作簿中添加一个CommandButton,因此我将使用Do-Loop宏。这是循环代码供参考。

Sub Macro2() 

Application.ScreenUpdating = False

Dim sFile As String
Dim wb As Workbook
Dim FileName1 As String
Dim FileName2 As String
Dim wksSource As Worksheet
Const scWkbSourceName As String = "theFILE 1.1.xlsm"

Set wkbSource = Workbooks(scWkbSourceName)
Set wksSource = wkbSource.Sheets("Sheet1") ' Replace 'Sheet1' w/ sheet name of SourceSheet

Const wsOriginalBook As String = "theFILE 1.1.xlsm"
Const sPath As String = "E:\ExampleFolder\"

SourceRow = 5

Do While Cells(SourceRow, "D").Value <> ""

    Sheets("Sheet1").Select

    FileName1 = wksSource.Range("A" & SourceRow).Value
    FileName2 = wksSource.Range("K" & SourceRow).Value

    sFile = sPath & FileName1 & "\" & FileName2 & ".xlsm"

    Set wb = Workbooks.Open(sFile)

        ''insert code for loop operation

    '''CLOSE WORKBOOK W/O BEFORE SAVE
    Application.EnableEvents = False
    ActiveWorkbook.Save
    ActiveWorkbook.Close
    Application.EnableEvents = True

SourceRow = SourceRow + 1

Loop

End Sub

我想将按钮设置为变量(我认为),因此我可以编辑格式/属性,并希望以后向该按钮添加宏。

    Dim buttonControl As MSForms.CommandButton

    Set buttonControl = _
        ActiveSheet.OLEObjects.Add(ClassType:="Forms.CommandButton.1", _
        Link:=False, _
        DisplayAsIcon:=False, _
        Left:=1464, Top:=310, Width:=107.25, Height:=30)

    With buttonControl.Opject
        .Caption = "OPEN FOLDER"
        .Name = "cmd_OPEN_FOLDER"

    End With

我遇到“运行时错误13:类型不匹配”错误。我不确定为什么,因为在正确的位置创建了“ CommandButton1”。

1 个答案:

答案 0 :(得分:4)

OLEObjects.Add创建一个OLEObject并将其添加到OLEObjects集合中; Add函数返回的对象是OLEObject,而不是MSForm.CommandButton。这是OLEObject.Object的基础类型-因此,请将您的buttonControl设置为返回对象的.Object属性:

Set buttonControl = _
    ActiveSheet.OLEObjects.Add(ClassType:="Forms.CommandButton.1", _
    Link:=False, _
    DisplayAsIcon:=False, _
    Left:=1464, Top:=310, Width:=107.25, Height:=30).Object

在正确的位置创建了按钮,因为Add函数可以工作并返回-类型不匹配的失败之处在于将返回的OLEObject分配给了CommandButton变量,该操作结束后。

随后的With块可以只是With buttonControl