VBE 中的 ReplaceLine 方法仅替换部分行

ReplaceLine method in VBE only replacing part of line

在回答 问题时,我认为编写一个 VBE 宏会很有趣,它会自动替换看起来像

的行
DimAll a, b, c, d As Integer

来自

Dim a As Integer, b As Integer, c As Integer, d As Integer

在我的初稿中,我只想修改选定的一行。在建立了适当的引用以获取 VBE 对象模型(参见 http://www.cpearson.com/excel/vbe.aspx )并进行了一些尝试后,我想出了:

Function ExpandDim(codeLine As String) As String
    Dim fragments As Variant
    Dim i As Long, n As Long, myType As String
    Dim last As Variant
    Dim expanded As String

    If UCase(codeLine) Like "*DIMALL*AS*" Then
        codeLine = Replace(codeLine, "dimall", "Dim", , , vbTextCompare)
        fragments = Split(codeLine, ",")
        n = UBound(fragments)
        last = Split(Trim(fragments(n)))
        myType = last(UBound(last))
        For i = 0 To n - 1 'excludes last fragment
            expanded = expanded & IIf(i = 0, "", ",") & fragments(i) & " As " & myType
        Next i
        expanded = expanded & IIf(n > 0, ",", "") & fragments(n)
        ExpandDim = expanded
    Else
        ExpandDim = codeLine
    End If
End Function

Sub DimAll()
    Dim myVBE As VBE
    Dim startLine As Long, startCol As Long
    Dim endLine As Long, endCol As Long
    Dim myLine As String
    Set myVBE = Application.VBE
    myVBE.ActiveCodePane.GetSelection startLine, startCol, endLine, endCol
    myLine = myVBE.ActiveCodePane.CodeModule.Lines(startLine, 1)
    Debug.Print ExpandDim(myLine)
    myVBE.ActiveCodePane.CodeModule.ReplaceLine startLine, ExpandDim(myLine)
End Sub

在我的另一个代码模块中:

Sub test()
    DimAll a, b, c, d As Integer
    Debug.Print TypeName(a)
    Debug.Print TypeName(b)
    Debug.Print TypeName(c)
    Debug.Print TypeName(d)
End Sub

这是奇怪的部分。当我突出显示以 DimAll a 开头的行并调用我笨拙命名的子 DimAll 时,立即 window 我看到

Dim a As Integer, b As Integer, c As Integer, d As Integer

符合预期,但在代码模块本身中,该行更改为

Dim a, b, c, d As Integer

DimAll 已替换为 Dim -- 但该行的其余部分未修改。我怀疑逗号混淆了 ReplaceLine 方法。关于如何解决此问题的任何想法?

当我 运行 使用调试器时,myLine 会在两次调用之间更改值。 DimAll 在第二次通过时变为 Dim

这是因为在 ExpandDim Function.

中输入主要 If 条件后,您将替换 codeLine 的值

在该函数中创建一个新变量,您应该没问题...或者传递它 ByVal,您就可以了:

Function ExpandDim(ByVal codeLine As String) As String