VBA-无效或不合格的参考错误

Srpic 发表于 Dev

斯皮克

我正在尝试创建excel模板（数据量因情况而异），看起来像这样：

在每偶数行中都是“客户”，我想在每奇数行中都输入“账本”。基本上，应该将“ Ledger”放入每个奇数行，直到C列中有数据为止。我有以下代码：

'========================================================================
' INSERTING LEDGERS for every odd row (below Customer)
'========================================================================

Sub Ledgers()

    Dim rng As Range
    Dim r As Range
    Dim LastRow As Long

    LastRow = .Cells(.Rows.Count, "C").End(xlUp).Row
    Set rng = .Range("C5:C" & LastRow)

    For i = 1 To rng.Rows.Count
        Set r = rng.Cells(i, -2)
        If i Mod 2 = 1 Then
            r.Value = "Ledger"
        End If

    Next i

End Sub

但是它给我一个错误msg无效或不合格的引用。您能告诉我我哪里有错误吗？

非常感谢！

ᴇʜ

如果一个命令开始.喜欢.Cells它希望成为一个内with像的语句...

With Worksheets("MySheetName")
    LastRow = .Cells(.Rows.Count, "C").End(xlUp).Row
    Set rng = .Range("C5:C" & LastRow)
End With

因此，您需要指定期望单元格所在的工作表的名称。

并不是最好Option Explicit在模块顶部使用它来强制声明每个变量（您未声明i As Long）。

您的代码可以简化为...

Option Explicit 

Public Sub Ledgers()
    Dim LastRow As Long
    Dim i As Long

    With Worksheets("MySheetName") 
        LastRow = .Cells(.Rows.Count, "C").End(xlUp).Row

        'make sure i starts with a odd number
        'here we start at row 5 and loop to the last row
        'step 2 makes it overstep the even numbers if you start with an odd i
        'so there is no need to proof for even/odd
        For i = 5 To LastRow Step 2 
            .Cells(i, "A") = "Ledger" 'In column A
            '^ this references the worksheet of the with-statement because it starts with a `.`
        Next i
    End With
End Sub

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