Question

我在VBA中遇到了一个奇怪的问题，因为它似乎正在丢失变量的值。有什么想法吗？如果删除循环，debug.print将显示“ test”，否则为空（除非我在循环中打印“ dupa”的值）...似乎很奇怪。

Function carbon_copy(indeks As String) As String

Dim emails(1 To 3) As String
Dim i As Integer
Dim dupa As String

emails(1) = "abc@wp.pl"
emails(2) = "pabc@wp.pl"
emails(3) = "rabc@wp.pl"

i = 1
dupa = "test"

Do While emails(i) <> ""

    If i = indeks Then
         GoTo NextIteration
    End If

    dupa = dupa & ";" & emails(i)


    NextIteration:
    i = i + 1

Loop

Debug.Print dupa

carbon_copy = dupa

End Function

Answer 1

您应该收到运行时错误9，因为在遍历电子邮件的String数组后，索引i的值为4。尝试将emails(4)与""的值进行比较时，由于您已将Array定义为只有3个元素长，因此它应该会产生“索引超出范围”。

为澄清起见，请尝试以下示例代码，它会产生相同的错误：

Function littleTest()
    Dim teststr(1 To 3) As String
    Dim i As Integer

    teststr(1) = "abc"
    teststr(2) = "def"
    teststr(3) = "ghi"

    i = 1

    Do While teststr(i) <> ""
        Debug.Print "i do it for the " & i & " time!"
        i = i + 1
    Loop

End Function

自UBound()返回数组的实际长度（在您的情况下为3）以来，您就已经找到了解决方案。

Answer 2

实际上，我已经通过使用其他循环类型（对于i = 1到UBound（emails），下一个i）解决了该问题，但是为什么上一个循环不起作用对我来说还是很神秘的...如果任何人都可以解释，我会很感激，因为我更喜欢理解事物，而不是感谢他们正确地做它们。

W。

Answer 3

您正在索引超出数组范围。给定数组，条件Do While emails(i) <> ""始终为true，因此在emails(4)上失败。只需测试数组范围并在其上循环即可：

For i = LBound(emails) To UBound(emails)
    If emails(i) <> "" And i = indeks Then
        dupa = dupa & ";" & emails(i)
    End If
Next

Answer 4

这应该有效（注释中的解释）：

Function carbon_copy(indeks As Long) As String
    Dim emails(1 To 3) As String
    Dim i As Long
    Dim dupa As String

    emails(1) = "abc@wp.pl"
    emails(2) = "pabc@wp.pl"
    emails(3) = "rabc@wp.pl"

    i = 1
    Do While emails(i) <> ""
        If i <> indeks Then dupa = dupa & ";" & emails(i) ' update 'dupa' if current index doesn't natch passed 'indeks'
        i = i + 1
        If i > UBound(emails, 1) Then Exit Do ' be sure to exit upon exceeding 'emails()' array size
    Loop

    carbon_copy = dupa
End Function

VBA：循环丢失变量值

4 个答案: