变量包含相同的值（VBA）

为了生成邮件列表，我发现变量“To”中包含相同的值

test@test.com

。邮件列表是在Visual Basic for Applications（VBA）中定义的。嗯，我在考虑如何定义一个要检查的语句，当变量具有相同的值时，然后修剪所有重复项。这意味着我需要变量在邮件列表中只出现一次

例如：

Dim objMail As Object
Dim objOutlook As Object
Set objOutlook = CreateObject("Outlook.Application")
Set objMail = objOutlook.CreateItem(0)

With objMail
  .To = test@test.com; name1@test.com; name2@test.com; name3@test.com; test@test.com; name4@test.com
  ...
End With

---

有人知道吗？

您可以使用字典删除重复项：

Sub Test()

    Dim EmailAddresses As String

    EmailAddresses = "test@test.com; name1@test.com; name2@test.com; name3@test.com; test@test.com; name4@test.com"
    EmailAddresses = RemoveDuplicates(EmailAddresses)

    Debug.Print EmailAddresses

End Sub

Public Function RemoveDuplicates(sTo As String) As String

    Dim dict As Object
    Dim vEmails As Variant
    Dim x As Long
    Dim sTemp As String

    vEmails = Split(Replace(sTo, " ", ""), ";")
    If UBound(vEmails) > 0 Then
        'Remove duplicates.
        Set dict = CreateObject("Scripting.Dictionary")
        For x = LBound(vEmails) To UBound(vEmails)
            If Not dict.exists(vEmails(x)) Then
                dict.Add vEmails(x), 1
                sTemp = sTemp & vEmails(x) & ";"
            End If
        Next x
        sTemp = Left(sTemp, Len(sTemp) - 1) 'Remove the final ;
        RemoveDuplicates = sTemp
    Else
        'There's only 1 address.
        RemoveDuplicates = sTo
    End If

End Function

如果你喜欢的话，上述内容实际上也可以通过几种方式简化

对于这样简单的重复数据消除，无需使用.code>Exists方法或

.Add

方法，因为字典项是惰性创建的。这意味着，如果一个项目不存在，那么简单地引用它就会创建它，如果它存在，就会覆盖它

您可以在dict的

键上执行
Join
函数，而不是与字典并行手动构造字符串

以下是修订版：

Public Function RemoveDuplicates2(sTo As String) As String
    Dim dict As Object
    Dim vEmails As Variant
    Dim x As Long

    vEmails = Split(Replace(sTo, " ", ""), ";")
    Set dict = CreateObject("Scripting.Dictionary")
    For x = LBound(vEmails) To UBound(vEmails)
       dict(vEmails(x)) = dict(vEmails(x)) 'Keep track of how many occurrences, in case you want to do something with it later
    Next

    RemoveDuplicates = Join(dict.Keys(), "; ")

End Function

---

您可以使用字典删除重复项：

Sub Test()

    Dim EmailAddresses As String

    EmailAddresses = "test@test.com; name1@test.com; name2@test.com; name3@test.com; test@test.com; name4@test.com"
    EmailAddresses = RemoveDuplicates(EmailAddresses)

    Debug.Print EmailAddresses

End Sub

Public Function RemoveDuplicates(sTo As String) As String

    Dim dict As Object
    Dim vEmails As Variant
    Dim x As Long
    Dim sTemp As String

    vEmails = Split(Replace(sTo, " ", ""), ";")
    If UBound(vEmails) > 0 Then
        'Remove duplicates.
        Set dict = CreateObject("Scripting.Dictionary")
        For x = LBound(vEmails) To UBound(vEmails)
            If Not dict.exists(vEmails(x)) Then
                dict.Add vEmails(x), 1
                sTemp = sTemp & vEmails(x) & ";"
            End If
        Next x
        sTemp = Left(sTemp, Len(sTemp) - 1) 'Remove the final ;
        RemoveDuplicates = sTemp
    Else
        'There's only 1 address.
        RemoveDuplicates = sTo
    End If

End Function

如果你喜欢的话，上述内容实际上也可以通过几种方式简化

对于这样简单的重复数据消除，无需使用.code>Exists方法或

.Add

方法，因为字典项是惰性创建的。这意味着，如果一个项目不存在，那么简单地引用它就会创建它，如果它存在，就会覆盖它

您可以在dict的

键上执行
Join
函数，而不是与字典并行手动构造字符串

以下是修订版：

Public Function RemoveDuplicates2(sTo As String) As String
    Dim dict As Object
    Dim vEmails As Variant
    Dim x As Long

    vEmails = Split(Replace(sTo, " ", ""), ";")
    Set dict = CreateObject("Scripting.Dictionary")
    For x = LBound(vEmails) To UBound(vEmails)
       dict(vEmails(x)) = dict(vEmails(x)) 'Keep track of how many occurrences, in case you want to do something with it later
    Next

    RemoveDuplicates = Join(dict.Keys(), "; ")

End Function

---

“To”变量是什么数据类型？它是一个数组吗？还是连接字符串？将列表转储到字典中。“它确实是为这类事情而设计的。”风暴莎草我编辑了我的帖子。现在您可以看到变量是如何定义的。“To”变量是什么数据类型？它是一个数组吗？还是连接字符串？将列表转储到字典中。“它确实是为这类事情而设计的。”风暴莎草我编辑了我的帖子。现在您可以看到变量是如何定义的；我在你的答案中添加了一个稍微简化的方法（&解释），希望没问题：）是的，很好，谢谢@David。我自己才刚刚开始使用这本词典，所以我仍然在想它是如何运作的。@DarrenBartrup Cook感谢Darren的帖子。我可以把你的例子和变量结合起来吗，我在我的文章中额外添加了什么；我在你的答案中添加了一个稍微简化的方法（&解释），希望没问题：）是的，很好，谢谢@David。我自己才刚刚开始使用这本词典，所以我仍然在想它是如何运作的。@DarrenBartrup Cook感谢Darren的帖子。我可以把你的例子和我在帖子中添加的变量结合起来吗？