赋值左侧的函数调用必须返回 Variant 或 Object，在两个数组中查找公共值答案

【问题标题】：Function call on left hand side of the assignment must return Variant or Object, finding common values in two arrays赋值左侧的函数调用必须返回 Variant 或 Object，在两个数组中查找公共值
【发布时间】：2013-04-15 22:41:14
【问题描述】：

我有两个数组，我正在尝试提取其中的公共值，函数 findUniques 应该返回由逗号连接并分隔的字符串（它们是名称）。我想我已经很接近了，但我找不到问题所在。第 10 行出现错误

Function findUniques(astrArray1() As String, astrArray2() As String) As String

    Dim blnMP5 As Boolean
    blnMP5 = False
    Dim counter1 As Long
    Dim counter2 As Long

    For counter1 = LBound(astrArray1) To UBound(astrArray1)
        For counter2 = LBound(astrArray1) To UBound(astrArray2)
            If astrArray1(counter1) = astrArray2(counter2) Then
                blnMP5 = False
                If blnMP5 = True Then
                    findUniques() = findUniques & "," & "astrArray1()"
                End If
            End If
        Next counter2
    Next counter1

End Function

【问题讨论】：

标签： arrays string vba excel for-loop

【解决方案1】：

你的错误在这一行：

findUniques() = findUniques & "," & "astrArray1()"

您正在尝试为函数调用分配一个值，这没有意义。

您应该在函数的开头声明一个变量并继续添加该变量并最终返回它：

Function findUniques(astrArray1() As String, astrArray2() As String) As String

    Dim blnMP5 As Boolean
    blnMP5 = False    
    Dim counter1 As Long
    Dim counter2 As Long  
    Dim uniquesString as String = ""

    For counter1 = LBound(astrArray1) To UBound(astrArray1)
        For counter2 = LBound(astrArray1) To UBound(astrArray2)
            If astrArray1(counter1) = astrArray2(counter2) Then
               blnMP5 = False
               If blnMP5 = True Then
                   uniquesString = uniquesString & "," & "astrArray1()"
               End If
             End If
        Next counter2
     Next counter1

End Function

【讨论】：

【解决方案2】：

声明一个变量来保存你正在构建的字符串

    Dim tempString as string

并在第 10 行添加此内容

    tempString = tempString & "," &  astrArray1(counter1)

最后

    findUniques = tempString 
End function

【讨论】：

完美！感谢您的帮助
Kenneth 的回答中可能已经确定了问题的原因。虽然目前的答案解决了 OP 的问题，但很可能是由于删除了导致错误的行。本说明旨在说明不必强制使用辅助字符串变量来解决问题。