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从调用函数返回数组并分配给数组

vba

surendra · 技术社区 · 6 年前

Public Sub DoSomeThing()
    Dim dict As Object
    Dim arr2(5)
    Set arr2() = aaa()

   For m = LBound(arr2) To UBound(arr2)
        Set dict = aaa()(m)
        Dim key As Variant
       For Each key In dict.Keys
           Debug.Print dict(key)
       Next key
  Next
End Sub

Public Function aaa() As Variant
    Dim arr(5)
   Dim dict_123 As Object
   For k = 1 To 2
       If k = 1 Then
           val1 = 300
           val2 = 500
       ElseIf k = 2 Then
           val1 = 600
           val2 = 1200
      End If
       Set dict_123 = CreateObject("Scripting.Dictionary")
       dict_123.Add "first", val1
       dict_123.Add "Second", val2
       Set arr(k) = dict_123
   Next
  aaa = arr
End Function

在这里我想从返回数组 aaa 致 DoSomething 并从中处理该数组 DoSomeThing .我该怎么做?

我收到无法分配给数组的错误

1 回复 | 直到 6 年前

QHarr 6 年前

有很多错误,我不确定你到底想实现什么。您的代码在下面的“固定”上方。

笔记:

arr2 = aaa 将一个数组设置为另一个数组(不将set关键字设置为not object)。不标注尺寸 arr2 第一。
测试当前阵列( 2号跑道 )项是在尝试设置之前的字典。您只在基于0的数组中的索引1和2处添加了字典。不那么健壮 If m = 1 Or m = 2
使用 Option Explicit 并声明所有变量
我喜欢A Select Case 在函数中 If 语句,特别是如果要添加更多的条件,而您可能需要多个条件的相同结果。

代码:

Option Explicit

Public Sub DoSomeThing()
    Dim dict As Object, arr2, m As Long, key As Variant
    arr2 = aaa  '<==Set one array equal to the other (no set keyword as not object)

    For m = LBound(arr2) To UBound(arr2)
       If TypeName(arr2(m)) = "Dictionary"  ' <== We can test if current array item is a dictionary before attempting the set. You have only added dictionaries at position 1 and 2 in the array. Less robust would be If m = 1 Or m = 2
            Set dict = arr2(m)   '<==index into your arr2 array
            For Each key In dict.Keys
                Debug.Print dict(key)
            Next key
        End If
    Next
End Sub

Public Function aaa() As Variant
   Dim arr(5), k As Long, val1 As Long, val2 As Long, dict_123 As Object
   For k = 1 To 2
       Select Case k '<== Use select statement 
       Case 1
           val1 = 300
           val2 = 500
       Case 2
           val1 = 600
           val2 = 1200
      End Select
      Set dict_123 = CreateObject("Scripting.Dictionary")
      dict_123.Add "first", val1
      dict_123.Add "Second", val2
      Set arr(k) = dict_123 'K starts at 1 so position 0 is empty; as are positions after 2.
   Next k
   aaa = arr

End Function