Question

我正在尝试使用VBA中的index命令运行宏。它是关于风险矩阵中的风险变量，基于“机会”和“效果”变量。我正在创建一个工具，帮助我的组织进行风险评估。

用户提供机会和效果变量（X和Y轴）的输入。应根据此输入从矩阵中选择总和。 Matrix是在不同的工作表上创建的。

到目前为止，我得到了这个：

Dim ws2 As Worksheet
Set ws2 = Worksheets("RisicoMatrix")

Dim effectIniVar As Long
Dim waarschijnlijkheidIniVar As Long
Dim risicoIniVar As Long

effectIniVar = WorksheetFunction.Match(comboIniEffect, ws2.Range("A3:A10"), 0)

waarschijnlijkheidIniVar = WorksheetFunction.Match(comboIniWaarschijnlijkheid, ws2.Range("B2:I2"), 0)

risicoIniVar = Application.WorksheetFunction.Index(ws2.Range("b3:I10"), effectIniVar, waarschijnlijkheidIniVar)

前两个匹配命令工作正常并给我正确的值，但索引行不起作用。当我运行宏时，我得到运行时错误13，类型不匹配错误，并且在调试时突出显示'索引'行。我知道这条线可能不太正确，但我不明白我在论坛上找到的例子。

我想在VBA中完成此操作，因为这是从表单字段运行的宏。风险矩阵只是其中的一个要素。

非常感谢任何帮助。

提前感谢您的时间和精力。

Answer 1

我和罗里一起关于返回值并不能隐式转换为Long类型。

您可以单步执行代码并查看Index(ws2.Range("b3:I10"), effectIniVar, waarschijnlijkheidIniVar)

返回的值

或者你放置了一个错误处理程序，主要是（我相信）用于调试目的

此外Match()功能可以有一些＆＃34; side＆＃34;效果只要＆＃34;数字＆＃34; vs＆＃34; strings＆＃34;匹配，可以使用Find()对象的Range方法最有效地处理

所以你可以这样：

Option Explicit

Sub main2()
    Dim effectIniVar As Long
    Dim waarschijnlijkheidIniVar As Long
    Dim risicoIniVar As Long
    Dim comboIniEffect As String, comboIniWaarschijnlijkheid As String '<--| just a guess
    Dim myVal As Variant '<--| variable to hold value returned by Index function

    comboIniEffect = 3 '<--| following on my guess
    comboIniWaarschijnlijkheid = "D" '<--| following on my guess

    With Worksheets("RisicoMatrix")
        If Not FindIndex(.Range("A3:A10"), comboIniEffect, "R", effectIniVar) Then Exit Sub '<--| try finding the row match, otherwise exit
        If Not FindIndex(.Range("B2:I2"), comboIniWaarschijnlijkheid, "C", waarschijnlijkheidIniVar) Then Exit Sub '<--| try finding the column match, otherwise exit
        myVal = Application.WorksheetFunction.index(.Range("b3:I10"), effectIniVar, waarschijnlijkheidIniVar) '<--| get value by Index function
        On Error GoTo errorhandler '<--| set your error handler
        risicoIniVar = CLng(myVal) '<--| try forcing value returned by Index into a Long type
        On Error GoTo 0 '<--| resume standard error handling
    End With

    MsgBox effectIniVar & ", " & waarschijnlijkheidIniVar & ", " & risicoIniVar '<--| if everything has worked, then show: row, column and value
    Exit Sub

errorhandler:
    MsgBox "Error: " & Err.Number & vbCrLf & Err.Description & vbCrLf & "while forcing '" & myVal & "' into a Long" '<--| show error conditions
End Sub


Function FindIndex(rng As Range, val As Variant, rowOrCol As String, index As Long) As Boolean
    Dim found As Range

    With rng
        Set found = .Find(what:=val, LookIn:=xlValues, lookat:=xlWhole, MatchCase:=False)
        If Not found Is Nothing Then
            FindIndex = True
            index = IIf(UCase(rowOrCol) = "R", found.Row - .Rows(1).Row + 1, found.Column - .Columns(1).Column + 1)
        End If
    End With
End Function

让'Worksheetfunction.Index'工作的困难（Excel 2010）

1 个答案: