Question

我正在尝试使用this answer解决MATCH函数的字符限制：

= MATCH（TRUE，INDEX（ISNUMBER（SEARCH（A2，Dict！A：A）），0），0））

我尝试了完整版：

Sub test

  Dim KEY_CONTRACT_ROW As Variant
  KEY_CONTRACT_ROW = _
            Application.WorksheetFunction.Match(True, _
Application.WorksheetFunction.Index(Application.WorksheetFunction.IsNumber(Application.WorksheetFunction.Search("testing string", _
                ThisWorkbook.Worksheets("Sheet1").Range("A:A"))), 0), 0)

End Sub

它不起作用-Type mismatch error。

所以我尝试了较短的版本：

Sub test

  MsgBox (Application.WorksheetFunction.IsNumber(Application.WorksheetFunction.Search("testing string", _
                ThisWorkbook.Worksheets("Sheet1").Range("A:A"))))

End Sub

达到相同的结果。同时，当我在工作表上手动使用此公式时，效果很好。

Answer 1

WorksheetFunction会在发生错误时产生中断错误。而是使用Application对象的Match方法。实际上，您可以完全避免使用WorksheetFunction。

Sub test()

    Dim KEY_CONTRACT_ROW As Variant
    Dim rngSearchRange As Range

    'Define the search range as to avoid a whole column reference
    With ThisWorkbook.Worksheets("Sheet1")
        Set rngSearchRange = Application.Intersect(.UsedRange, .Range("A:A"))
    End With

    'Search for the specified search string
    With Application
        KEY_CONTRACT_ROW = .Match(True, .IsNumber(.Search("testing string", rngSearchRange)), 0)
    End With

    'Check whether a match was found
    If IsError(KEY_CONTRACT_ROW) Then
        MsgBox "Match not found!", vbExclamation
    Else
        MsgBox "Match found!", vbExclamation
    End If

End Sub

Application.WorksheetFunction.Search和.IsNumber在VBA中不起作用：类型不匹配错误；固定匹配功能字符限制

1 个答案: