Vba как найти номер элементов

Dim pos, arr, val

arr=Array(1,2,4,5)
val = 4

pos=Application.Match(val, arr, False)

if not iserror(pos) then
   Msgbox val & " is at position " & pos
else
   Msgbox val & " not found!"
end if

Updated to show using Match (with .Index) to find a value in a dimension of a two-dimensional array:

Dim arr(1 To 10, 1 To 2)
Dim x

For x = 1 To 10
    arr(x, 1) = x
    arr(x, 2) = 11 - x
Next x

Debug.Print Application.Match(3, Application.Index(arr, 0, 1), 0)
Debug.Print Application.Match(3, Application.Index(arr, 0, 2), 0)

EDIT: it’s worth illustrating here what @ARich pointed out in the comments – that using Index() to slice an array has horrible performance if you’re doing it in a loop.

In testing (code below) the Index() approach is almost 2000-fold slower than using a nested loop.

Sub PerfTest()

    Const VAL_TO_FIND As String = "R1800:C8"
    Dim a(1 To 2000, 1 To 10)
    Dim r As Long, c As Long, t

    For r = 1 To 2000
        For c = 1 To 10
            a(r, c) = "R" & r & ":C" & c
        Next c
    Next r

    t = Timer
    Debug.Print FindLoop(a, VAL_TO_FIND), Timer - t
    ' >> 0.00781 sec

     t = Timer
    Debug.Print FindIndex(a, VAL_TO_FIND), Timer - t
    ' >> 14.18 sec

End Sub

Function FindLoop(arr, val) As Boolean
    Dim r As Long, c As Long
    For r = 1 To UBound(arr, 1)
    For c = 1 To UBound(arr, 2)
        If arr(r, c) = val Then
            FindLoop = True
            Exit Function
        End If
    Next c
    Next r
End Function

Function FindIndex(arr, val)
    Dim r As Long
    For r = 1 To UBound(arr, 1)
        If Not IsError(Application.Match(val, Application.Index(arr, r, 0), 0)) Then
            FindIndex = True
            Exit Function
        End If
    Next r
End Function

Dim pos, arr, val

arr=Array(1,2,4,5)
val = 4

pos=Application.Match(val, arr, False)

if not iserror(pos) then
   Msgbox val & " is at position " & pos
else
   Msgbox val & " not found!"
end if

Обновлено, чтобы показать, используя Match (с .Index), чтобы найти значение в измерении двумерного массива:

Dim arr(1 To 10, 1 To 2)
Dim x

For x = 1 To 10
    arr(x, 1) = x
    arr(x, 2) = 11 - x
Next x

Debug.Print Application.Match(3, Application.Index(arr, 0, 1), 0)
Debug.Print Application.Match(3, Application.Index(arr, 0, 2), 0)

EDIT: стоит ли здесь упомянуть то, что @ARich указал в комментариях – использование Index() для среза массива имеет ужасную производительность, если вы делаете это в цикле.

При тестировании (код ниже) подход Index() почти в 2000 раз медленнее, чем использование вложенного цикла.

Sub PerfTest()

    Const VAL_TO_FIND As String = "R1800:C8"
    Dim a(1 To 2000, 1 To 10)
    Dim r As Long, c As Long, t

    For r = 1 To 2000
        For c = 1 To 10
            a(r, c) = "R" & r & ":C" & c
        Next c
    Next r

    t = Timer
    Debug.Print FindLoop(a, VAL_TO_FIND), Timer - t
    ' >> 0.00781 sec

     t = Timer
    Debug.Print FindIndex(a, VAL_TO_FIND), Timer - t
    ' >> 14.18 sec

End Sub

Function FindLoop(arr, val) As Boolean
    Dim r As Long, c As Long
    For r = 1 To UBound(arr, 1)
    For c = 1 To UBound(arr, 2)
        If arr(r, c) = val Then
            FindLoop = True
            Exit Function
        End If
    Next c
    Next r
End Function

Function FindIndex(arr, val)
    Dim r As Long
    For r = 1 To UBound(arr, 1)
        If Not IsError(Application.Match(val, Application.Index(arr, r, 0), 0)) Then
            FindIndex = True
            Exit Function
        End If
    Next r
End Function