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
更新为显示使用 Match(带有 .Index)在二维数组的某个维度中查找值:
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)
编辑:这里值得说明 @ARich 在评论中指出的内容 - 使用Index()
如果您在循环中执行此操作,则对数组进行切片会产生可怕的性能。
在测试中(下面的代码),Index() 方法几乎比使用嵌套循环慢 2000 倍。
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