我有一张 Excel 工作表,其中单元格 A1-C20=
=INT(RAND()*10)
。这是我的数据范围。单元格 E1=1、E2=2、E3=3 等。这些是我试图找到的值。我设置单元格 F1==MATCH(E1,A:C,0)
,F2==MATCH(E1,A:C,0)
,等等
但是,所有
MATCH
函数都会返回 #N/A
,因为输入范围是多维的。如何测试给定值(1、2、3、4 等)是否存在于多维范围(A1-C20)中?
/编辑:此功能有效,但超出了我的需要。有没有办法让它只返回 TRUE 或 FALSE,具体取决于查找值是否在范围内?
Function OzgridLookup(Find_Val As Variant, Occurrence As Long, Table_Range As Range, _
Offset_Cols As Long, Optional Column_Lookin As Long, Optional Row_Offset As Long) As Variant
Dim lLoop As Long
Dim FoundCell As Range
If Column_Lookin = 0 Then 'No column # specified
With Table_Range
'Top left cell has Find_Val & Occurrence is 1
If Table_Range.Cells(1, 1) = Find_Val And Occurrence = 1 Then
OzgridLookup = .Cells(1, 1)(1, Offset_Cols + 1)
Exit Function 'All done :)
Else 'No column # specified so search all for _
nth Occurrence reading left to right
Set FoundCell = .Cells(1, 1) 'Set cell variable for Find start
For lLoop = 1 To Occurrence 'Loop as many times as Occurrences _
and each time Set "FoundCell" to start next Find from
Set FoundCell = _
Table_Range.Find(What:=Find_Val, After:=FoundCell, _
LookIn:=xlValues, LookAt:=xlWhole, _
SearchOrder:=xlRows, SearchDirection:=xlNext)
Next lLoop
End If
End With
Else 'column # specified
With Table_Range.Columns(Column_Lookin) 'Work with column # specified
Set FoundCell = .Cells(1, 1) 'Set cell variable for Find start
For lLoop = 1 To Occurrence 'Loop as many times as Occurrences _
and each time Set "FoundCell" to start next Find from
Set FoundCell = _
Table_Range.Find(What:=Find_Val, After:=FoundCell, _
LookIn:=xlValues, LookAt:=xlWhole, _
SearchOrder:=xlRows, SearchDirection:=xlNext)
Next lLoop
End With
End If
OzgridLookup = FoundCell.Offset(Row_Offset, Offset_Cols)
End Function
您可以使用 COUNTIF 来执行此操作,因为您只想知道该号码是否存在(而不是其位置)。
=COUNTIF(A:C,E1)>0
如果存在则返回“TRUE”,如果不存在则返回“FALSE”。
只是为了好玩,这里有一个工作表函数解决方案,它返回与查找值匹配的单元格地址。它利用了您仅在 3 列中搜索的事实。
=IF(ISERROR(MATCH(E1,A:A,0)),IF(ISERROR(MATCH(E1,B:B,0)),IF(ISERROR(MATCH(E1,C:C,0)),"Not found.","C"&MATCH(E1,C:C,0)),"B"&MATCH(E1,B:B,0)),"A"&MATCH(E1,A:A,0))
我想我还应该引入一个 VBA 解决方案,它可以返回(连续)范围内的匹配位置。它从左到右一次查看一列,并返回找到的第一个匹配项的地址。
Public Function MDMATCH(srchfor As String, lookin As Range) As String
Application.Volatile
Dim RngArray() As Variant
Dim topleft As String
Dim tmpval As String
topleft = lookin.Address
topleft = Left(topleft, InStr(topleft, ":") - 1)
tmpval = "Not found."
RngArray = lookin
For i = 1 To UBound(RngArray, 2)
If tmpval = "Not found." Then
For j = 1 To UBound(RngArray, 1)
If RngArray(j, i) = srchfor Then
tmpval = Range(topleft).Offset(j - 1, i - 1).Address
Exit For
End If
Next j
Else
Exit For
End If
Next i
MDMATCH = tmpval
End Function
这与答案有关 2011 年 8 月 24 日 17:47 经过 非常好
如果你有
Option Explicit
除非您包含,否则您会收到错误
Dim i As Long, j As Long
如果你没有
Option Explicit
那么如果不同的宏使用不同的假设,你就会遇到麻烦。
如果您输入一系列整列,例如“A:C”,则该行
tmpval = Range(topleft).Offset(j - 1, i - 1).address
将会失败,因为
topleft = Left(topleft, InStr(topleft, ":") - 1)
只是“A”,但如果您只想要 A 列,则
Range(topleft)
需要 topleft
为“A:A”
同样,对于整行的输入,它也会在同一点失败。
分配后应该有一些案例测试逻辑
topleft
。
示例: 检查
topleft
的第一个字符是否是数字,如果是,那么你知道它是一个完整的行范围。如果失败,请检查 topleft
的最后一个字符是否是数字,如果是数字,则您知道它是一个单元格的地址。如果失败,您知道它只是一个列范围,因此您可以将 topleft
扩展到 topleft & ":" & topleft
。