使用 Excel VBA 检查值是否在数组中

Question

我在下面有一些代码，应该是检查一个值是否在一个数组中。

Sub test()
    vars1 = Array("Examples")
    vars2 = Array("Example")
    If IsInArray(Range("A1").Value, vars1) Then
        x = 1
    End If

    If IsInArray(Range("A1").Value, vars2) Then
        x = 1
    End If
End Sub

Function IsInArray(stringToBeFound As String, arr As Variant) As Boolean
  IsInArray = (UBound(Filter(arr, stringToBeFound)) > -1)
End Function

如果单元格A1 包含单词Examples，由于某种原因，两个IsInArray 都检测到它对于两个数组都存在，而它应该只发现它存在于vars1 数组中

我需要进行哪些更改才能使我的 IsInArray 函数完全匹配？

Answer 1

你可以像这样暴力破解它：

Public Function IsInArray(stringToBeFound As String, arr As Variant) As Boolean
    Dim i
    For i = LBound(arr) To UBound(arr)
        If arr(i) = stringToBeFound Then
            IsInArray = True
            Exit Function
        End If
    Next i
    IsInArray = False

End Function

使用喜欢

IsInArray("example", Array("example", "someother text", "more things", "and another"))

Answer 2

这个问题在这里被问到：VBA Arrays - Check strict (not approximative) match

Sub test()
    vars1 = Array("Examples")
    vars2 = Array("Example")
    If IsInArray(Range("A1").value, vars1) Then
        x = 1
    End If

    If IsInArray(Range("A1").value, vars2) Then
        x = 1
    End If
End Sub

Function IsInArray(stringToBeFound As String, arr As Variant) As Boolean
    IsInArray = Not IsError(Application.Match(stringToBeFound, arr, 0))
End Function

Answer 3

我搜索了这个问题，当我看到答案时，我最终创建了一些不同的东西（因为大多数时候我喜欢更少的代码而不是大多数其他东西），这些东西在绝大多数情况下都应该有效。基本上是把数组变成一个字符串，数组元素用某个分隔符分隔，然后将搜索值包裹在分隔符中，通过instr。

Function is_in_array(value As String, test_array) As Boolean
    If Not (IsArray(test_array)) Then Exit Function
    If InStr(1, "'" & Join(test_array, "'") & "'", "'" & value & "'") > 0 _
        Then is_in_array = True
End Function

你会像这样执行函数：

test = is_in_array(1, array(1, 2, 3))

Answer 4

使用 Excel VBA 中的 Match() 函数来检查值是否存在于数组中。

Sub test()
    Dim x As Long

    vars1 = Array("Abc", "Xyz", "Examples")
    vars2 = Array("Def", "IJK", "MNO")

    If IsNumeric(Application.Match(Range("A1").Value, vars1, 0)) Then
        x = 1
    ElseIf IsNumeric(Application.Match(Range("A1").Value, vars2, 0)) Then
        x = 1
    End If

    MsgBox x
End Sub

Answer 5

如果没有匹配，以下函数将返回“0”，如果匹配则返回“正整数”：

---

Function IsInArray(stringToBeFound As String, arr As Variant) As Integer IsInArray = InStr(Join(arr, ""), stringToBeFound) End Function _______________________________________________________________________________

注意：该函数首先使用 'Join' 将整个数组内容连接到一个字符串（不确定连接方法是否在内部使用循环），然后使用InStr.

Answer 6

虽然这基本上只是@Brad 的回答，但我认为可能值得包含一个稍微修改的函数，如果它存在于数组中，它将返回您正在搜索的项目的索引。如果该项目不在数组中，则返回-1。

这个的输出可以像“in string”函数一样检查，If InStr(...) > 0 Then，所以我在它下面做了一个小测试函数作为例子。

Option Explicit

Public Function IsInArrayIndex(stringToFind As String, arr As Variant) As Long

    IsInArrayIndex = -1

    Dim i As Long
    For i = LBound(arr, 1) To UBound(arr, 1)
        If arr(i) = stringToFind Then
            IsInArrayIndex = i
            Exit Function
        End If
    Next i

End Function

Sub test()

    Dim fruitArray As Variant
    fruitArray = Array("orange", "apple", "banana", "berry")

    Dim result As Long
    result = IsInArrayIndex("apple", fruitArray)

    If result >= 0 Then
        Debug.Print chr(34) & fruitArray(result) & chr(34) & " exists in array at index " & result
    Else
        Debug.Print "does not exist in array"
    End If

End Sub

然后我有点过分了，为二维数组充实了一个，因为当你 generate an array based on a range 它通常是这种形式。

它返回一个只有两个值的单维变量数组，数组的两个索引用作输入（假设找到了值）。如果未找到该值，则返回(-1, -1) 的数组。

Option Explicit

Public Function IsInArray2DIndex(stringToFind As String, arr As Variant) As Variant

    IsInArray2DIndex= Array(-1, -1)

    Dim i As Long
    Dim j As Long

    For i = LBound(arr, 1) To UBound(arr, 1)
        For j = LBound(arr, 2) To UBound(arr, 2)
            If arr(i, j) = stringToFind Then
                IsInArray2DIndex= Array(i, j)
                Exit Function
            End If
        Next j
    Next i

End Function

这是我为测试设置的数据图片，然后是测试：

Sub test2()

    Dim fruitArray2D As Variant
    fruitArray2D = sheets("Sheet1").Range("A1:B2").value

    Dim result As Variant
    result = IsInArray2DIndex("apple", fruitArray2D)

    If result(0) >= 0 And result(1) >= 0 Then
        Debug.Print chr(34) & fruitArray2D(result(0), result(1)) & chr(34) & " exists in array at row: " & result(0) & ", col: " & result(1)
    Else
        Debug.Print "does not exist in array"
    End If

End Sub

Answer 7

我想提供另一个应该性能强大的变体，因为

• 它不使用sometimes slower Match)
• 支持String、Integer、Boolean 等 (not String-only)
• 返回搜索项的索引
• 支持第n次出现

...

'-1 if not found
'https://stackoverflow.com/a/56327647/1915920
Public Function IsInArray( _
  item As Variant, _
  arr As Variant, _
  Optional nthOccurrence As Long = 1 _
  ) As Long

    IsInArray = -1

    Dim i As Long:  For i = LBound(arr, 1) To UBound(arr, 1)
        If arr(i) = item Then
            If nthOccurrence > 1 Then
                nthOccurrence = nthOccurrence - 1
                GoTo continue
            End If
            IsInArray = i
            Exit Function
        End If
continue:
    Next i

End Function

像这样使用它：

Sub testInt()
  Debug.Print IsInArray(2, Array(1, 2, 3))  '=> 1
End Sub

Sub testString1()
  Debug.Print IsInArray("b", Array("a", "b", "c", "a"))  '=> 1
End Sub

Sub testString2()
  Debug.Print IsInArray("b", Array("a", "b", "c", "b"), 2)  '=> 3
End Sub

Sub testBool1()
  Debug.Print IsInArray(False, Array(True, False, True))  '=> 1
End Sub

Sub testBool2()
  Debug.Print IsInArray(True, Array(True, False, True), 2)  '=> 2
End Sub

Answer 8

你想检查 Example 是否存在于 Range("A1").Value 中，如果失败，那么检查 Example 对吗？我认为 mycode 会完美运行。请检查。

Sub test()
Dim string1 As String, string2 As String
string1 = "Examples"
string2 = "Example"
If InStr(1, Range("A1").Value, string1) > 0 Then
    x = 1
ElseIf InStr(1, Range("A1").Value, string2) > 0 Then
    x = 2
End If

结束子