假设我有一个由空格/连字符分隔的字符串。
例如。带上这些折断的翅膀,学会飞翔。
什么vba函数可以找到破字的位置,应该返回3,而不是字符位置11。
一种解决方案(可能有一种更有效的方法)是拆分字符串并遍历返回的数组:
Function wordPosition(sentence As String, searchWord As String) As Long
Dim words As Variant
Dim i As Long
words = Split(sentence, " ")
For i = LBound(words, 1) To UBound(words, 1)
If words(i) = searchWord Then Exit For
Next i
'return -1 if not found
wordPosition = IIf(i > UBound(words, 1), -1, i + 1)
End Function
你可以这样称呼它:
Sub AnExample()
Dim s As String
Dim sought As String
s = "Take these broken wings and learn to fly"
sought = "broken"
MsgBox sought & " is in position " & wordPosition(s, sought)
End Sub
assylias 提出的解决方案非常好,只需要稍作修改即可解决多次出现的问题:
Function wordPosition(sentence As String, searchWord As String) As Long()
Dim words As Variant
Dim i As Long
words = Split(sentence, " ")
Dim matchesCount As Long: matchesCount = 0
ReDim matchesArray(UBound(words) + 1) As Long
For i = LBound(words, 1) To UBound(words, 1)
If words(i) = searchWord Then
matchesCount = matchesCount + 1
matchesArray(matchesCount) = IIf(i > UBound(words, 1), -1, i + 1)
End If
Next i
If (matchesCount > 0) Then
matchesArray(0) = matchesCount
End If
wordPosition = matchesArray
End Function
Sub AnExample()
Dim s As String
Dim sought As String
s = "Take these broken wings and learn to fly and broken again"
sought = "broken"
Dim matches() As Long: matches = wordPosition(s, sought)
If (matches(0) > 0) Then
Dim count As Integer: count = 0
Do
count = count + 1
MsgBox "Match No. " & count & " for " & sought & " is in position " & matches(count)
Loop While (count < matches(0))
End If
End Sub