如题，如何自定义一个函数来实现查找指定字符串在另一字符串中第N个位置？
我是初学者，我知道寻找字符串的位置可以使用Instr函数，但是如果我寻找的字符在一个字符串中出现n次，我想随便找到其中的一次的位置，怎么自己编写一个函数来实现呢？请大家指教！！～～

我试着写了下，请大家看下这样写有没有问题？
Public Function InStrL(n As Integer,
inString As String, srchString As String) As Integer

Dim iCurPos As Integer

If Len(srchString)<>0 Then
If n=1 Then

iCurPos = InStr(inString, srchString)
Else
for iCurPos = 1 to n-1

iCurPos = InStr(iCurPos + 1, inString, srchString)
Next iCurPos

End If
End If
InStrL = iCurPos

End Function

On Error Goto ErrorOccurred
Dim i,k
k=1
For i=1 to n
k=instr(k,instr,sechstr)
next i
return k
ErrorOccurred:

根据楼上的我把原来写的改了下，看这样可以吗？
Public Function InStrL(n As Integer,
inString As String, srchString As String) As Integer

Dim iCurPos ，i As Integer

If Len(srchString)<>0 Then
iCurPos = 1
for i = 1 to n
iCurPos = InStr(iCurPos, inString, srchString)
Next iCurPos
End If

InStrL = iCurPos
End Function

[CODE]Public Function InStrL(n As Integer,
inString As String, srchString As String) As Integer

Dim iCurPos ，i As Integer

If Len(srchString)<>0 Then
iCurPos = 1
for i = 1 to n
iCurPos = InStr(iCurPos, inString, srchString)
Next iCurPos ' 再想想
End If

InStrL = iCurPos
End Function[/CODE]

！！！ srch=""时没有返回值！！！

！！！ 抱歉我笔误了一下，instr中带入的k应当为上次的instr结果+1

修改后的代码如下：（修改处加粗了）

[CODE]Public Function InStrL(n As Integer, inString As String, srchString As String) As Long
Dim iCurPos, i As Integer
If Len(srchString) = 0 Then InStrL = -1: Exit Function

iCurPos = 0
For i = 1 To n
iCurPos = InStr(iCurPos + 1, inString, srchString)
Next i

InStrL = iCurPos
End Function[/CODE]

关于为什么As Long：

VB中字符串的长度最大为约2,147,483,647字节，所以要用Long类型来存储。

[QUOTE]String 数据类型
字符串有两种：变长与定长的字符串。
变长字符串最多可包含大约 20 亿 ( 2^31)个字符。
定长字符串可包含 1 到大约 64K ( 2^16 ) 个字符。
注意 Public 定长字符串不能在类模块中使用。
String 之字符码的范围是 0 到 255。字符集的前 128 个字符（0 到 127）对应于标准的 U.S. 键盘上的字符与符号。这前 128 个字符与 ASCII 字符集中所定义的相同。后 128 个字符（128 到 255）则代表特殊字符，例如国际字符，重音符号，货币符号及分数。String 的类型声明字符为美元号 ($)。[/QUOTE]

谢谢，但是我不明白“If Len(srchString) = 0 Then InStrL = -1: Exit Function”
为什么要赋值为-1。

Private Function fncGetStrNum(InString As String, InSeach As String, InCount As Integer) As Long
Dim i As Long, llCount As Long
Dim llLength As Long
i = 1: llCount = 0
llLength = Len(InString)
fncGetStrNum = -1
Do While i <= llLength
If InStr(i, InString, InSeach) > 0 Then
If llCount = InCount - 1 Then
fncGetStrNum = InStr(i, InString, InSeach)
Exit Do
End If
i = InStr(i, InString, InSeach) + Len(InSeach)
llCount = llCount + 1
End If
Loop
End Function

Private Sub Command1_Click()
MsgBox fncGetStrNum("abcdabcdabcdabcdabcd", "ab", 3)
End Sub

不好意思，楼上的方法不是很明白，可否稍微注解一下？？

没有测试效率，只是提供个另外的思路……

代码长度……

Private Function fncGetStrNum(InString As String, InSearch As String, InCount As Integer) As Long
Dim TempStr As String
TempStr = Replace(InString, InSearch, "", 1, InCount - 1)
fncGetStrNum = InStr(TempStr, InSearch)
If fncGetStrNum <> 0 Then fncGetStrNum = fncGetStrNum + (InCount - 1) * Len(InSearch)
End Function