大家讨论：如何在包含若干个数字的字符串中获取第任意个数

[z_liming]

例如：“12 34 45 56 67 78 122 －123”，间隔符也可是其它字符，如：“，”
如何提取第三个数，即：45
当然用instr，可以实现，但如何做才是高效的呢？

[ZOU_SEAFARER]

看我这样的方法可以不? 效率问题如何?
Option Explicit
Dim str As String

Function GetNumber(ByVal Text As String, ByVal Index As Long) As Long
Dim i As Integer
Dim j As Integer
Dim str_Temp As String * 1
Dim Str_Text As String
Dim str_Arr() As String
Dim str_Arr1() As String
Str_Text = vbNullString
For i = 1 To Len(Text)
str_Temp = Mid(Text, i, 1)
If Asc(str_Temp) < 48 Or Asc(str_Temp) > 57 Then
Str_Text = Str_Text & ","
Else
Str_Text = Str_Text & str_Temp
End If
Next
str_Arr = Split(Str_Text, ",")

'下面也可以使用数组移位来达到删除非数字元素
'***********************************************************************
For i = 0 To UBound(str_Arr)
If IsNumeric(str_Arr(i)) = True Then
ReDim Preserve str_Arr1(j) As String
str_Arr1(j) = str_Arr(i)
j = j + 1
End If
Next
**************************************************************************
If Index <= UBound(str_Arr1) + 1 And Index > 0 Then
GetNumber = str_Arr1(Index - 1)
Else
GetNumber = vbNullString
End If
End Function


Private Sub Form_Load()
str = "12 34 45 56 67 78 122 －123"
Debug.Print GetNumber(str, 3)
End Sub

[ZOU_SEAFARER]

Option Explicit
Private Declare Function timeGetTime Lib "winmm.dll" () As Long

Dim str As String

Function GetNumber(ByVal Text As String, ByVal Index As Long) As Long
Dim i As Integer
Dim j As Integer
Dim str_Temp As String * 1
Dim Str_Text As String
Dim str_Arr() As String
Dim str_Arr1() As String
Str_Text = vbNullString
For i = 1 To Len(Text)
str_Temp = Mid(Text, i, 1)
If Asc(str_Temp) < 48 Or Asc(str_Temp) > 57 Then
Str_Text = Str_Text & ","
Else
Str_Text = Str_Text & str_Temp
End If
Next
str_Arr = Split(Str_Text, ",")
For i = 0 To UBound(str_Arr)
If IsNumeric(str_Arr(i)) = True Then
ReDim Preserve str_Arr1(j) As String
str_Arr1(j) = str_Arr(i)
j = j + 1
End If
Next

If Index <= UBound(str_Arr1) + 1 And Index > 0 Then
GetNumber = str_Arr1(Index - 1)
Else
GetNumber = vbNullString
End If
End Function


Private Sub Command1_Click()
Dim lng_Loop As Long
Dim lng_Start As Long
Dim lng_End As Long
lng_Start = timeGetTime
For lng_Loop = 1 To 500
Call GetNumber(str, 3)
Next
lng_End = timeGetTime
Debug.Print lng_End - lng_Start
End Sub

Private Sub Form_Load()
str = "12 34 45 56 67 78 122 －123"
End Sub

运行500次,时间30MS

[GoldFox]

用Split函数把字符串拆分到数组,再直接取就可以了

[迈克揉索芙特]

Dim str As String
str = "12 34 45 56 67 78 122 －123"
Debug.Print GetNumder(str, 3)

Function GetNumder(ByVal Text As String, ByVal Index As Long) As Long
Dim str1 As String, str2 As String
Dim ascX As Integer, i As Long
Dim str() As String

str1 = Text
For i = 1 To Len(str1)
str2 = Mid(str1, i, 1)
ascX = Asc(str2)
If ascX > 97 Or ascX < 48 Then
str1 = Replace(str1, str2, Space(1))
End If
Next

Do While InStr(1, str1, Space(2))
str1 = Replace(str1, Space(2), Space(1))
Loop

str = Split(str1, Space(1))

GetNumder = str(Index - 1)
End Function

[TechnoFantasy]

Dim s As String
'在字符串最后要加一个非数字字符
s = "12 34 45 56 67 78 122 －123 "

Dim c() As Byte

c = StrConv(s, vbFromUnicode)

Dim ss(100) As String
Dim count As Integer
count = 0

Dim prv As Boolean
Dim prv1 As Boolean

prv = False
prv1 = False

For i = LBound(c) To UBound(c)
If ((c(i) >= 48) And (c(i) <= 59)) Then
prv1 = True
Else
prv1 = False
End If

If (prv1) Then
ss(count) = ss(count) + Chr(c(i))
End If

If ((prv = True) And (prv1 = False)) Then
count = count + 1
End If

prv = prv1
Next i

For i = LBound(ss) To count
Debug.Print ss(i)
Next i

[迈克揉索芙特]

更正：
Dim str As String
str = "12 34 45 56 67 78 122 －123"
Debug.Print GetNumder(str, 3)

Function GetNumder(ByVal Text As String, ByVal Index As Long) As Long
Dim str1 As String, str2 As String
Dim ascX As Integer, i As Long
Dim str() As String

str1 = Text
For i = 1 To Len(str1)
str2 = Mid(str1, i, 1)
ascX = Asc(str2)
If ascX > 59 Or ascX < 48 Then
str1 = Replace(str1, str2, Space(1))
End If
Next

Do While InStr(1, str1, Space(2))
str1 = Replace(str1, Space(2), Space(1))
Loop

str = Split(str1, Space(1))

GetNumder = str(Index - 1)
End Function

[迈克揉索芙特]

GetNumder<----------GetNumber
粗心啊

老大的方法也不错

[TechnoFantasy]

上面我的代码测试了一下，对一个包含40个数字的字符串做500次访问，时间在200毫秒左右。

[迈克揉索芙特]

老大的速度比我的快多了
佩服！！！

[DengXingJie]

熱鬧一下

[rotartsinimdA]

使用数组。
例如：
dim vN() as string '定义一个数组
dim vS as string ' 定义一个字符串
vs= "12,34,45,56,67,78,122,123"
vn = split(vs,",")
msgbox vn(2)
看看结果就知道了。

原理：
把按照字符串的分隔符","把字符串vS分割，然后存放入数组vN
vN 的第三个元素就是你要的45
分割使用split函数分割
更详细的方法，建议你安装一个MSDN看看。。。
上面是例子。这样的方法在VB中来看是比较高效的（可能也是唯一的快捷途径）

[rotartsinimdA]

不是吧，我就去厨房转了一圈，回来点了一下回复，就这样了。。
还以为是沙发那。。 －.－！
so fast !