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自动编号问题 00,02,...99,A1,A2,...,A9,B1,B2,...,B9,...,1A,1B,..1Z,2A,..2Z,...AA,AB,...,AZ,BA,BB,...BZ,..ZA,...ZZ

JoyceSong 2008-03-06 06:34:04

还是求助高手来帮忙
目前需要自动产生流水码,编码规则如下:
00,02,...99,A1,A2,...,A9,B1,B2,...,B9,...,1A,1B,..1Z,2A,..2Z,...AA,AB,...,AZ,BA,BB,...BZ,..ZA,...ZZ
还是没有想出来该怎么写循环

例如判断目前已经产生号码"5C"
则通过循环计算出下一个是"5D"

...全文

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JoyceSong 2008-03-10

??没注意,我的贴子在论坛主页"头条推荐"显示了,看来还是有点技术含量,呵呵
还是采用了字典方法,其他计算没空研究了.

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b哈利路亚d 2008-03-10

要是我的话，就先建立字典库，然后依次+1取编号。。
实在犯不着用逻辑判断来取，效率不高，出错率高。

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vbman2003 2008-03-09

这样的编号规则真是奇怪，赞成of123的观点：如果能说服客户采用纯36进制最好.....

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Luoyulianshen 2008-03-08

Sub work()
Dim i As Integer, base As Integer
Dim s1 As String, s2 As String, Param As String * 2

Param = "9Z"

s1 = Left(Param, 1): s2 = Right(Param, 1)
Select Case Asc(s2)
Case 48 To 57
base = 49: i = 9
Case 65 To 90
base = 65: i = 26
Case Else
MsgBox "wrong Enter!"
Exit Sub
End Select
s1 = Chr$(Asc(s1) + Int((Asc(s2) - base + 1) / i))
If s1 > "Z" Then
If base = 49 Then
s1 = "1": base = 65: i = 26: s2 = "Z"
Else
MsgBox "Overflow!"
Exit Sub
End If
ElseIf s1 > "9" And s1 < "A" Then
s1 = "A"
End If
s2 = Chr$(base + (Asc(s2) - base + 1) Mod i)

Debug.Print s1 & s2
End Sub

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嗷嗷叫的老马 2008-03-07

例如"S9"的结果是"SA" 正确的是"T1"

//

晕

你这里的9到底是进位到A还是0呀????

如果是0,那前面的S又从何而来?

如果是A,那怎么又会是T1????

我看主要问题在于你那100大分~~~嘿嘿.

不过算了,不想再写代码了...

我发的是EXCEL的标志规律,楼主看着办...

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CommandButton 2008-03-07

有点像气车牌子升位，但又怕号码重复的那种工作。而且保留了一些特殊码禁用，

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CommandButton 2008-03-07

01,02,...99,A1,A2,...,A9,B1,B2,...,B9,...,Z1,Z2,...,Z9,
1A,1B,..1Z,2A,2B,..2Z,...,9A,..9Z,
AA,AB,...,AZ,BA,BB,...BZ,..ZA,...ZZ

实际上是这样的：
起初是两位十进制。
但到后来99不够用了，就在高位上采用36进制。低位保持10进制
到Z9后，有不够用了，高位重新循环36进制（但0不要了），底位36进制（但0-9不要）

CBM666说的好，题意理解起来很困难

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of123 2008-03-07

楼主在北京吗？能请你喝咖啡吗？

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of123 2008-03-07

Const strChars As String = "0123456789ABCDEFGHIJKLMNOPQRSTUVWXYZ"

Public Function Num2Str(ByVal x As Integer) As Integer
Dim tmp As String, n As Integer

n = x Mod (36 * 36)
If n < 360 Then
tmp = Mid(strChars, (n \ 10) + 1, 1)
tmp = tmp & Mid(strChars, (n Mod 10) + 1, 1)
Else
n = n - 360
tmp = Mid(strChars, (n \ 26) + 1, 1)
tmp = tmp & Mid(strChars, (n Mod 26) + 11, 1)
End If
Num2Str = tmp
End Function

Public Function Str2Num(ByVal s As String) As String
Dim tmp As String, n As Integer

tmp = Right("0" & s, 2)

If Right(tmp, 1) <= "9" Then
n = InStr(1, strChars, Left(tmp, 1)) - 1
n = n * 10
n = n + Val(Right(tmp, 1))
Else
n = InStr(1, strChars, Left(tmp, 1)) - 1
n = n * 26
n = n + InStr(1, strChars, Right(tmp, 1)) - 11
n = n + 360
End If
Str2Num = n
End Function

Public Function GetNextNum(ByVal s As String) As String
GetNextNum = Num2Str(Str2Num(s) + 1)
End Function

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mostice 2008-03-07

你可以测试下，我测试到 ZS 后出现的结果是：[0。

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danjiewu 2008-03-07

[Quote=引用 37 楼 mostice 的回复:]
但是循环到 ZW 是就不正确了。出现[4。
[/Quote]

这个不太可能吧，应该到ZZ才会越界。

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of123 2008-03-07

大概明白楼主的意思了。实际上是个36进制码转换问题。
不过，客户的排序方式有点奇怪，实际上将数字标记法分成了两段：
1 低位数采用 10 进制，高位数采用 36 进制：01,02,...99,A1,A2,...,A9,B1,B2,...,B9,...,Z1,Z2,...,Z9
2 低位数采用 26 进制，高位数采用 36 进制：1A,1B,..1Z,2A,2B,..2Z,...,9A,..9Z, AA,AB,...,AZ,BA,BB,...BZ,..ZA,...ZZ

如果能说服客户采用纯 36 进制最好，可以简化运算。

否则：
Option Explicit
Const strChars As String = "0123456789ABCDEFGHIJKLMNOPQRSTUVWXYZ"

Private Sub Form_Load()
Dim i As Integer, n As Integer

For i = 0 To 36 * 36 - 1
List1.AddItem Num2Str(i)
Next i
End Sub

Public Function Num2Str(ByVal x As Integer)
Dim tmp As String, n As Integer

n = x Mod (36 * 36)
If n < 360 Then
tmp = Mid(strChars, (n \ 10) + 1, 1)
tmp = tmp & Mid(strChars, (n Mod 10) + 1, 1)
Else
n = n - 360
tmp = Mid(strChars, (n \ 26) + 1, 1)
tmp = tmp & Mid(strChars, (n Mod 26) + 11, 1)
End If
Num2Str = tmp
End Function

Public Function Str2Num(ByVal s As String)
Dim tmp As String, n As Integer

tmp = Right("0" & s, 2)

If Right(tmp, 1) <= "9" Then
n = InStr(1, strChars, Left(tmp, 1)) - 1
n = n * 10
n = n + Val(Right(tmp, 1))
Else
n = InStr(1, strChars, Left(tmp, 1)) - 1
n = n * 26
n = n + InStr(1, strChars, Right(tmp, 1)) - 11
n = n + 360
End If
Str2Num = n
End Function

Private Sub List1_Click()
MsgBox List1.ListIndex & ": " & Str2Num(List1.List(List1.ListIndex))
End Sub

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mostice 2008-03-07

但是循环到 ZW 是就不正确了。出现[4。

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mostice 2008-03-07

danjiewu 的方法很好。用了一下。呵呵

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cbm6666 2008-03-07

'呵呵...早说清楚嘛

'添加 Command1 List1

Dim i%, j%, aa$, bb$, s, ss
Private Sub Command1_Click()
s = Array("0 99", "65 90 1 9", "1 9 65 90", "65 90 65 90")
List1.Clear
For suzu = 0 To 3
ss = Split(s(suzu), " ")
If suzu = 0 Then
For i = Val(ss(0)) To Val(ss(1))
List1.AddItem Format(CStr(i), "00")
Next i
Else
For i = Val(ss(0)) To Val(ss(1))
aa = IIf(i <= 9, CStr(i), Chr(i))
For j = Val(ss(2)) To Val(ss(3))
bb = IIf(j <= 9, CStr(j), Chr(j))
List1.AddItem aa & bb
Next j
Next i
End If
Next suzu
End Sub

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danjiewu 2008-03-07

错了，26应该是36



string GetNextCode(string preCode)

{

     int num = (preCode[0]-'0') * 36 + preCode[1] - '0' + 1;

     return ""+ (char)(num / 36 +'0') + (char)(num % 36 + '0');

}

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danjiewu 2008-03-07



string GetNextCode(string preCode)

{

     int num = (preCode[0]-'0') * 26 + preCode[1] - '0' + 1;

     return ""+ (char)(num / 26 +'0') + (char)(num % 26 + '0');

}

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