在c＃中返回一个8位字母数字递增的值 [英] Return a 8 digit alphanumeric incremented value in c#

问题描述

我想增加一个8位数的字母数字。

I want to increment a 8 digit alphanumeric.

我的8位数将是这样的东西

My 8 digit will some thing like this

前3个字符为ABC，其余5个数字应增加

first 3 characters will be ABC and rest 5 digits should increment

示例

ABC00001
ABC00002
ABC00003
.......
ABC00008
ABC00009
ABC0000A
ABC0000B
ABC0000C
ABC0000D
and soo on
........
........
ABC0000Y
ABC0000Z
ABC00010
ABC00012
and so on
.......

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PS.Shakeer Hussain

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PS.Shakeer Hussain

推荐答案

有趣的问题。你基本上想要在再次回绕之前从1到9再到A到Z计数。如果是ASCII字符集是一致然后我说从1开始并通过代码枚举，直到你到达Z，但你必须跳过非字母数字。

Interesting problem. You basically want to count from 1 to 9 and then A to Z before wrapping around again. If the ASCII char set was consistent then I'd say start at 1 and enumerate through the codes until you get to Z but you'd have to skip the non-alphanumerics.

你真正想要的是转换不幸的是，标准的ToString方法仅支持"正常"基数。但有人超过

SO 编写了一个快速函数来执行此操作。调用DecimalToArbitrarySystem（20,36）得到K，其中我相信是正确的。所以你可以使用常规计数器变量，但在将它连接到你的ABC值之前调用转换例程。

What you really want is to convert the number to base 36. Unfortunately the standard ToString method only supports the "normal" bases. However someone over on SO wrote a quick function to do it. Calling DecimalToArbitrarySystem(20, 36) yields K, which I believe is correct. So you could use a regular counter variable but call the conversion routine before concatenating it to your ABC value.

for (var index = 0; index < 40; ++index)
   Console.WriteLine(

" ABC {DecimalToArbitrarySystem（index，36）.PadLeft（5，'0'）}"）;

"ABC{DecimalToArbitrarySystem(index, 36).PadLeft(5, '0')}");

Michael Taylor 

http://www.michaeltaylorp3.net

Michael Taylor 
http://www.michaeltaylorp3.net

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