我如何在 C# 中打印像 AAA001 到 zzz999 这样的字母

Question

我想使用 c#

在控制台中打印以下系列

AAA000
AAA001
AAA002
.
.
.
AAA999
AAB000
AAB001
.
.
.UPTO
ZZZ999

亲切的问候。

Answer 1

一个简单的方法是：

for (int i1 = 'A'; i1 <= 'Z'; i1++)
{
    for (int i2 = 'A'; i2 <= 'Z'; i2++)
    {
        for (int i3 = 'A'; i3 <= 'Z'; i3++)
        {
            for (int i4 = '0'; i4 <= '9'; i4++)
            {
                for (int i5 = '0'; i5 <= '9'; i5++)
                {
                    for (int i6 = '0'; i6 <= '9'; i6++)
                    {
                        Console.WriteLine(new string(new Char[] { (Char)i1, (Char)i2, (Char)i3, (Char)i4, (Char)i5, (Char)i6 }));
                    }
                }
            }
        }
    }
}

重点在'A'到'Z'和'0'到'9'的顺序和int与Char的转换

Answer 2

Reza Aghaei 很好，我通过减少两个循环来缩短代码，如下所示：

for (int i1 = 'A'; i1 <= 'Z'; i1++)
    {
      for (int i2 = 'A'; i2 <= 'Z'; i2++)
        {
           for (int i3 = 'A'; i3 <= 'Z'; i3++)
              {
                 for (int i4 = 0; i4 <= 999; i4++)
                   { Console.WriteLine(new string(new Char[] { (Char)i1, (Char)i2, (Char)i3 }) + i4.ToString("000"));}
               }
         }
    }

{} 用于提高可读性，这也不是必需的：只需使用以下内容即可获得预期的输出：

for (int i1 = 'A'; i1 <= 'Z'; i1++)
for (int i2 = 'A'; i2 <= 'Z'; i2++)
for (int i3 = 'A'; i3 <= 'Z'; i3++)
for (int i4 = 0; i4 <= 999; i4++)
Console.WriteLine(new string(new Char[] { (Char)i1, (Char)i2, (Char)i3 }) + i4.ToString("000"));

Answer 3

var letters = Enumerable.Range('A', 'Z' - 'A' + 1).Select(i => (char)i);
var numbers = Enumerable.Range(0, 10);
var series =
    from l1 in letters
    from l2 in letters
    from l3 in letters
    from n1 in numbers
    from n2 in numbers
    from n3 in numbers
    select l1.ToString() + l2 + l3 + n1 + n2 + n3;

foreach (string s in series)
    Console.WriteLine(s);

我如何在 C# 中打印像 AAA001 到 zzz999 这样的字母

how do i print letters like AAA001 to zzz999 in C#

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