生成 9 位随机数，包括前导零 [英] Generation 9 digits random number including leading zero

问题描述

我想生成随机数，如果数字小于 9 位，它是 9 位包括前导零，比如 123 将是 000000123.我有以下代码不包括前导零:

I want to generate random number, which is 9 digits including leading zero if the number is less than 9 digits, say 123 will be 000000123. I have the following code which doesn't include leading zero :

Dim RandomClass As New Random()
Dim RandomNumber = RandomClass.Next(1, 999999999)

谢谢.

推荐答案

虽然我仍然很喜欢下面的单个数字"方法，但有一个更简单的方法 - 只需给出自定义数字格式:

While I still quite like my "individual digits" approach below, there's an easier way - just give a custom number format:

C#:

Random rng = new Random();
int number = rng.Next(1, 1000000000);
string digits = number.ToString("000000000");
Console.WriteLine(digits);

VB:

Dim rng As New Random
Dim number As Integer = rng.Next(1, 1000000000)
Dim digits As String = number.ToString("000000000") 
Console.WriteLine(digits)

正如评论中所指出的，D9 的格式字符串也可以完成这项工作:

As has been pointed out in the comments, a format string of D9 will also do the job:

Dim digits As String = number.ToString("D9") 

就我个人而言，我必须确切地查找它会做什么，而我对自定义数字格式感到满意 - 但这更多地说明了我而不是代码:)

Personally I'd have to look up exactly what that would do, whereas I'm comfortable with custom number formats - but that says more about me than about the code :)

我不会生成 1 到 999999999 之间的单个数字，而是生成 0 到 9 之间的 9 个数字.基本上，您生成的是 字符串 而不是数字(因为数字 000000000 和 0 是等价的，但你不想要第一个).

Rather than generating a single number between 1 and 999999999, I would just generate 9 numbers between 0 and 9. Basically you're generating a string rather than a number (as numerically 000000000 and 0 are equivalent, but you don't want the first).

因此在字符数组中生成 9 个字符 '0' 到 '9'，然后从中创建一个字符串.

So generate 9 characters '0' to '9' in a Character array, and then create a string from that.

这是一些示例 C# 代码:

Here's some sample C# code:

using System;

class Test
{
    static void Main(string[] args)
    {
        Random rng = new Random();
        string digits = GenerateDigits(rng, 9);
        Console.WriteLine(digits);
    }

    static string GenerateDigits(Random rng, int length)
    {
        char[] chars = new char[length];
        for (int i = 0; i < length; i++)
        {
            chars[i] = (char)(rng.Next(10) + '0');            
        }
        return new string(chars);
    }
}

...并将其转换为 VB:

... and converting it to VB:

Public Class Test

    Public Shared Sub Main()
        Dim rng As New Random
        Dim digits As String = Test.GenerateDigits(rng, 9)
        Console.WriteLine(digits)
    End Sub

    Private Shared Function GenerateDigits(ByVal rng As Random, _
                     ByVal length As Integer) As String
        Dim chArray As Char() = New Char(length  - 1) {}
        Dim i As Integer
        For i = 0 To length - 1
            chArray(i) = Convert.ToChar(rng.Next(10) + &H30)
        Next i
        Return New String(chArray)
    End Function

End Class

需要注意的一点:此代码可以生成000000000"，而您的原始代码的最小值为 1.您实际上希望最小值是多少?

One point to note: this code can generate "000000000" whereas your original code had a minimum value of 1. What do you actually want the minimum to be?

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