具有给定长度的Java随机数 [英] Java random number with given length

问题描述

我需要在Java中使用正好6位的随机数进行生成。我知道我可以在随机数上循环6次但在标准Java SE中有没有其他方法可以做到这一点？

I need to genarate a random number with exactly 6 digits in Java. I know i could loop 6 times over a randomicer but is there a nother way to do this in the standard Java SE ?

编辑 - 跟进问题：

现在我可以生成我的6位数字，我遇到了一个新问题，我正在尝试创建的整个ID的语法为123456-A1B45。那么我如何随机收集最后5个可能是A-Z或0-9的字符？我正在考虑使用char值和randomice一个介于48 - 90之间的数字，并简单地删除任何得到代表58-64的数字的值。这是要走的路还是有更好的解决方案？

Now that I can generate my 6 digits i got a new problem, the whole ID I'm trying to create is of the syntax 123456-A1B45. So how do i randomice the last 5 chars that can be either A-Z or 0-9? I'm thinking of using the char value and randomice a number between 48 - 90 and simply drop any value that gets the numbers that represent 58-64. Is this the way to go or is there a better solution?

编辑2：

这是我的最终解决方案。感谢所有帮助人员！

This is my final solution. Thanks for all the help guys!

protected String createRandomRegistryId(String handleId)
{
    // syntax we would like to generate is DIA123456-A1B34      
    String val = "DI";      

    // char (1), random A-Z
    int ranChar = 65 + (new Random()).nextInt(90-65);
    char ch = (char)ranChar;        
    val += ch;      

    // numbers (6), random 0-9
    Random r = new Random();
    int numbers = 100000 + (int)(r.nextFloat() * 899900);
    val += String.valueOf(numbers);

    val += "-";
    // char or numbers (5), random 0-9 A-Z
    for(int i = 0; i<6;){
        int ranAny = 48 + (new Random()).nextInt(90-65);

        if(!(57 < ranAny && ranAny<= 65)){
        char c = (char)ranAny;      
        val += c;
        i++;
        }

    }

    return val;
}

推荐答案

生成范围内的数字从 100000 到 999999 。

Generate a number in the range from 100000 to 999999.

// pseudo code
int n = 100000 + random_float() * 900000;

我很确定你已经阅读过例如随机，可以自己弄清楚其余部分。

I’m pretty sure you have already read the documentation for e.g. Random and can figure out the rest yourself.

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