具有给定长度的Java随机数[复制]

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How do I generate random integers within a specific range in Java? 57 answers
如何在Java的特定范围内生成随机整数?57个答案

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 ?

我需要在Java中使用精确的6位数的随机数字。我知道我可以在一个随机数上循环6次但是在标准的Java SE中有什么方法可以做到这一点吗?

EDIT: Follow up question: 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:接下来的问题:现在我可以生成我的6位数字了，我有了一个新问题，我想要创建的整个ID是语法123456-A1B45。那么我如何随机化最后5个可以是A-Z或0-9的字符?我考虑的是在48 - 90之间使用char值和随机数，然后简单地删除任何表示58-64的值。这是去的路还是有更好的解决方案?

EDIT 2: This is my final solution. Thanks for all the help guys!

编辑2:这是我的最终解决方案。谢谢大家的帮助!

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;
}

8 个解决方案

#1

Generate a number in the range from 100000 to 999999.

从100000到999999之间生成一个数字。

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

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

我很确定你已经读过这个文档了，你可以自己弄清楚。

#2

To generate a 6-digit number:

生成6位数字:

Use Random and nextInt as follows:

使用随机和nextInt如下:

Random rnd = new Random();
int n = 100000 + rnd.nextInt(900000);

Note that n will never be 7 digits (1000000) since nextInt(900000) can at most return 899999.

注意，自nextInt(900000)最多可以返回899999时，n永远不会是7位(1000000)。

So how do I randomize the last 5 chars that can be either A-Z or 0-9?

那么我如何随机化最后的5个字符可以是A-Z还是0-9?

Here's a simple solution:

这里有一个简单的解决方案:

// Generate random id, for example 283952-V8M32
char[] chars = "ABCDEFGHIJKLMNOPQRSTUVWXYZ0123456789".toCharArray();
Random rnd = new Random();
StringBuilder sb = new StringBuilder((100000 + rnd.nextInt(900000)) + "-");
for (int i = 0; i < 5; i++)
    sb.append(chars[rnd.nextInt(chars.length)]);

return sb.toString();

#3

If you need to specify the exact charactor length, we have to avoid values with 0 in-front.

如果您需要指定准确的角色长度，我们必须避免值为0的值。

Final String representation must have that exact character length.

最后的字符串表示必须具有精确的字符长度。

String GenerateRandomNumber(int charLength) {
        return String.valueOf(charLength < 1 ? 0 : new Random()
                .nextInt((9 * (int) Math.pow(10, charLength - 1)) - 1)
                + (int) Math.pow(10, charLength - 1));
    }

#4

try this:

试试这个:

public int getRandomNumber(int min, int max) {
    return (int) Math.floor(Math.random() * (max - min + 1)) + min;
}

#5

int rand = (new Random()).getNextInt(900000) + 100000;

EDIT: Fixed off-by-1 error and removed invalid solution.

编辑:修正错误并删除无效的解决方案。

#6

For the follow-up question, you can get a number between 36^5 and 36^6 and convert it in base 36

后续问题,你可以得到一个36 36 ^ ^ 5和6之间数量和转化基地36

UPDATED:

更新:

using this code

使用这个代码

http://javaconfessions.com/2008/09/convert-between-base-10-and-base-62-in_28.html

http://javaconfessions.com/2008/09/convert -基地- 10 -和-基础- 62 in_28.html

It's written BaseConverterUtil.toBase36(60466176+r.nextInt(2116316160))

这是写BaseConverterUtil.toBase36(60466176 + r.nextInt(2116316160))

but in your use case, it can be optimized by using a StringBuilder and having the number in the reverse order ie 71 should be converted in Z1 instead of 1Z

但是在您的用例中，它可以通过使用StringBuilder来优化，并且在相反的顺序ie 71应该转换为Z1而不是1Z。

EDITED:

编辑:

#7

-1

Would that work for you?

这对你有用吗?

public class Main {

public static void main(String[] args) {
    Random r = new Random(System.currentTimeMillis());
    System.out.println(r.nextInt(100000) * 0.000001);
}

}

result e.g. 0.019007

结果如0.019007

#8

-2

Generate a random number (which is always between 0-1) and multiply by 1000000

生成一个随机数(它总是在0-1之间)并乘以1000000。

Math.round(Math.random()*1000000);

#1