在 Java 中创建一个带有 A-Z 和 0-9 的随机字符串 [英] Creating a random string with A-Z and 0-9 in Java

问题描述

正如标题所暗示的，我需要创建一个随机的、长度为 17 个字符的 ID.类似于AJB53JHS232ERO0H1".字母和数字的顺序也是随机的.我想创建一个包含字母 A-Z 的数组和一个随机到 1-2 的检查"变量.并且在一个循环中；

As the title suggest I need to create a random, 17 characters long, ID. Something like "AJB53JHS232ERO0H1". The order of letters and numbers is also random. I thought of creating an array with letters A-Z and a 'check' variable that randoms to 1-2. And in a loop;

Randomize 'check' to 1-2.
If (check == 1) then the character is a letter.
Pick a random index from the letters array.
else
Pick a random number.

但我觉得有一种更简单的方法可以做到这一点.有吗?

But I feel like there is an easier way of doing this. Is there?

推荐答案

这里可以使用我的方法生成随机字符串

Here you can use my method for generating Random String

protected String getSaltString() {
        String SALTCHARS = "ABCDEFGHIJKLMNOPQRSTUVWXYZ1234567890";
        StringBuilder salt = new StringBuilder();
        Random rnd = new Random();
        while (salt.length() < 18) { // length of the random string.
            int index = (int) (rnd.nextFloat() * SALTCHARS.length());
            salt.append(SALTCHARS.charAt(index));
        }
        String saltStr = salt.toString();
        return saltStr;

    }

我包中的上述方法用于生成用于登录目的的盐字符串.

The above method from my bag using to generate a salt string for login purpose.

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