在 Java 中创建一个带有 AZ 和 0-9 的随机字符串
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Creating a random string with A-Z and 0-9 in Java
提问by cbt
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;
正如标题所暗示的，我需要创建一个随机的 17 个字符长的 ID。类似“ AJB53JHS232ERO0H1”的东西。字母和数字的顺序也是随机的。我想创建一个带有字母 AZ 的数组和一个随机到1-2. 并且在一个循环中；
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.
But I feel like there is an easier way of doing this. Is there?
但我觉得有一种更简单的方法可以做到这一点。在那儿？
采纳答案by Suresh Atta
Here you can use my method for generating Random String
在这里你可以使用我的方法来生成随机字符串
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;
}
The above method from my bag using to generate a salt string for login purpose.
我包中的上述方法用于生成用于登录目的的盐字符串。
回答by cbt
RandomStringUtilsfrom Apache commons-lang might help:
RandomStringUtils来自 Apache commons-lang 可能有帮助：
RandomStringUtils.randomAlphanumeric(17).toUpperCase()
2017 update: RandomStringUtilshas been deprecated, you should now use RandomStringGenerator.
2017 更新：RandomStringUtils已弃用，您现在应该使用RandomStringGenerator。
回答by Masudul
You can easily do that with a for loop,
您可以使用 for 循环轻松做到这一点，
public static void main(String[] args) {
String aToZ="ABCD.....1234"; // 36 letter.
String randomStr=generateRandom(aToZ);
}
private static String generateRandom(String aToZ) {
Random rand=new Random();
StringBuilder res=new StringBuilder();
for (int i = 0; i < 17; i++) {
int randIndex=rand.nextInt(aToZ.length());
res.append(aToZ.charAt(randIndex));
}
return res.toString();
}
回答by MouseLearnJava
Three steps to implement your function:
实现功能的三个步骤：
Step#1You can specify a string, including the chars A-Z and 0-9.
步骤#1您可以指定一个字符串，包括字符 AZ 和 0-9。
Like.
喜欢。
String candidateChars = "ABCDEFGHIJKLMNOPQRSTUVWXYZ1234567890";
Step#2Then if you would like to generate a random char from this candidate string. You can use
步骤#2然后，如果您想从这个候选字符串生成一个随机字符。您可以使用
candidateChars.charAt(random.nextInt(candidateChars.length()));
Step#3At last, specify the length of random string to be generated (in your description, it is 17). Writer a for-loop and append the random chars generated in step#2 to StringBuilder object.
Step#3最后，指定要生成的随机字符串的长度（在您的描述中，它是17）。编写一个 for 循环并将步骤#2 中生成的随机字符附加到 StringBuilder 对象。
Based on this, here is an example public class RandomTest {
基于此，这里有一个示例 public class RandomTest {
public static void main(String[] args) {
System.out.println(generateRandomChars(
"ABCDEFGHIJKLMNOPQRSTUVWXYZ1234567890", 17));
}
/**
*
* @param candidateChars
* the candidate chars
* @param length
* the number of random chars to be generated
*
* @return
*/
public static String generateRandomChars(String candidateChars, int length) {
StringBuilder sb = new StringBuilder();
Random random = new Random();
for (int i = 0; i < length; i++) {
sb.append(candidateChars.charAt(random.nextInt(candidateChars
.length())));
}
return sb.toString();
}
}
