我一直在寻找一个简单的Java算法来生成伪随机的字母数字字符串。在我的情况下,它将被用作唯一的会话/键标识符,可能在500K+次生成中是唯一的(我的需求并不需要更复杂的东西)。
理想情况下,我希望能够根据我的唯一性需求指定长度。例如,长度为12的生成字符串可能看起来像"AEYGF7K0DM1X"
。
我一直在寻找一个简单的Java算法来生成伪随机的字母数字字符串。在我的情况下,它将被用作唯一的会话/键标识符,可能在500K+次生成中是唯一的(我的需求并不需要更复杂的东西)。
理想情况下,我希望能够根据我的唯一性需求指定长度。例如,长度为12的生成字符串可能看起来像"AEYGF7K0DM1X"
。
import java.util.*;
import javax.swing.*;
public class alphanumeric {
public static void main(String args[]) {
String nval, lenval;
int n, len;
nval = JOptionPane.showInputDialog("Enter number of codes you require: ");
n = Integer.parseInt(nval);
lenval = JOptionPane.showInputDialog("Enter code length you require: ");
len = Integer.parseInt(lenval);
find(n, len);
}
public static void find(int n, int length) {
String str1 = "0123456789ABCDEFGHIJKLMNOPQRSTUVWXYZ";
StringBuilder sb = new StringBuilder(length);
Random r = new Random();
System.out.println("\n\t Unique codes are \n\n");
for(int i=0; i<n; i++) {
for(int j=0; j<length; j++) {
sb.append(str1.charAt(r.nextInt(str1.length())));
}
System.out.println(" " + sb.toString());
sb.delete(0, length);
}
}
}
以下是由abacus-common提供的一行代码:
String.valueOf(CharStream.random('0', 'z').filter(c -> N.isLetterOrDigit(c)).limit(12).toArray())
随机并不意味着必须是唯一的。要获取唯一的字符串,请使用:
N.uuid() // E.g.: "e812e749-cf4c-4959-8ee1-57829a69a80f". length is 36.
N.guid() // E.g.: "0678ce04e18945559ba82ddeccaabfcd". length is 32 without '-'
private static final String NUMBERS = "0123456789";
private static final String UPPER_ALPHABETS = "ABCDEFGHIJKLMNOPQRSTUVWXYZ";
private static final String LOWER_ALPHABETS = "abcdefghijklmnopqrstuvwxyz";
private static final String SPECIALCHARACTERS = "@#$%&*";
private static final int MINLENGTHOFPASSWORD = 8;
public static String getRandomPassword() {
StringBuilder password = new StringBuilder();
int j = 0;
for (int i = 0; i < MINLENGTHOFPASSWORD; i++) {
password.append(getRandomPasswordCharacters(j));
j++;
if (j == 3) {
j = 0;
}
}
return password.toString();
}
private static String getRandomPasswordCharacters(int pos) {
Random randomNum = new Random();
StringBuilder randomChar = new StringBuilder();
switch (pos) {
case 0:
randomChar.append(NUMBERS.charAt(randomNum.nextInt(NUMBERS.length() - 1)));
break;
case 1:
randomChar.append(UPPER_ALPHABETS.charAt(randomNum.nextInt(UPPER_ALPHABETS.length() - 1)));
break;
case 2:
randomChar.append(SPECIALCHARACTERS.charAt(randomNum.nextInt(SPECIALCHARACTERS.length() - 1)));
break;
case 3:
randomChar.append(LOWER_ALPHABETS.charAt(randomNum.nextInt(LOWER_ALPHABETS.length() - 1)));
break;
}
return randomChar.toString();
}
Long.toString(Math.abs( UUID.randomUUID().getLeastSignificantBits(), 36));
这将产生一个长度最多为13个字符的字符串。我们使用Math.abs()确保没有负号混入。
random.nextLong()
?或者甚至是Double.doubleToLongBits(Math.random())
? - ericksonpublic static String randomSeriesForThreeCharacter() {
Random r = new Random();
String value = "";
char random_Char ;
for(int i=0; i<10; i++)
{
random_Char = (char) (48 + r.nextInt(74));
value = value + random_char;
}
return value;
}
以下是一个Scala解决方案:
(for (i <- 0 until rnd.nextInt(64)) yield {
('0' + rnd.nextInt(64)).asInstanceOf[Char]
}) mkString("")
import org.apache.commons.lang.RandomStringUtils;
RandomStringUtils.randomAlphanumeric(64);
public class Utils {
private final Random RANDOM = new SecureRandom();
private final String ALPHABET = "0123456789QWERTYUIOPASDFGHJKLZXCVBNMqwertyuiopasdfghjklzxcvbnm";
private String generateRandomString(int length) {
StringBuffer buffer = new StringBuffer(length);
for (int i = 0; i < length; i++) {
buffer.append(ALPHABET.charAt(RANDOM.nextInt(ALPHABET.length())));
}
return new String(buffer);
}
}
我认为这是最小的解决方案之一:
这个方案非常紧凑。 public String generateRandomString(int length) {
String randomString = "";
final char[] chars = "ABCDEFGHIJKLMNOPQRSTUVWXYZabcdefghijklmnopqrstuvwxyz01234567890".toCharArray();
final Random random = new Random();
for (int i = 0; i < length; i++) {
randomString = randomString + chars[random.nextInt(chars.length)];
}
return randomString;
}
这段代码运行良好。如果您使用此方法,建议使用超过10个字符。在5个字符/30362次迭代时发生碰撞。这需要9秒钟。
Long.toHexString(Double.doubleToLongBits(Math.random()));
生成一个16进制的随机字符串。UUID.randomUUID().toString();
生成一个随机唯一标识符。RandomStringUtils.randomAlphanumeric(12);
生成一个包含12个字符的随机字母数字组合。 - Manindar