我一直在寻找一种简单的Java算法来生成伪随机字母数字字符串。在我的情况下,它将被用作唯一的会话/密钥标识符,“很可能”在超过50万代的时间内是唯一的(我的需求实际上不需要更复杂的东西)。
理想情况下,我可以根据我的独特性需求指定长度。例如,生成的长度为12的字符串可能看起来像“AEYGF7K0DM1X”。
我一直在寻找一种简单的Java算法来生成伪随机字母数字字符串。在我的情况下,它将被用作唯一的会话/密钥标识符,“很可能”在超过50万代的时间内是唯一的(我的需求实际上不需要更复杂的东西)。
理想情况下,我可以根据我的独特性需求指定长度。例如,生成的长度为12的字符串可能看起来像“AEYGF7K0DM1X”。
当前回答
我使用的是一个非常简单的Java8解决方案。只需根据您的需求进行定制。
...
import java.security.SecureRandom;
...
//Generate a random String of length between 10 to 20.
//Length is also randomly generated here.
SecureRandom random = new SecureRandom();
String sampleSet = "ABCDEFGHIJKLMNOPQRSTUVWXYZabcdefghijklmnopqrstuvwxyz0123456789_";
int stringLength = random.ints(1, 10, 21).mapToObj(x -> x).reduce((a, b) -> a).get();
String randomString = random.ints(stringLength, 0, sampleSet.length() - 1)
.mapToObj(x -> sampleSet.charAt(x))
.collect(Collector
.of(StringBuilder::new, StringBuilder::append,
StringBuilder::append, StringBuilder::toString));
我们可以使用它生成如下的字母数字随机字符串(返回的字符串将强制包含一些非数字字符以及一些数字字符):
public String generateRandomString() {
String sampleSet = "ABCDEFGHIJKLMNOPQRSTUVWXYZabcdefghijklmnopqrstuvwxyz_";
String sampleSetNumeric = "0123456789";
String randomString = getRandomString(sampleSet, 10, 21);
String randomStringNumeric = getRandomString(sampleSetNumeric, 10, 21);
randomString = randomString + randomStringNumeric;
//Convert String to List<Character>
List<Character> list = randomString.chars()
.mapToObj(x -> (char)x)
.collect(Collectors.toList());
Collections.shuffle(list);
//This is needed to force a non-numeric character as the first String
//Skip this for() if you don't need this logic
for(;;) {
if(Character.isDigit(list.get(0))) Collections.shuffle(list);
else break;
}
//Convert List<Character> to String
randomString = list.stream()
.map(String::valueOf)
.collect(Collectors.joining());
return randomString;
}
//Generate a random number between the lower bound (inclusive) and upper bound (exclusive)
private int getRandomLength(int min, int max) {
SecureRandom random = new SecureRandom();
return random.ints(1, min, max).mapToObj(x -> x).reduce((a, b) -> a).get();
}
//Generate a random String from the given sample string, having a random length between the lower bound (inclusive) and upper bound (exclusive)
private String getRandomString(String sampleSet, int min, int max) {
SecureRandom random = new SecureRandom();
return random.ints(getRandomLength(min, max), 0, sampleSet.length() - 1)
.mapToObj(x -> sampleSet.charAt(x))
.collect(Collector
.of(StringBuilder::new, StringBuilder::append,
StringBuilder::append, StringBuilder::toString));
}
其他回答
在一行中:
Long.toHexString(Double.doubleToLongBits(Math.random()));
来源:Java-生成随机字符串
您可以为此使用Apache Commons库RandomStringUtils:
RandomStringUtils.randomAlphanumeric(20).toUpperCase();
Java提供了一种直接实现这一点的方法。如果你不想要破折号,它们很容易去掉。只需使用uuid.replace(“-”,“”)
import java.util.UUID;
public class randomStringGenerator {
public static void main(String[] args) {
System.out.println(generateString());
}
public static String generateString() {
String uuid = UUID.randomUUID().toString();
return "uuid = " + uuid;
}
}
输出
uuid = 2d7428a6-b58c-4008-8575-f05549f16316
public 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;
}
我使用的是一个非常简单的Java8解决方案。只需根据您的需求进行定制。
...
import java.security.SecureRandom;
...
//Generate a random String of length between 10 to 20.
//Length is also randomly generated here.
SecureRandom random = new SecureRandom();
String sampleSet = "ABCDEFGHIJKLMNOPQRSTUVWXYZabcdefghijklmnopqrstuvwxyz0123456789_";
int stringLength = random.ints(1, 10, 21).mapToObj(x -> x).reduce((a, b) -> a).get();
String randomString = random.ints(stringLength, 0, sampleSet.length() - 1)
.mapToObj(x -> sampleSet.charAt(x))
.collect(Collector
.of(StringBuilder::new, StringBuilder::append,
StringBuilder::append, StringBuilder::toString));
我们可以使用它生成如下的字母数字随机字符串(返回的字符串将强制包含一些非数字字符以及一些数字字符):
public String generateRandomString() {
String sampleSet = "ABCDEFGHIJKLMNOPQRSTUVWXYZabcdefghijklmnopqrstuvwxyz_";
String sampleSetNumeric = "0123456789";
String randomString = getRandomString(sampleSet, 10, 21);
String randomStringNumeric = getRandomString(sampleSetNumeric, 10, 21);
randomString = randomString + randomStringNumeric;
//Convert String to List<Character>
List<Character> list = randomString.chars()
.mapToObj(x -> (char)x)
.collect(Collectors.toList());
Collections.shuffle(list);
//This is needed to force a non-numeric character as the first String
//Skip this for() if you don't need this logic
for(;;) {
if(Character.isDigit(list.get(0))) Collections.shuffle(list);
else break;
}
//Convert List<Character> to String
randomString = list.stream()
.map(String::valueOf)
.collect(Collectors.joining());
return randomString;
}
//Generate a random number between the lower bound (inclusive) and upper bound (exclusive)
private int getRandomLength(int min, int max) {
SecureRandom random = new SecureRandom();
return random.ints(1, min, max).mapToObj(x -> x).reduce((a, b) -> a).get();
}
//Generate a random String from the given sample string, having a random length between the lower bound (inclusive) and upper bound (exclusive)
private String getRandomString(String sampleSet, int min, int max) {
SecureRandom random = new SecureRandom();
return random.ints(getRandomLength(min, max), 0, sampleSet.length() - 1)
.mapToObj(x -> sampleSet.charAt(x))
.collect(Collector
.of(StringBuilder::new, StringBuilder::append,
StringBuilder::append, StringBuilder::toString));
}