java随机字符串
❶ java如何实现随机输出一个字符串，由[a-zA-Z0-9]组成 长度为4,并以数字结尾
如图所示即可。
package test;
import java.util.Random;
public class Test{
public static void main(String[] args) {
String str = getRandomString(4);
System.out.println(str);
}
public static String getRandomString(int length) {
// 定义一个字符串（A-Z，a-z，0-9）即62位；
String str = "";
// 由Random生成随机数
Random random = new Random();
StringBuffer sb = new StringBuffer();
// 长度为几就循环几减一次
for (int i = 0; i < length - 1; ++i) {
// 产生0-61的数字
int number = random.nextInt(62);
// 将产生的数字通过length次承载到sb中
sb.append(str.charAt(number));
}
//补全最后一个数字位
sb.append(random.nextInt(9));
// 将承载的字符转换成字符串
return sb.toString();
}
}
❷ 在java语言中如何随机地生成一个字符串
可以配合UUID或者GUID来实现
GUID是一个128位长的数字，一般用16进制表示。算法的核心思想是结合机器的网卡、当地时间、一个随机数来生成GUID。从理论上讲，如果一台机器每秒产生10000000个GUID，则可以保证（概率意义上）3240年不重复。
UUID是1.5中新增的一个类，在java.util下，用它可以产生一个号称全球唯一的ID
import java.util.UUID;
public class Test {
public static void main(String[] args) {
UUID uuid = UUID.randomUUID();
System.out.println (uuid);
}
}
编译运行输出：
07ca3dec-b674-41d0-af9e-9c37583b08bb
两种方式生成guid 与uuid
需要comm log 库
/**
* @author Administrator
*
* TODO To change the template for this generated type comment go to
* Window - Preferences - Java - Code Style - Code Templates
*/
import java.net.InetAddress;
import java.net.UnknownHostException;
import java.security.MessageDigest;
import java.security.NoSuchAlgorithmException;
import java.security.SecureRandom;
import java.util.Random;
public class RandomGUID extends Object {
protected final org.apache.commons.logging.Log logger = org.apache.commons.logging.LogFactory
.getLog(getClass());
public String valueBeforeMD5 = "";
public String valueAfterMD5 = "";
private static Random myRand;
private static SecureRandom mySecureRand;
private static String s_id;
private static final int PAD_BELOW = 0x10;
private static final int TWO_BYTES = 0xFF;
/*
* Static block to take care of one time secureRandom seed.
* It takes a few seconds to initialize SecureRandom. You might
* want to consider removing this static block or replacing
* it with a "time since first loaded" seed to rece this time.
* This block will run only once per JVM instance.
*/
static {
mySecureRand = new SecureRandom();
long secureInitializer = mySecureRand.nextLong();
myRand = new Random(secureInitializer);
try {
s_id = InetAddress.getLocalHost().toString();
} catch (UnknownHostException e) {
e.printStackTrace();
}
}
/*
* Default constructor. With no specification of security option,
* this constructor defaults to lower security, high performance.
*/
public RandomGUID() {
getRandomGUID(false);
}
/*
* Constructor with security option. Setting secure true
* enables each random number generated to be cryptographically
* strong. Secure false defaults to the standard Random function seeded
* with a single cryptographically strong random number.
*/
public RandomGUID(boolean secure) {
getRandomGUID(secure);
}
/*
* Method to generate the random GUID
*/
private void getRandomGUID(boolean secure) {
MessageDigest md5 = null;
StringBuffer sbValueBeforeMD5 = new StringBuffer(128);
try {
md5 = MessageDigest.getInstance("MD5");
} catch (NoSuchAlgorithmException e) {
logger.error("Error: " + e);
}
try {
long time = System.currentTimeMillis();
long rand = 0;
if (secure) {
rand = mySecureRand.nextLong();
} else {
rand = myRand.nextLong();
}
sbValueBeforeMD5.append(s_id);
sbValueBeforeMD5.append(":");
sbValueBeforeMD5.append(Long.toString(time));
sbValueBeforeMD5.append(":");
sbValueBeforeMD5.append(Long.toString(rand));
valueBeforeMD5 = sbValueBeforeMD5.toString();
md5.update(valueBeforeMD5.getBytes());
byte[] array = md5.digest();
StringBuffer sb = new StringBuffer(32);
for (int j = 0; j < array.length; ++j) {
int b = array[j] & TWO_BYTES;
if (b < PAD_BELOW)
sb.append('0');
sb.append(Integer.toHexString(b));
}
valueAfterMD5 = sb.toString();
} catch (Exception e) {
logger.error("Error:" + e);
}
}
/*
* Convert to the standard format for GUID
* (Useful for SQL Server UniqueIdentifiers, etc.)
* Example: C2FEEEAC-CFCD-11D1-8B05-00600806D9B6
*/
public String toString() {
String raw = valueAfterMD5.toUpperCase();
StringBuffer sb = new StringBuffer(64);
sb.append(raw.substring(0, 8));
sb.append("-");
sb.append(raw.substring(8, 12));
sb.append("-");
sb.append(raw.substring(12, 16));
sb.append("-");
sb.append(raw.substring(16, 20));
sb.append("-");
sb.append(raw.substring(20));
return sb.toString();
}
// Demonstraton and self test of class
public static void main(String args[]) {
for (int i=0; i< 100; i++) {
RandomGUID myGUID = new RandomGUID();
System.out.println("Seeding String=" + myGUID.valueBeforeMD5);
System.out.println("rawGUID=" + myGUID.valueAfterMD5);
System.out.println("RandomGUID=" + myGUID.toString());
}
}
}
❸ java中，我想随机获取不重复的字符串应该怎么办
1.先确定字符串的长度；
2.再确定字符串包括哪些字符；//这样可以确定字符的ASCII范围
3.然后就有很多方法去随机获取字符：
把字符存放在一个Map中的value中，如：map.put(1,"字符1");map.put(2,"字符2");
map.put(3,"字符3");一次类推，通过Math.random()*字符的个数，取整就可以得到key值，
map.get(key)，value不为null,然后在map中删除这个key-value，可以设置map.put(key,null);
循环到字符串的长度为需要的长度。
具体一种方法的代码：
import java.util.HashMap;
import java.util.Map;
public class RandomStr {
public static void main(String[] args) {
for(int i=0; i< 10; i++){//测试10次产生随机不重复字符串
generateRandomStr(5);
}
}
/**
* 产生不重复的随机字符串
* @param len 生成字符串的长度
* @return
*/
public static String generateRandomStr(int len){
String strRange = "";//字符串范围，根据自己的需求确定
Map<String, String> tmp = new HashMap<String, String>();
for(int i=0; i<strRange.length(); i++){
tmp.put(i+"",strRange.charAt(i)+"");
}
StringBuffer result=new StringBuffer();
for(int i=0; i<len; i++){
String key = (int)(Math.random()*(strRange.length()-i))+"";
result.append(tmp.get(key));
tmp.remove(key);
tmp.put(key, tmp.get((strRange.length()-i-1)+""));//拿最后一个字符填充删除的位置key-value
}
System.out.println(result);
return result.toString();
}
}
❹ Java如何生成随机字符串
import java.util.Random;
public class Stat {
public static void main(String[] args) {
// Math.random() 获取0～1之间的double类型数值
int num = (int)(Math.random()*100); // 获取0～100之间的整数
System.out.println(num);
System.out.println(getRandomString(12));
}
// 获取随机字符串
public static String getRandomString(int length) { // length 字符串长度
StringBuffer buffer = new StringBuffer("");
StringBuffer sb = new StringBuffer();
Random r = new Random();
int range = buffer.length();
for (int i = 0; i < length; i ++) {
sb.append(buffer.charAt(r.nextInt(range)));
}
return sb.toString();
}
}
❺ java随机生成字符串
代码如下：
import java.util.Random;
public class Demo {
public static void main(String args[]){
Demo demo = new Demo();
System.out.println(demo.getRandomString(20));
}
public static String getRandomString(int length) {
StringBuffer buffer = new StringBuffer("");
StringBuffer sb = new StringBuffer();
Random random = new Random();
int range = buffer.length();
for (int i = 0; i < length; i ++) {
sb.append(buffer.charAt(random.nextInt(range)));
}
return sb.toString();
}
}
代码解读：
调用getRandomString（）方法时的length参数为要返回的随机字符串的长度。
StringBuffer buffer = new StringBuffer("");此段字符串代表要出现的字符还可以是其他字符
如：- + = * 。.等等
❻ java实现输出指定长度的随机字符（大小写字母和数字）遇到问题
程序输出错误的原因(见图)
❼ java中如何产生一个16位数字组成的随机字符串谢谢各位了
方法如下：
Random ran=new random();
boolean flag=true;
while(flag){
int a=ran.nextInt(99999999);
int b=ran.nextInt(99999999);
long c=a*100000000L+b;
if(c>1000000000000000L&&c<9999999999999999L){
flag=false;
c1=c;
String num=String.valueOf(c1);
}
｝
❽ java怎么随机生成一个字符串
/**
* 产生随机字符串
* */
private static Random randGen = null;
private static char[] numbersAndLetters = null;
public static final String randomString(int length) {
if (length < 1) {
return null;
}
if (randGen == null) {
randGen = new Random();
numbersAndLetters = ("" +
"").toCharArray();
//numbersAndLetters = ("").toCharArray();
}
char [] randBuffer = new char[length];
for (int i=0; i<randBuffer.length; i++) {
randBuffer[i] = numbersAndLetters[randGen.nextInt(71)];
//randBuffer[i] = numbersAndLetters[randGen.nextInt(35)];
}
return new String(randBuffer);
}
//调用此方法randomString(int),int是字符串的长度，即可产生指定长度的随机字符串。
❾ java如何定义一个以字母开头的随机字符串
需要创建生成器的实例，我们可以使用RandomStringGenerator.Builder()类build()方法。生成器类还可以帮助我们配置生成器的属性。在调用build()方法之前，我们可以使用以下方法设置构建器的属性:withinRange()指定在生成的字符串中允许的最小和最大代码点。filteredBy()将生成的字符串中的字符限制为与至少提供的谓词之一匹配的字符。这样即可将java定义一个以字母开头的随机字符串。