blocks|key|1031376|text|生成一个随机的字符串很容易-只需使用java.util.Random和一个包含您想要的所有字符的字符串，例如|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|1031377|public+static+String+generateString(Random+rng,+String+characters,+int+length)
{
++++char[]+text+=+new+char[length];
++++for+(int+i+=+0;+i+<+length;+i%2B%2B)
++++{
++++++++text[i]+=+characters.charAt(rng.nextInt(characters.length()));
++++}
++++return+new+String(text);
}|code-block|syntax|javascript|1031378|现在，为了保持唯一性，您需要将生成的字符串存储在某个地方。如何做到这一点将真正取决于您的应用程序的其余部分。|1031379|entityMap^0|I|G|0|0|0^^$0|@$1|2|3|4|5|6|7|O|8|@$9|P|A|Q|B|C]]|D|@]|E|$]]|$1|F|3|G|5|H|7|R|8|@]|D|@]|E|$I|J]]|$1|K|3|L|5|6|7|S|8|@]|D|@]|E|$]]|$1|M|3|-4|5|6|7|T|8|@]|D|@]|E|$]]]|N|$]]

Generating a random string of characters is easy - just use <code>java.util.Random</code> and a string containing all the characters you want to be available, e.g.

<pre><code>public static String generateString(Random rng, String characters, int length)
{
 char[] text = new char[length];
 for (int i = 0; i &lt; length; i++)
 {
 text[i] = characters.charAt(rng.nextInt(characters.length()));
 }
 return new String(text);
}
</code></pre>

Now, for uniqueness you'll need to store the generated strings somewhere. How you do that will really depend on the rest of your application.

blocks|key|1031455|text|这非常好用：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1031456|http://commons.apache.org/proper/commons-lang/apidocs/org/apache/commons/lang3/RandomStringUtils.html+-类似于RandomStringUtils.randomNumeric(7)。|offset|length|style|CODE|1031457|有10%5E7个等概率(如果java.util.Random未损坏)不同的值，因此唯一性可能是个问题。|1031458|entityMap|0|LINK|mutability|MUTABLE|url|http://commons.apache.org/proper/commons-lang/apidocs/org/apache/commons/lang3/RandomStringUtils.html^0|0|2Y|Y|0|2T|0|0|C|G|0^^$0|@$1|2|3|4|5|6|7|R|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|S|8|@$D|T|E|U|F|G]]|9|@$D|V|E|W|1|X]]|A|$]]|$1|H|3|I|5|6|7|Y|8|@$D|Z|E|10|F|G]]|9|@]|A|$]]|$1|J|3|-4|5|6|7|11|8|@]|9|@]|A|$]]]|K|$L|$5|M|N|O|A|$P|Q]]]]

This is very nice: 

<a href="http://commons.apache.org/proper/commons-lang/apidocs/org/apache/commons/lang3/RandomStringUtils.html" rel="nofollow noreferrer">http://commons.apache.org/proper/commons-lang/apidocs/org/apache/commons/lang3/RandomStringUtils.html</a> - something like <code>RandomStringUtils.randomNumeric(7)</code>. 

There are 10^7 equiprobable (if <code>java.util.Random</code> is not broken) distinct values so uniqueness may be a concern.

blocks|key|1246546|text|您也可以使用java.util包中的UUID类，它返回32位字符串的随机uuid。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1246547|java.util.UUID.randomUUID().toString()|offset|length|style|CODE|1246548|http://java.sun.com/j2se/1.5.0/docs/api/java/util/UUID.html|1246549|entityMap|0|LINK|mutability|MUTABLE|url^0|0|0|12|0|0|1N|0|0^^$0|@$1|2|3|4|5|6|7|Q|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|R|8|@$D|S|E|T|F|G]]|9|@]|A|$]]|$1|H|3|I|5|6|7|U|8|@]|9|@$D|V|E|W|1|X]]|A|$]]|$1|J|3|-4|5|6|7|Y|8|@]|9|@]|A|$]]]|K|$L|$5|M|N|O|A|$P|I]]]]

You can also use UUID class from java.util package, which returns random uuid of 32bit characters String. 

<code>java.util.UUID.randomUUID().toString()</code>

<a href="http://java.sun.com/j2se/1.5.0/docs/api/java/util/UUID.html" rel="nofollow noreferrer">http://java.sun.com/j2se/1.5.0/docs/api/java/util/UUID.html</a>

blocks|key|1246609|text|Random+ran+=+new+Random();
int+top+=+3;
char+data+=+'+';
String+dat+=+"";

for+(int+i=0;+i<=top;+i%2B%2B)+{
++data+=+(char)(ran.nextInt(25)%2B97);
++dat+=+data+%2B+dat;
}

System.out.println(dat);|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|1246610|unstyled|entityMap^0|0^^$0|@$1|2|3|4|5|6|7|G|8|@]|9|@]|A|$B|C]]|$1|D|3|-4|5|E|7|H|8|@]|9|@]|A|$]]]|F|$]]

<pre><code>Random ran = new Random();
int top = 3;
char data = ' ';
String dat = "";

for (int i=0; i&lt;=top; i++) {
 data = (char)(ran.nextInt(25)+97);
 dat = data + dat;
}

System.out.println(dat);
</code></pre>

blocks|key|1246594|text|我认为下面的类代码会对你有所帮助。它支持多线程，但你可以做一些改进，比如删除同步块和同步到getRandomId()方法。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1246595|public+class+RandomNumberGenerator+{

private+static+final+Set<String>+generatedNumbers+=+new+HashSet<String>();

public+RandomNumberGenerator()+{
}

public+static+void+main(String[]+args)+{
++++final+int+maxLength+=+7;
++++final+int+maxTry+=+10;

++++for+(int+i+=+0;+i+<+10;+i%2B%2B)+{
++++++++System.out.println(i+%2B+".+studentId="+%2B+RandomNumberGenerator.getRandomId(maxLength,+maxTry));
++++}
}

public+static+String+getRandomId(final+int+maxLength,+final+int+maxTry)+{
++++final+Random+random+=+new+Random(System.nanoTime());
++++final+int+max+=+(int)+Math.pow(10,+maxLength);
++++final+int+maxMin+=+(int)+Math.pow(10,+maxLength-1);
++++int+i+=+0;
++++boolean+unique+=+false;
++++int+randomId+=+-1;
++++while+(i+<+maxTry)+{
++++++++randomId+=+random.nextInt(max+-+maxMin+-+1)+%2B+maxMin;

++++++++synchronized+(generatedNumbers)+{
++++++++++++if+(generatedNumbers.contains(randomId)+==+false)+{
++++++++++++++++unique+=+true;
++++++++++++++++break;
++++++++++++}
++++++++}
++++++++i%2B%2B;
++++}
++++if+(unique+==+false)+{
++++++++throw+new+RuntimeException("Cannot+generate+unique+id!");
++++}

++++synchronized+(generatedNumbers)+{
++++++++generatedNumbers.add(String.valueOf(randomId));
++++}

++++return+String.valueOf(randomId);
}

}|code-block|syntax|javascript|1246596|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|J|8|@]|9|@]|A|$E|F]]|$1|G|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|H|$]]

I think the following class code will help you. It supports multithreading but you can do some improvement like remove sync block and and sync to getRandomId() method.

<pre><code>public class RandomNumberGenerator {

private static final Set&lt;String&gt; generatedNumbers = new HashSet&lt;String&gt;();

public RandomNumberGenerator() {
}

public static void main(String[] args) {
 final int maxLength = 7;
 final int maxTry = 10;

 for (int i = 0; i &lt; 10; i++) {
 System.out.println(i + ". studentId=" + RandomNumberGenerator.getRandomId(maxLength, maxTry));
 }
}

public static String getRandomId(final int maxLength, final int maxTry) {
 final Random random = new Random(System.nanoTime());
 final int max = (int) Math.pow(10, maxLength);
 final int maxMin = (int) Math.pow(10, maxLength-1);
 int i = 0;
 boolean unique = false;
 int randomId = -1;
 while (i &lt; maxTry) {
 randomId = random.nextInt(max - maxMin - 1) + maxMin;

 synchronized (generatedNumbers) {
 if (generatedNumbers.contains(randomId) == false) {
 unique = true;
 break;
 }
 }
 i++;
 }
 if (unique == false) {
 throw new RuntimeException("Cannot generate unique id!");
 }

 synchronized (generatedNumbers) {
 generatedNumbers.add(String.valueOf(randomId));
 }

 return String.valueOf(randomId);
}

}
</code></pre>

blocks|key|1031422|text|您需要问的第一个问题是，您是否真的需要ID是随机的。有时，连续ID就足够好了。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1031423|现在，如果你确实需要它是随机的，我们首先注意到一个生成的不包含重复的数字序列不能被称为随机。:p现在我们解决了这个问题，最快的方法是创建一个包含所有已经生成的is的Hashtable或HashMap。每当生成新的ID时，根据哈希表检查它，如果该ID已经出现，则重新生成。如果学生的数量远远少于ID的范围，这通常会很好地工作。否则，当需要重新生成ID的概率增加时，您的麻烦就更大了，P(生成新ID)+=+number_of_id_already_generated+/+number_of_all_possible_ids。在这种情况下，检查第一段(您是否需要ID是随机的？)。|offset|length|style|CODE|1031424|希望这能有所帮助。|1031425|entityMap^0|0|2A|9|2K|7|0|0^^$0|@$1|2|3|4|5|6|7|L|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|M|8|@$D|N|E|O|F|G]|$D|P|E|Q|F|G]]|9|@]|A|$]]|$1|H|3|I|5|6|7|R|8|@]|9|@]|A|$]]|$1|J|3|-4|5|6|7|S|8|@]|9|@]|A|$]]]|K|$]]

The first question you need to ask is whether you really need the ID to be random. Sometime, sequential IDs are good enough.

Now, if you do need it to be random, we first note a generated sequence of numbers that contain no duplicates can not be called random. :p Now that we get that out of the way, the fastest way to do this is to have a <code>Hashtable</code> or <code>HashMap</code> containing all the IDs already generated. Whenever a new ID is generated, check it against the hashtable, re-generate if the ID already occurs. This will generally work well if the number of students is much less than the range of the IDs. If not, you're in deeper trouble as the probability of needing to regenerate an ID increases, P(generate new ID) = number_of_id_already_generated / number_of_all_possible_ids. In this case, check back the first paragraph (do you need the ID to be random?).

Hope this helps.

blocks|key|1031565|text|有很多可能性。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1031566|你知道如何随机生成一个整数，对吧？因此，您可以从它生成一个字符...(ex+65+->+A)|1031567|这取决于你需要什么，随机性的程度，涉及的安全性...但对于学校项目，我想获取UUID子字符串将适合:)|1031568|entityMap^0|0|0|0^^$0|@$1|2|3|4|5|6|7|H|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|I|8|@]|9|@]|A|$]]|$1|D|3|E|5|6|7|J|8|@]|9|@]|A|$]]|$1|F|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|G|$]]

Many possibilities...

You know how to generate randomly an integer right?
You can thus generate a char from it... (ex 65 -> A)

It depends what you need, the level of randomness, the security involved... but for a school project i guess getting UUID substring would fit :)

I have an object called <code>Student</code>, and it has <code>studentName</code>, <code>studentId</code>, <code>studentAddress</code>, etc. For the <code>studentId</code>, I have to generate random string consist of seven numeric charaters,
eg.

<pre><code>studentId = getRandomId();
studentId = "1234567" &lt;-- from the random generator.
</code></pre>

And I have to make sure that there is no duplicate id.

How to generate a random String in Java

我有一个名为Student的对象，它有studentName，studentId，studentAddress等。对于studentId，我必须生成由七个数字字符组成的随机字符串，例如。studentId = getRandomId();studentId = "1234567" <-- from the random generator.我必须确保没有重复的id。

问如何在Java中生成随机字符串
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