Java 生成 6 位随机数
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Generate 6 digit random number
提问by Kraken
I just want to generate 6 digit random number, and the range should be start from 000000 to 999999.
我只想生成 6 位随机数，范围应该从 000000 到 999999。
new Random().nextInt(999999)is returning me number but it is not in 6 digit.
new Random().nextInt(999999)正在返回我的号码，但它不是 6 位数字。
采纳答案by Dev Sabby
Its as simple as that, you can use your code and just do one thing extra here
就这么简单，你可以使用你的代码，在这里只做一件事
String.format("%06d", number);
this will return your number in string format, so the "0" will be "000000".
这将以字符串格式返回您的号码，因此“0”将是“000000”。
Here is the code.
这是代码。
public static String getRandomNumberString() {
// It will generate 6 digit random Number.
// from 0 to 999999
Random rnd = new Random();
int number = rnd.nextInt(999999);
// this will convert any number sequence into 6 character.
return String.format("%06d", number);
}
回答by Karol Dowbecki
If you need a six digit number it has to start at 100000
如果你需要一个六位数的号码，它必须从 100000
int i = new Random().nextInt(900000) + 100000;
Leading zeros do not have effect, 000000is the same as 0. You can further simplify it with ThreadLocalRandomif you are on Java 7+:
前导零没有作用，000000与0. ThreadLocalRandom如果您使用的是 Java 7+，您可以进一步简化它：
int i = ThreadLocalRandom.current().nextInt(100000, 1000000)
回答by Codebender
1 + nextInt(2) shall always give 1 or 2. You then multiply it by 10000 to satisfy your requirement and then add a number between [0..9999].
1 + nextInt(2) 应始终给出 1 或 2。然后将其乘以 10000 以满足您的要求，然后在 [0..9999] 之间添加一个数字。
already solved here
public int gen()
{
Random r = new Random( System.currentTimeMillis() );
return ((1 + r.nextInt(2)) * 10000 + r.nextInt(10000));
}
回答by Bonfra
i know it's very difficult but you can do something like this: create a class for BinaryNumber; create a constructor that generate a char[] of 6 character where every single one is generated with a randomiser from 0 to 1 override the toStrig() method so that it will return the digits char[] as a string if you want to display it. then crate a method toInt() that esaminate the string char by char with a for and turn it in a decimal base number by multiplying current digit to 10 to the pow of i:
我知道这很困难，但您可以这样做：为 BinaryNumber 创建一个类；创建一个生成 6 个字符的 char[] 的构造函数，其中每个字符都使用从 0 到 1 的随机生成器生成，覆盖 toSrig() 方法，以便如果要显示它，它将以字符串形式返回数字 char[] . 然后创建一个 toInt() 方法，该方法使用 for 逐个字符化字符串，并通过将当前数字乘以 10 到 i 的 pow 将其转换为十进制基数：
char[] digits = {‘1' , ‘0' , ‘1' , ‘1' , ‘0' , ‘1'};
//random
int result = 0;
for( int i = 0; i < digits.length; i++) {
result += Integer.parseInt(digits[i]) * Math.pow(10, i);
}
return result;
回答by Abhishek Tiwari
This is the code in java which generate a 6 digit random code.
这是java中生成6位随机代码的代码。
import java.util.*;
public class HelloWorld{
public static void main(String []args)
{
Random r=new Random();
HashSet<Integer> set= new HashSet<Integer>();
while(set.size()<1)
{
int ran=r.nextInt(99)+100000;
set.add(ran);
}
int len = 6;
String random=String.valueOf(len);
for(int random1:set)
{
System.out.println(random1);
random=Integer.toString(random1);
}
}
}
