java 生成 5 位随机引脚
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Generating a Random Pin of 5 Digits
提问by Craig
I would like to generate Unique Pins based on a random number I have found this on Stack Overflow How to generate a random five digit number Java. That Question uses time to generate the numbers so therefore you get a lot of duplicates.
我想根据我在 Stack Overflow How to generate a random 5 digit number Java上发现的随机数生成唯一 Pin 图。该问题使用时间来生成数字，因此您会得到很多重复项。
Here is my code
这是我的代码
public int createRandomPin() {
int k = random.nextInt(Integer.SIZE);
k = (k + 1) * 9999;
if (9999 > k || k > 99999) {
//then regenerate
} else {
return k;
}
}
My Question
我的问题
Java Compiler then gives a warning missing "return". As well I need to restructure the code so that if it isn't a 5 digit pin it generates again before it "returns k".
Java Compiler 然后给出一个缺少“return”的警告。同样，我需要重组代码，以便如果它不是 5 位引脚，它会在“返回 k”之前再次生成。
回答by Naruto
I would suggest to generate random number using SecureRandom class and then as it needs to be of 5 digits create a random number between 0 and 99999 using random.nextInt(100000) , here 0 is inclusive and 100000 is exclusive and then format it into 5 digit by appending zero.
我建议使用 SecureRandom 类生成随机数，然后因为它需要是 5 位数字，所以使用 random.nextInt(100000) 创建一个介于 0 和 99999 之间的随机数，这里 0 是包含的，100000 是不包括的，然后将其格式化为 5数字通过附加零。
SecureRandom random = new SecureRandom();
int num = random.nextInt(100000);
String formatted = String.format("%05d", num);
System.out.println(formatted);
I hope this solves your problem
我希望这能解决你的问题
Edit: This post incorrectly said 10,000, has been edited to say 100,000
编辑：此贴错误地表示 10,000，已编辑为 100,000
回答by Hyman
Integer.SIZEyields the number in bits of the intdata type. This nothing has to do with the range span for a random value.
Integer.SIZE产生int数据类型的位数。这与随机值的范围跨度无关。
Using it as the argument of Random.nextIntdoesn't make any sense (actually it generates a random value in range [0,32)).
将它用作 of 的参数Random.nextInt没有任何意义（实际上它会在 range 中生成一个随机值[0,32)）。
Just generate a int value = random.nextInt(100000)so that you will obtain a value in [0,99999].
只需生成一个 int 即可value = random.nextInt(100000)获得[0,99999].
Now your definition of 5 digits pin is not precise, 40could be interpreted as 00040so it's still 5 digits if you pad it. You must take this thing into account, since forcing 5 "visible" digits implies generating a number in range [10000,99999]but this is just an awkward solution.
现在您对 5 位数 pin 的定义不准确，40可以解释为00040，如果您填充它，它仍然是 5 位数。您必须考虑到这一点，因为强制 5 个“可见”数字意味着在范围内生成一个数字，[10000,99999]但这只是一个尴尬的解决方案。
回答by Rohit
RandomUtils.randomNumeric(int count)or RandomUtils.randomNumeric(int inLength, int maxLength)from apache common lang3 can also be used for the same purpose.
RandomUtils.randomNumeric(int count)或者RandomUtils.randomNumeric(int inLength, int maxLength)来自 apache 的 common lang3 也可以用于相同的目的。
