我想要一个5个字符的字符串,由从集合[a-zA-Z0-9]中随机选取的字符组成。
用JavaScript实现这一点的最佳方法是什么?
我想要一个5个字符的字符串,由从集合[a-zA-Z0-9]中随机选取的字符组成。
用JavaScript实现这一点的最佳方法是什么?
当前回答
假设您使用underscorejs,就可以在两行中优雅地生成随机字符串:
var possible = 'ABCDEFGHIJKLMNOPQRSTUVWXYZabcdefghijklmnopqrstuvwxyz0123456789';
var random = _.sample(possible, 5).join('');
其他回答
以下代码将使用npm包加密随机字符串生成大小为[a-zA-Z0-9]的加密安全随机字符串。使用以下方法安装:
npm install crypto-random-string
要在集合[a-zA-Z0-9]中获得30个字符的随机字符串:
const cryptoRandomString = require('crypto-random-string');
cryptoRandomString({length: 100, type: 'base64'}).replace(/[/+=]/g,'').substr(-30);
摘要:我们正在替换一个大的随机base64字符串中的/,+,=,并获取最后N个字符。
PS:在子字符串中使用-N
我没有找到支持小写和大写字符的干净解决方案。
仅小写支持:
Math.random().toString(36).substr(2,5)
基于该解决方案,支持小写和大写:
Math.random().toString(36).substr(2,5).split(“”).map(c=>Math.randm()<0.5?c.toUpperCase():c).jjoin(“”);
更改substr(2,5)中的5以调整到所需的长度。
带有es6排列运算符的更新版本:
[…数组(30)].map(()=>Math.random().toString(36)[2]).join(“”)
30是一个任意数字,您可以选择所需的任何令牌长度36是可以传递给numeric.toString()的最大基数,表示所有数字和a-z小写字母2用于从如下所示的随机字符串中选择第三个索引:“0.mfbiohx64i”,我们可以取0之后的任何索引。
function generate(length) {
var letters = ["a","b","c","d","e","f","g","h","i","j","k","l","m","n","o","p","q","r","s","t","u","v","w","x","y","z","A","B","C","D","E","F","G","H","I","J","K","L","M","N","O","P","Q","R","S","T","U","V","W","X","Y","Z","0","1","2","3","4","5","6","7","8","9"];
var IDtext = "";
var i = 0;
while (i < length) {
var letterIndex = Math.floor(Math.random() * letters.length);
var letter = letters[letterIndex];
IDtext = IDtext + letter;
i++;
}
console.log(IDtext)
}
加密强
如果您想获得满足您要求的加密强字符串(我看到的答案使用了这个,但给出了无效答案),请使用
let pass = n=> [...crypto.getRandomValues(new Uint8Array(n))]
.map((x,i)=>(i=x/255*61|0,String.fromCharCode(i+(i>9?i>35?61:55:48)))).join``
let pass=n=>[…crypto.getRandomValues(新Uint8Array(n))].map((x,i)=>(i=x/255*61|0,String.fromCharCode(i+(i>9?i>35?61:55:48))).join``console.log(通过(5));
更新:感谢Zibri评论,我更新代码以获得任意长密码