numeric string in JavaScript?
This question already has an answer here:
If you only want to allow specific characters, you could also do it like this:
function randomString(length, chars) {
var result = '';
for (var i = length; i > 0; --i) result += chars[Math.floor(Math.random() * chars.length)];
return result;
}
var rString = randomString(32, '0123456789abcdefghijklmnopqrstuvwxyzABCDEFGHIJKLMNOPQRSTUVWXYZ');
Here's a jsfiddle to demonstrate: http://jsfiddle.net/wSQBx/
Another way to do it could be to use a special string that tells the function what types of characters to use. You could do that like this:
function randomString(length, chars) {
var mask = '';
if (chars.indexOf('a') > -1) mask += 'abcdefghijklmnopqrstuvwxyz';
if (chars.indexOf('A') > -1) mask += 'ABCDEFGHIJKLMNOPQRSTUVWXYZ';
if (chars.indexOf('#') > -1) mask += '0123456789';
if (chars.indexOf('!') > -1) mask += '~`!@#$%^&*()_+-={}[]:";'<>?,./|';
var result = '';
for (var i = length; i > 0; --i) result += mask[Math.floor(Math.random() * mask.length)];
return result;
}
console.log(randomString(16, 'aA'));
console.log(randomString(32, '#aA'));
console.log(randomString(64, '#A!'));
Fiddle: http://jsfiddle.net/wSQBx/2/
Alternatively, to use the base36 method as described below you could do something like this:
function randomString(length) {
return Math.round((Math.pow(36, length + 1) - Math.random() * Math.pow(36, length))).toString(36).slice(1);
}
我刚刚认识到这是一个非常好的优雅解决方案:
Math.random().toString(36).slice(2)
Another variation of answer suggested by JAR.JAR.beans
(Math.random()*1e32).toString(36)
By changing multiplicator 1e32
you can change length of random string.
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