在Perl中,我可以使用以下语法多次重复一个字符:

$a = "a" x 10; // results in "aaaaaaaaaa"

有没有一种简单的方法在Javascript中实现这一点?我显然可以使用一个函数,但我想知道是否有内置的方法,或者其他一些巧妙的技术。


当前回答

也可用作一个衬垫:

function repeat(str, len) {
    while (str.length < len) str += str.substr(0, len-str.length);
    return str;
}

其他回答

以下是我使用的:

function repeat(str, num) {
        var holder = [];
        for(var i=0; i<num; i++) {
            holder.push(str);
        }
        return holder.join('');
    }

Lodash提供了与Javascript repeat()函数类似的功能,这在所有浏览器中都不可用。它名为_.repeat,从3.0.0版开始提供:

_.repeat('a', 10);

截至目前,96.39%的浏览器支持String.repeat()。

function pad(text, maxLength){ 
  return text + "0".repeat(maxLength - text.length);
}
console.log(pad('text', 7)); //text000

最有效的方法是https://developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/Global_Objects/String/repeat

下面是简短版本。

  String.prototype.repeat = function(count) {
    if (count < 1) return '';
    var result = '', pattern = this.valueOf();
    while (count > 1) {
      if (count & 1) result += pattern;
      count >>>= 1, pattern += pattern;
    }
    return result + pattern;
  };
  var a = "a";
  console.debug(a.repeat(10));

Mozilla的Polyfill:

if (!String.prototype.repeat) {
  String.prototype.repeat = function(count) {
    'use strict';
    if (this == null) {
      throw new TypeError('can\'t convert ' + this + ' to object');
    }
    var str = '' + this;
    count = +count;
    if (count != count) {
      count = 0;
    }
    if (count < 0) {
      throw new RangeError('repeat count must be non-negative');
    }
    if (count == Infinity) {
      throw new RangeError('repeat count must be less than infinity');
    }
    count = Math.floor(count);
    if (str.length == 0 || count == 0) {
      return '';
    }
    // Ensuring count is a 31-bit integer allows us to heavily optimize the
    // main part. But anyway, most current (August 2014) browsers can't handle
    // strings 1 << 28 chars or longer, so:
    if (str.length * count >= 1 << 28) {
      throw new RangeError('repeat count must not overflow maximum string size');
    }
    var rpt = '';
    for (;;) {
      if ((count & 1) == 1) {
        rpt += str;
      }
      count >>>= 1;
      if (count == 0) {
        break;
      }
      str += str;
    }
    // Could we try:
    // return Array(count + 1).join(this);
    return rpt;
  }
}
var stringRepeat = function(string, val) {
  var newString = [];
    for(var i = 0; i < val; i++) {
      newString.push(string);
  }
  return newString.join('');
}

var repeatedString = stringRepeat("a", 1);