所以，我的问题已经被其他人以Java形式在这里问了:Java -创建一个新的字符串实例，具有指定的长度和填充特定的字符。最好的解决方案?
……但我正在寻找它的JavaScript等效。
基本上，我想动态填充文本字段与“#”字符，基于每个字段的“maxlength”属性。因此，如果一个输入有maxlength="3"，那么该字段将被"###"填充。
理想情况下，应该有类似Java StringUtils的东西。repeat("#"， 10);，但是，到目前为止，我能想到的最好的选择是循环遍历和追加"#"字符，一次一个，直到达到最大长度。我总觉得还有比这更有效的方法。
什么好主意吗?
供您参考-我不能简单地在输入中设置默认值，因为“#”字符需要在焦点上清除，并且，如果用户没有输入值，将需要在模糊上“重新填充”。我所关心的是“补充”步骤
当前回答
For Evergreen browsers, this will build a staircase based on an incoming character and the number of stairs to build.
function StairCase(character, input) {
let i = 0;
while (i < input) {
const spaces = " ".repeat(input - (i+1));
const hashes = character.repeat(i + 1);
console.log(spaces + hashes);
i++;
}
}
//Implement
//Refresh the console
console.clear();
StairCase("#",6);
您还可以为旧浏览器添加重复填充
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;
}
}
其他回答
试一试:P
S = '#'.repeat(10) document.body.innerHTML = s
(我所见过的)最好的方法是
var str = new Array(len + 1).join( character );
这将创建一个具有给定长度的数组，然后将其与给定的字符串连接以重复操作。.join()函数的作用是，不管元素是否赋值，数组的长度都是不变的，未定义的值会呈现为空字符串。
您必须在所需的长度上加上1，因为分隔符字符串位于数组元素之间。
For Evergreen browsers, this will build a staircase based on an incoming character and the number of stairs to build.
function StairCase(character, input) {
let i = 0;
while (i < input) {
const spaces = " ".repeat(input - (i+1));
const hashes = character.repeat(i + 1);
console.log(spaces + hashes);
i++;
}
}
//Implement
//Refresh the console
console.clear();
StairCase("#",6);
您还可以为旧浏览器添加重复填充
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;
}
}
根据霍根和零戏法小马的回答。我认为这应该足够快速和灵活，以处理大多数用例:
var hash = '####################################################################'
function build_string(length) {
if (length == 0) {
return ''
} else if (hash.length <= length) {
return hash.substring(0, length)
} else {
var result = hash
const half_length = length / 2
while (result.length <= half_length) {
result += result
}
return result + result.substring(0, length - result.length)
}
}
如果你喜欢，你可以使用函数的第一行作为一行:
function repeat(str, len) {
while (str.length < len) str += str.substr(0, len-str.length);
return str;
}
