正如许多人提到的，原型应该是最好的方法。确保原型出现的时间早于它被使用的时间。

String.prototype.insert = function (x, str) {
    return (x > 0) ? this.substring(0, x) + str + this.substr(x) : str + this;
};

下面是我写的一个像所有其他编程语言一样的方法:

string .prototype.insert =函数(索引，字符串){ If (index > 0) ｛ 返回。子字符串(0,index) +字符串+ this。substring(指数、this.length); ｝ 返回string + this; }; //使用示例: var something =“你好吗?”; 某事=某事。插入(3，“are”); console.log(东西)

参考:

http://coderamblings.wordpress.com/2012/07/09/insert-a-string-at-a-specific-index/

我们可以同时使用子字符串和切片方法。

String.prototype.customSplice = function (index, absIndex, string) {
    return this.slice(0, index) + string+ this.slice(index + Math.abs(absIndex));
};


String.prototype.replaceString = function (index, string) {
    if (index > 0)
        return this.substring(0, index) + string + this.substr(index);

    return string + this;
};


console.log('Hello Developers'.customSplice(6,0,'Stack ')) // Hello Stack Developers
console.log('Hello Developers'.replaceString(6,'Stack ')) //// Hello Stack Developers

子字符串方法的唯一问题是它不能与负索引一起工作。它总是从第0位开始取字符串下标。

您可以将自己的splice()原型化为String。

Polyfill

if (!String.prototype.splice) {
    /**
     * {JSDoc}
     *
     * The splice() method changes the content of a string by removing a range of
     * characters and/or adding new characters.
     *
     * @this {String}
     * @param {number} start Index at which to start changing the string.
     * @param {number} delCount An integer indicating the number of old chars to remove.
     * @param {string} newSubStr The String that is spliced in.
     * @return {string} A new string with the spliced substring.
     */
    String.prototype.splice = function(start, delCount, newSubStr) {
        return this.slice(0, start) + newSubStr + this.slice(start + Math.abs(delCount));
    };
}

例子

String.prototype.splice = function（idx， rem， str） { 返回 this.slice（0， idx） + str + this.slice（idx + Math.abs（rem））; }; var result = “foo baz”.splice（4， 0， “bar ”）; document.body.innerHTML = result;“嘟”

EDIT:修改它，以确保rem是一个绝对值。

另一种解决办法是，把绳子切成两半，在中间放一根绳子。

var str = jQuery('#selector').text();

var strlength = str.length;

strf = str.substr(0 , strlength - 5);
strb = str.substr(strlength - 5 , 5);

jQuery('#selector').html(strf + 'inserted' + strb);

我想比较使用substring的方法和使用slice的方法分别来自Base33和user113716，为此我写了一些代码

还可以看看这个性能比较，子字符串，切片

我使用的代码创建了巨大的字符串，并将字符串“bar”多次插入到巨大的字符串中

if (!String.prototype.splice) { /** * {JSDoc} * * The splice() method changes the content of a string by removing a range of * characters and/or adding new characters. * * @this {String} * @param {number} start Index at which to start changing the string. * @param {number} delCount An integer indicating the number of old chars to remove. * @param {string} newSubStr The String that is spliced in. * @return {string} A new string with the spliced substring. */ String.prototype.splice = function (start, delCount, newSubStr) { return this.slice(0, start) + newSubStr + this.slice(start + Math.abs(delCount)); }; } String.prototype.splice = function (idx, rem, str) { return this.slice(0, idx) + str + this.slice(idx + Math.abs(rem)); }; String.prototype.insert = function (index, string) { if (index > 0) return this.substring(0, index) + string + this.substring(index, this.length); return string + this; }; function createString(size) { var s = "" for (var i = 0; i < size; i++) { s += "Some String " } return s } function testSubStringPerformance(str, times) { for (var i = 0; i < times; i++) str.insert(4, "bar ") } function testSpliceStringPerformance(str, times) { for (var i = 0; i < times; i++) str.splice(4, 0, "bar ") } function doTests(repeatMax, sSizeMax) { n = 1000 sSize = 1000 for (var i = 1; i <= repeatMax; i++) { var repeatTimes = n * (10 * i) for (var j = 1; j <= sSizeMax; j++) { var actualStringSize = sSize * (10 * j) var s1 = createString(actualStringSize) var s2 = createString(actualStringSize) var start = performance.now() testSubStringPerformance(s1, repeatTimes) var end = performance.now() var subStrPerf = end - start start = performance.now() testSpliceStringPerformance(s2, repeatTimes) end = performance.now() var splicePerf = end - start console.log( "string size =", "Some String ".length * actualStringSize, "\n", "repeat count = ", repeatTimes, "\n", "splice performance = ", splicePerf, "\n", "substring performance = ", subStrPerf, "\n", "difference = ", splicePerf - subStrPerf // + = splice is faster, - = subStr is faster ) } } } doTests(1, 100)

性能上的一般差异充其量是边际的，两种方法都工作得很好(即使在长度~~ 12000000的字符串上)