对于你当前的例子，你可以用任何一种方法来达到这个结果

var txt2 = txt1.split(' ').join(' bar ')

or

var txt2 = txt1.replace(' ', ' bar ');

但既然你可以做出这样的假设，你不妨直接跳过葛伦的例子。

在这种情况下，除了基于字符索引之外，您真的不能做出任何假设，那么我真的会选择子字符串解决方案。

使用切片

你可以使用slice(0,index) + str + slice(index)。或者您可以为它创建一个方法。

String.prototype.insertAt =函数(索引，str){ 返回this.slice(0,index) + STR + this.slice(index) ｝ console.log (" foo栏”。insertAt(4，'baz ')) //foo baz bar

字符串的拼接方法

你可以split()主字符串并添加，然后使用普通的splice()

String.prototype.splice = function(index,del,...newStrs){ let str = this.split(''); str.splice(index,del,newStrs.join('') || ''); return str.join(''); } var txt1 = "foo baz" //inserting single string. console.log(txt1.splice(4,0,"bar ")); //foo bar baz //inserting multiple strings console.log(txt1.splice(4,0,"bar ","bar2 ")); //foo bar bar2 baz //removing letters console.log(txt1.splice(1,2)) //f baz //remving and inseting atm console.log(txt1.splice(1,2," bar")) //f bar baz

在多个索引上应用splice()

该方法接受一个数组，数组中的每个元素表示一个splice()。

String.prototype.splice = function(index,del,...newStrs){ let str = this.split(''); str.splice(index,del,newStrs.join('') || ''); return str.join(''); } String.prototype.mulSplice = function(arr){ str = this let dif = 0; arr.forEach(x => { x[2] === x[2] || []; x[1] === x[1] || 0; str = str.splice(x[0] + dif,x[1],...x[2]); dif += x[2].join('').length - x[1]; }) return str; } let txt = "foo bar baz" //Replacing the 'foo' and 'bar' with 'something1' ,'another' console.log(txt.splice(0,3,'something')) console.log(txt.mulSplice( [ [0,3,["something1"]], [4,3,["another"]] ] ))

在特定索引处插入(而不是在第一个空格字符处)必须使用字符串切片/子字符串:

var txt2 = txt1.slice(0, 3) + "bar" + txt1.slice(3);

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;
};

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)

对于你当前的例子，你可以用任何一种方法来达到这个结果

var txt2 = txt1.split(' ').join(' bar ')

or

var txt2 = txt1.replace(' ', ' bar ');

但既然你可以做出这样的假设，你不妨直接跳过葛伦的例子。

在这种情况下，除了基于字符索引之外，您真的不能做出任何假设，那么我真的会选择子字符串解决方案。