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。

更新2016:下面是另一个基于一行RegExp方法的原型函数(在未定义或负索引上提供prepend支持):

/**
 * Insert `what` to string at position `index`.
 */
String.prototype.insert = function(what, index) {
    return index > 0
        ? this.replace(new RegExp('.{' + index + '}'), '$&' + what)
        : what + this;
};

console.log( 'foo baz'.insert('bar ', 4) );  // "foo bar baz"
console.log( 'foo baz'.insert('bar ')    );  // "bar foo baz"

之前(回到2012年)只是为了好玩的解决方案:

var index = 4,
    what  = 'bar ';

'foo baz'.replace(/./g, function(v, i) {
    return i === index - 1 ? v + what : v;
});  // "foo bar baz"

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)

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);

这基本上是@Base33所做的，除了我还提供了使用负号从末尾开始计数的选项。有点像substr方法所允许的。

// use a negative index to insert relative to the end of the string.

String.prototype.insert = function (index, string) {
  var ind = index < 0 ? this.length + index  :  index;
  return  this.substring(0, ind) + string + this.substr(ind);
};

例子: 假设您有使用命名约定的全尺寸图像，但不能更新数据以同时提供缩略图url。

var url = '/images/myimage.jpg';
var thumb = url.insert(-4, '_thm');

//    result:  '/images/myimage_thm.jpg'

my_string          = "hello world";
my_insert          = " dear";
my_insert_location = 5;

my_string = my_string.split('');  
my_string.splice( my_insert_location , 0, my_insert );
my_string = my_string.join('');

https://jsfiddle.net/gaby_de_wilde/wz69nw9k/