在上面adamJLev的精彩回答的基础上，下面是TypeScript版本:

// Extending String prototype
interface String {
    format(...params: any[]): string;
}

// Variable number of params, mimicking C# params keyword
// params type is set to any so consumer can pass number
// or string, might be a better way to constraint types to
// string and number only using generic?
String.prototype.format = function (...params: any[]) {
    var s = this,
        i = params.length;

    while (i--) {
        s = s.replace(new RegExp('\\{' + i + '\\}', 'gm'), params[i]);
    }

    return s;
};

创建了一个以集合或数组作为参数的格式函数

用法:

format("i can speak {language} since i was {age}",{language:'javascript',age:10});

format("i can speak {0} since i was {1}",'javascript',10});

代码:

var format = function (str, col) {
    col = typeof col === 'object' ? col : Array.prototype.slice.call(arguments, 1);

    return str.replace(/\{\{|\}\}|\{(\w+)\}/g, function (m, n) {
        if (m == "{{") { return "{"; }
        if (m == "}}") { return "}"; }
        return col[n];
    });
};

我有一个活塞，将其添加到字符串原型: string.format 它不仅像其他一些例子一样简短，而且灵活得多。

用法类似于c#版本:

var str2 = "Meet you on {0}, ask for {1}";
var result2 = str2.format("Friday", "Suzy"); 
//result: Meet you on Friday, ask for Suzy
//NB: also accepts an array

另外，增加了对使用名称和对象属性的支持

var str1 = "Meet you on {day}, ask for {Person}";
var result1 = str1.format({day: "Thursday", person: "Frank"}); 
//result: Meet you on Thursday, ask for Frank

虽然不完全是Q所要求的，但我已经构建了一个类似的，但使用命名占位符而不是编号。我个人更喜欢使用命名参数，并将对象作为参数发送给它(更冗长，但更容易维护)。

String.prototype.format = function (args) {
    var newStr = this;
    for (var key in args) {
        newStr = newStr.replace('{' + key + '}', args[key]);
    }
    return newStr;
}

下面是一个用法示例……

alert("Hello {name}".format({ name: 'World' }));

虽然赛季末已经过去了，但我一直在看给出的答案，并得到了我的两便士:

用法:

var one = strFormat('"{0}" is not {1}', 'aalert', 'defined');
var two = strFormat('{0} {0} {1} {2}', 3.14, 'a{2}bc', 'foo');

方法:

function strFormat() {
    var args = Array.prototype.slice.call(arguments, 1);
    return arguments[0].replace(/\{(\d+)\}/g, function (match, index) {
        return args[index];
    });
}

结果:

"aalert" is not defined
3.14 3.14 a{2}bc foo

如果你正在使用验证插件，你可以使用:

jQuery.validator。Format ("{0} {1}"， "cool"， "formatting") = 'cool formatting'

http://docs.jquery.com/Plugins/Validation/jQuery.validator.format templateargumentargumentN……