虽然不是世界上最推荐的功能，但它确实有效。

如果需要sprintf，只需复制并粘贴相同的函数，并将return console.log（sb）更改为只返回sb。

printf=函数（s，/*args…*/）{a=参数；al=a.length；如果（al<=1）返回-2；如果（al>=2&&s.toLowerCase（）.search（/%[a-z]/）==-1）返回-1；sb=s；对于（i=1；i<=al-1；i++）{sb=某人替换（/%[a-z]/，a[i]）；}返回控制台日志（sb）；}var someString=“您好%s\n现在是%s：%s%s。\n今天是%s\n”；printf（someString，“StackOverflowUser”，“5”，“48”，“PM”，“漂亮”）；

此方法适用于｛0｝、｛1｝和｛｝。

String.prototype.format = function format()
{                                                                                                               
  var msg = this;
  for(var i in arguments)
    msg = msg.replace(/\{\}/,arguments[i]).replace(new RegExp('\\{'+i+'\\}','g'),arguments[i]);
  return msg;
}

+1 Zippo，但函数体需要如下所示，否则它会在每次迭代时附加当前字符串：

String.prototype.format = function() {
    var formatted = this;
    for (var arg in arguments) {
        formatted = formatted.replace("{" + arg + "}", arguments[arg]);
    }
    return formatted;
};

arg函数：

/**
 * Qt stil arg()
 * var scr = "<div id='%1' class='%2'></div>".arg("mydiv").arg("mydivClass");
 */
String.prototype.arg = function() {
    var signIndex = this.indexOf("%");
    var result = this;
    if (signIndex > -1 && arguments.length > 0) {
        var argNumber = this.charAt(signIndex + 1);
        var _arg = "%"+argNumber;
        var argCount = this.split(_arg);
        for (var itemIndex = 0; itemIndex < argCount.length; itemIndex++) {
            result = result.replace(_arg, arguments[0]);
        }
    }
    return result;
}

String.prototype.repeat = function(n) { 
    return new Array(++n).join(this); 
};

String.prototype.pad = function(requiredLength, paddingStr, paddingType) {    
    var n = requiredLength - this.length; 

    if (n) {
        paddingType = paddingType ? paddingType.toLowerCase() : '';
        paddingStr = paddingStr || ' ';
        paddingStr = paddingStr.repeat( Math.ceil(n / paddingStr.length) ).substr(0, n);

        if (paddingType == 'both') {
            n /= 2;
            return paddingStr.substr( 0, Math.ceil(n) ) + this + paddingStr.substr( 0, Math.floor(n) );
        }   

        if (paddingType == 'left') {
            return paddingStr + this;
        }

        return this + paddingStr;
    } 

    return this; 
}; 

// синтаксис аналогичен printf
// 'Привет, %s!'.format('мир') -> "Привет, мир!"
// '%.1s.%.1s. %s'.format('Иван', 'Иванович', 'Иванов') -> "И.И. Иванов"
String.prototype.format = function() {
    var i = 0, 
        params = arguments;

    return this.replace(/%(?:%|(?:(|[+-]+)(|0|'.+?)([1-9]\d*)?(?:\.([1-9]\d*))?)?(s|d|f))/g, function(match, sign, padding, width, precision, type) {
        if (match == '%%') { 
            return '%'; 
        }

        var v = params[i++];

        if (type == 'd') { 
            v = Math.round(v); 
        }
        else if (type == 'f') {
            v = v.toFixed(precision ? precision : 6);
        }

        if (/\+/.test(sign) && v > 0) {
            v = '+' + v;
        }

        v += '';

        if (type != 'f' && precision) {
            v = v.substr(0, precision);
        }

        if (width) {
            v = v.pad(width, padding == '' ? ' ' : padding[0] == "'" ? padding.substr(1) : padding, /-/.test(sign) ? 'right' : 'left'); 
        }

        return v;
    });
};

// this.name = 'Вася';
// console.log( 'Привет, ${name}!'.template(this) );
// "Привет, Вася!"
String.prototype.template = function(context) {
    return this.replace(/\$\{(.*?)\}/g, function(match, name) {
        return context[name];
    });
};

bobjs可以做到这一点：

var sFormat = "My name is {0} and I am {1} years old."; 
var result = bob.string.formatString(sFormat, "Bob", 29); 
console.log(result); 
//output: 
//========== 
// My name is Bob and I am 29 years old.