/**
 * Format string by replacing placeholders with value from element with
 * corresponsing index in `replacementArray`.
 * Replaces are made simultaneously, so that replacement values like
 * '{1}' will not mess up the function.
 *
 * Example 1:
 * ('{2} {1} {0}', ['three', 'two' ,'one']) -> 'one two three'
 *
 * Example 2:
 * ('{0}{1}', ['{1}', '{0}']) -> '{1}{0}'
 */
function stringFormat(formatString, replacementArray) {
    return formatString.replace(
        /\{(\d+)\}/g, // Matches placeholders, e.g. '{1}'
        function formatStringReplacer(match, placeholderIndex) {
            // Convert String to Number
            placeholderIndex = Number(placeholderIndex);

            // Make sure that index is within replacement array bounds
            if (placeholderIndex < 0 ||
                placeholderIndex > replacementArray.length - 1
            ) {
                return placeholderIndex;
            }

            // Replace placeholder with value from replacement array
            return replacementArray[placeholderIndex];
        }
    );
}

这里是sprintf在JavaScript中的一个最小实现：它只执行“%s”和“%d”，但我为它保留了扩展空间。这对OP来说是无用的，但其他偶然发现这条来自谷歌的线索的人可能会从中受益。

function sprintf() {
    var args = arguments,
    string = args[0],
    i = 1;
    return string.replace(/%((%)|s|d)/g, function (m) {
        // m is the matched format, e.g. %s, %d
        var val = null;
        if (m[2]) {
            val = m[2];
        } else {
            val = args[i];
            // A switch statement so that the formatter can be extended. Default is %s
            switch (m) {
                case '%d':
                    val = parseFloat(val);
                    if (isNaN(val)) {
                        val = 0;
                    }
                    break;
            }
            i++;
        }
        return val;
    });
}

例子：

alert(sprintf('Latitude: %s, Longitude: %s, Count: %d', 41.847, -87.661, 'two'));
// Expected output: Latitude: 41.847, Longitude: -87.661, Count: 0

与之前回复中的类似解决方案相比，此解决方案一次性完成所有替换，因此不会替换先前替换值的部分。

当前JavaScript

从上的ES6可以使用模板字符串：

let soMany = 10;
console.log(`This is ${soMany} times easier!`);
// "This is 10 times easier!"

详见下文Kim的回答。

较旧的答案

试试JavaScript的sprintf（）。

如果你真的想自己做一个简单的格式化方法，不要连续替换，而是同时替换。

因为当先前替换的替换字符串也包含如下格式序列时，提到的大多数其他建议都会失败：

"{0}{1}".format("{1}", "{0}")

通常，您希望输出为｛1｝｛0｝，但实际输出为｛1｝{1｝｝。所以要像恐惧噬菌体的建议那样同时进行替换。

/**
 * Format string by replacing placeholders with value from element with
 * corresponsing index in `replacementArray`.
 * Replaces are made simultaneously, so that replacement values like
 * '{1}' will not mess up the function.
 *
 * Example 1:
 * ('{2} {1} {0}', ['three', 'two' ,'one']) -> 'one two three'
 *
 * Example 2:
 * ('{0}{1}', ['{1}', '{0}']) -> '{1}{0}'
 */
function stringFormat(formatString, replacementArray) {
    return formatString.replace(
        /\{(\d+)\}/g, // Matches placeholders, e.g. '{1}'
        function formatStringReplacer(match, placeholderIndex) {
            // Convert String to Number
            placeholderIndex = Number(placeholderIndex);

            // Make sure that index is within replacement array bounds
            if (placeholderIndex < 0 ||
                placeholderIndex > replacementArray.length - 1
            ) {
                return placeholderIndex;
            }

            // Replace placeholder with value from replacement array
            return replacementArray[placeholderIndex];
        }
    );
}

在typescript中，创建一个名为format.ts的文件，并导入您需要使用的任何格式。

// contents of format.ts

interface String {
  format(...args: any[]): string;
}

if (!String.prototype.format) {
  String.prototype.format = function() {
    let a = this;
    let b: any;
    // tslint:disable-next-line: forin
    for (b in arguments) {
      a = a.replace(/%[a-z]/, arguments[b]);
    }
    return a;
  };
}

要格式化字符串，请使用以下代码：

import './format';

console.log('Hello, %s!'.format('World'));

实例

String.prototype.format=函数（）{设a=此；设b；for（参数中的b）{a=a.replace（/%[a-z]/，自变量[b]）；}返回a；};console.log（“您好，%s！”.format（“世界”））；

如果只需要使用%s说明符格式化字符串

function _sprintf(message){
    const regexp = RegExp('%s','g');
    let match;
    let index = 1;
    while((match = regexp.exec(message)) !== null) {
        let replacement = arguments[index];
        if (replacement) {
            let messageToArray = message.split('');
            messageToArray.splice(match.index, regexp.lastIndex - match.index, replacement);
            message = messageToArray.join('');
            index++;
        } else {
            break;
        }
    }

    return message;
}

_sprintf("my name is %s, my age is %s", "bob", 50); // my name is bob, my age is 50