作为一个相对的JavaScript新手，公认的答案https://stackoverflow.com/a/494046/1904943值得注意/赞赏，但这不是很直观。

这里有一个更简单的解释，例如（使用一个简单的JavaScript IDE）。

myString = 'apple pie, banana loaf';

console.log(myString.replaceAll(/pie/gi, 'PIE'))
// apple PIE, banana loaf

console.log(myString.replaceAll(/\bpie\b/gi, 'PIE'))
// apple PIE, banana loaf

console.log(myString.replaceAll(/pi/gi, 'PIE'))
// apple PIEe, banana loaf

console.log(myString.replaceAll(/\bpi\b/gi, 'PIE'))
// [NO EFFECT] apple pie, banana loaf

const match_word = 'pie';

console.log(myString.replaceAll(/match_word/gi, '**PIE**'))
// [NO EFFECT] apple pie, banana loaf

console.log(myString.replaceAll(/\b`${bmatch_word}`\b/gi, '**PIE**'))
// [NO EFFECT] apple pie, banana loaf

// ----------------------------------------
// ... new RegExp(): be sure to \-escape your backslashes: \b >> \\b ...

const match_term = 'pie';
const match_re = new RegExp(`(\\b${match_term}\\b)`, 'gi')

console.log(myString.replaceAll(match_re, 'PiE'))
// apple PiE, banana loaf

console.log(myString.replace(match_re, '**PIE**'))
// apple **PIE**, banana loaf

console.log(myString.replaceAll(match_re, '**PIE**'))
// apple **PIE**, banana loaf

应用

例如：替换字符串/句子中的单词（颜色突出显示），如果搜索词与匹配单词的用户定义比例相匹配，则[可选]。

注：匹配术语的原始字符大小写保留。hl：高光；re:regex |正则表达式

mySentence = "Apple, boOk? BOoks; booKEd. BookMark, 'BookmarkeD', bOOkmarks! bookmakinG, Banana; bE, BeEn, beFore."

function replacer(mySentence, hl_term, hl_re) {
    console.log('mySentence [raw]:', mySentence)
    console.log('hl_term:', hl_term, '| hl_term.length:', hl_term.length)
    cutoff = hl_term.length;
    console.log('cutoff:', cutoff)

    // `.match()` conveniently collects multiple matched items
    // (including partial matches) into an [array]
    const hl_terms  = mySentence.toLowerCase().match(hl_re, hl_term);
    if (hl_terms == null) {
        console.log('No matches to hl_term "' + hl_term + '"; echoing input string then exiting ...')
        return mySentence;
    }
    console.log('hl_terms:', hl_terms)
    for (let i = 0;  i < hl_terms.length; i++) {
        console.log('----------------------------------------')
        console.log('[' + i + ']:', hl_terms[i], '| length:', hl_terms[i].length, '| parseInt(0.7(length)):', parseInt(0.7*hl_terms[i].length))
        // TEST: if (hl_terms[i].length >= cutoff*10) {
        if (cutoff >= parseInt(0.7 * hl_terms[i].length)) {
            var match_term = hl_terms[i].toString();

            console.log('matched term:', match_term, '[cutoff length:', cutoff, '| 0.7(matched term length):', parseInt(0.7 * hl_terms[i].length))

            const match_re = new RegExp(`(\\b${match_term}\\b)`, 'gi')

            mySentence = mySentence.replaceAll(match_re, '<font style="background:#ffe74e">$1</font>');
        }
        else {
            var match_term = hl_terms[i].toString();
            console.log('NO match:', match_term, '[cutoff length:', cutoff, '| 0.7(matched term length):', parseInt(0.7 * hl_terms[i].length))
        }
    }
    return mySentence;
}

// TESTS:
// const hl_term = 'be';
// const hl_term = 'bee';
// const hl_term = 'before';
// const hl_term = 'book';
const hl_term = 'bookma';
// const hl_term = 'Leibniz';

// This regex matches from start of word:
const hl_re = new RegExp(`(\\b${hl_term}[A-z]*)\\b`, 'gi')

mySentence = replacer(mySentence, hl_term, hl_re);
console.log('mySentence [processed]:', mySentence)

输出

mySentence [raw]: Apple, boOk? BOoks; booKEd. BookMark, 'BookmarkeD',
bOOkmarks! bookmakinG, Banana; bE, BeEn, beFore.

hl_term: bookma | hl_term.length: 6
cutoff: 6
hl_terms: Array(4) [ "bookmark", "bookmarked", "bookmarks", "bookmaking" ]

----------------------------------------
[0]: bookmark | length: 8 | parseInt(0.7(length)): 5
matched term: bookmark [cutoff length: 6 | 0.7(matched term length): 5
----------------------------------------
[1]: bookmarked | length: 10 | parseInt(0.7(length)): 7
NO match: bookmarked [cutoff length: 6 | 0.7(matched term length): 7
----------------------------------------
[2]: bookmarks | length: 9 | parseInt(0.7(length)): 6
matched term: bookmarks [cutoff length: 6 | 0.7(matched term length): 6
----------------------------------------
[3]: bookmaking | length: 10 | parseInt(0.7(length)): 7
NO match: bookmaking [cutoff length: 6 | 0.7(matched term length): 7

mySentence [processed]: Apple, boOk? BOoks; booKEd.
<font style="background:#ffe74e">BookMark</font>, 'BookmarkeD',
<font style="background:#ffe74e">bOOkmarks</font>! bookmakinG,
Banana; bE, BeEn, beFore.

您始终可以重复使用indexOf：

String.prototype.replaceAll = function(substring, replacement) {
    var result = '';
    var lastIndex = 0;

    while(true) {
        var index = this.indexOf(substring, lastIndex);
        if(index === -1) break;
        result += this.substring(lastIndex, index) + replacement;
        lastIndex = index + substring.length;
    }

    return result + this.substring(lastIndex);
};

当替换包含匹配项时，这不会进入无限循环。

String.prototype.replaceAll = function(a, b) {
    return this.replace(new RegExp(a.replace(/([.?*+^$[\]\\(){}|-])/ig, "\\$1"), 'ig'), b)
}

测试方式如下：

var whatever = 'Some [b]random[/b] text in a [b]sentence.[/b]'

console.log(whatever.replaceAll("[", "<").replaceAll("]", ">"))

对于任何希望在匹配方法中使用变量的人来说，这对我很有用：

var alpha = 'fig';
'food fight'.match(alpha + 'ht')[0]; // fight

如果$1对您不起作用，您可以使用此选项：

var pattern = new RegExp("amman", "i");
"abc Amman efg".replace(pattern, "<b>" + "abc Amman efg".match(pattern)[0] + "</b>");

您可以构造一个新的RegExp对象，而不是使用/regex\d/g语法：

var replace = "regex\\d";
var re = new RegExp(replace,"g");

您可以通过这种方式动态创建正则表达式对象。然后您将执行以下操作：

"mystring1".replace(re, "newstring");