我使用p来存储以前的回归替换结果:

function replaceAll(s, m, r, p) {
    return s === p || r.contains(m) ? s : replaceAll(s.replace(m, r), m, r, s);
}

它将取代链 s 的所有事件,直到它是可能的:

replaceAll('abbbbb', 'ab', 'a') → 'abbbb' → 'abbb' → 'abb' → 'ab' → 'a'

要避免无限旋转,我检查替代r是否包含一匹匹匹配m:

replaceAll('abbbbb', 'a', 'ab') → 'abbbbb'

'a cat is not a caterpillar'.replace(/\bcat\b/gi,'dog');
//"a dog is not a caterpillar"

這是一個簡單的雷格斯,避免在大多數情況下取代字的部分. 然而,一個<unk> - 仍然被認為是字的邊界. 所以條件可以用在這種情況下,以避免取代線,如冷貓:

'a cat is not a cool-cat'.replace(/\bcat\b/gi,'dog');//wrong
//"a dog is not a cool-dog" -- nips
'a cat is not a cool-cat'.replace(/(?:\b([^-]))cat(?:\b([^-]))/gi,'$1dog$2');
//"a dog is not a cool-cat"

Regexp 不是唯一的替代多种现象的方法,远离它,思考灵活,思考分裂!

var newText = "the cat looks like a cat".split('cat').join('dog');

否则,要防止替代词部分 - 批准的答案也会做什么! 你可以通过常规的表达方式来围绕这个问题,我承认,有点复杂,并且作为一个惊喜,一个缓慢的,也:

var regText = "the cat looks like a cat".replace(/(?:(^|[^a-z]))(([^a-z]*)(?=cat)cat)(?![a-z])/gi,"$1dog");

结果与接受的答案相同,但是,在这个行上使用 /cat/g 表达式:

var oops = 'the cat looks like a cat, not a caterpillar or coolcat'.replace(/cat/g,'dog');
//returns "the dog looks like a dog, not a dogerpillar or cooldog" ??

var caterpillar = 'the cat looks like a cat, not a caterpillar or coolcat'.replace(/(?:(^|[^a-z]))(([^a-z]*)(?=cat)cat)(?![a-z])/gi,"$1dog");
//return "the dog looks like a dog, not a caterpillar or coolcat"

RegExp(常规表达式)对象 Regular-Expressions.info

在这种情况下,它显著简化表达,并提供更多的灵活性,如用正确的资本化替换或在一个行中替换两只猫和猫:

'Two cats are not 1 Cat! They\'re just cool-cats, you caterpillar'
   .replace(/(^|.\b)(cat)(s?\b.|$)/gi,function(all,char1,cat,char2)
    {
       // Check 1st, capitalize if required
       var replacement = (cat.charAt(0) === 'C' ? 'D' : 'd') + 'og';
       if (char1 === ' ' && char2 === 's')
       { // Replace plurals, too
           cat = replacement + 's';
       }
       else
       { // Do not replace if dashes are matched
           cat = char1 === '-' || char2 === '-' ? cat : replacement;
       }
       return char1 + cat + char2;//return replacement string
    });
//returns:
//Two dogs are not 1 Dog! They're just cool-cats, you caterpillar

str = "Test abc test test abc test test test abc test test abc"

str.split(' ').join().replace(/abc/g,'').replace(/,/g, ' ')

我知道这不是最好的办法,但你可以尝试一下:

var annoyingString = "Test abc test test abc test test test abc test test abc";

while (annoyingString.includes("abc")) {
    annoyingString = annoyingString.replace("abc", "")
}

str = str.replace(/abc/g, '');

或者尝试替代所有方法,如本答案所建议:

str = str.replaceAll('abc', '');

或:

var search = 'abc';
str = str.replaceAll(search, '');

EDIT: 关于更换的澄清 所有可用性

替代All 方法将添加到 String 的原型,这意味着它将适用于所有 String 对象/字体。

例子:

var output = "test this".replaceAll('this', 'that'); // output is 'test that'.
output = output.replaceAll('that', 'this'); // output is 'test this'

说你想用“x”取代所有的“abc”:

let some_str = 'abc def def lom abc abc def'.split('abc').join('x')
console.log(some_str) //x def def lom x x def

我试图思考一些更简单的东西,而不是修改链条的原型。