所有答案都被接受,你可以以多种方式做到这一点。

const str = “测试测试测试测试测试测试测试测试测试测试测试测试测试测试测试测试测试测试测试测试测试测试测试测试测试测试测试测试测试测试测试测试测试测试测试测试测试测试测试测试测试测试测试测试测试测试测试测试测试测试测试测试测试测试测试测试测试测试测试测试测试测试测试测试测试测试测试测试测试测试测试测试测试测试测试测试测试测试测试测试测试测试测试测试测试测试测试测试测试测试测试测试测试测试测试测试测试测试

'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

这是最常见、最可读的方法。

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

方法1:

str = str.replace(/abc/g, "replaced text");

方法2:

str = str.split("abc").join("replaced text");

方法3:

str = str.replace(new RegExp("abc", "g"), "replaced text");

方法4:

while(str.includes("abc")){
   str = str.replace("abc", "replaced text");
}

出口:

console.log(str);
// Test replaced text test test replaced text test test test replaced text test test replaced text

您可以在没有Regex的情况下做到这一点,但如果替代文本包含搜索文本,则要小心。

吉。

replaceAll("nihIaohi", "hI", "hIcIaO", true)

因此,这里是一个合适的替代All 选项,包括字符串的原型:

function replaceAll(str, find, newToken, ignoreCase)
{
    let i = -1;

    if (!str)
    {
        // Instead of throwing, act as COALESCE if find == null/empty and str == null
        if ((str == null) && (find == null))
            return newToken;

        return str;
    }

    if (!find) // sanity check 
        return str;

    ignoreCase = ignoreCase || false;
    find = ignoreCase ? find.toLowerCase() : find;

    while ((
        i = (ignoreCase ? str.toLowerCase() : str).indexOf(
            find, i >= 0 ? i + newToken.length : 0
        )) !== -1
    )
    {
        str = str.substring(0, i) +
            newToken +
            str.substring(i + find.length);
    } // Whend 

    return str;
}

或者,如果你想有一个字符串原型功能:

String.prototype.replaceAll = function (find, replace) {
    let str = this;

    let i = -1;

    if (!str)
    {
        // Instead of throwing, act as COALESCE if find == null/empty and str == null
        if ((str == null) && (find == null))
            return newToken;

        return str;
    }

    if (!find) // sanity check 
        return str;

    ignoreCase = ignoreCase || false;
    find = ignoreCase ? find.toLowerCase() : find;

    while ((
        i = (ignoreCase ? str.toLowerCase() : str).indexOf(
            find, i >= 0 ? i + newToken.length : 0
        )) !== -1
    )
    {
        str = str.substring(0, i) +
            newToken +
            str.substring(i + find.length);
    } // Whend 

    return str;
};

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", "")
}

// Try this way

const str = "Test abc test test abc test test test abc test test abc";
const result = str.split('abc').join('');
console.log(result);