根據一條線:
s = "Test abc test test abc test test test abc test test abc";
这似乎只是在上面的行中删除ABC的第一次出现:
s = s.replace('abc', '');
如何替代所有事件?
根據一條線:
s = "Test abc test test abc test test test abc test test abc";
这似乎只是在上面的行中删除ABC的第一次出现:
s = s.replace('abc', '');
如何替代所有事件?
当前回答
'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
其他回答
与常规表达的i旗案例不敏感
console.log('App started'.replace(/a/g, '')) // Result: "App strted"
console.log('App started'.replace(/a/gi, '')) // Result: "pp strted"
在与主要答案相关的性能方面,这些是某些在线测试。
虽然以下是使用 console.time() 的某些性能测试(它们在自己的控制台上工作最好,因为时间很短,可以在下面的剪辑中看到)。
值得注意的是,如果你运行它们多次,结果总是不同的,尽管正常的表达解决方案似乎是最快的平均,而旋转解决方案是最慢的。
添加 /g
document.body.innerHTML = document.body.innerHTML.replace('hello', 'hi');
到
// Replace 'hello' string with /hello/g regular expression.
document.body.innerHTML = document.body.innerHTML.replace(/hello/g, 'hi');
G 意味着全球性
'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
以下功能为我工作:
String.prototype.replaceAllOccurence = function(str1, str2, ignore)
{
return this.replace(new RegExp(str1.replace(/([\/\,\!\\\^\$\{\}\[\]\(\)\.\*\+\?\|\<\>\-\&])/g,"\\$&"),(ignore?"gi":"g")),(typeof(str2)=="string")?str2.replace(/\$/g,"$$$$"):str2);
} ;
现在,请称这些功能如下:
"you could be a Project Manager someday, if you work like this.".replaceAllOccurence ("you", "I");
简单地复制并将此代码插入您的浏览器控制台进行测试。