在JavaScript中替换字符串/字符的所有实例的最快方法是什么?while, for循环,正则表达式?
当前回答
像这样使用正则表达式对象
var regex = new RegExp(' ' ', 'g'); STR = STR .replace(regex, '\ ");
它将取代所有出现的“into”。
其他回答
试试这个replaceAll: http://dumpsite.com/forum/index.php?topic=4.msg8#msg8
String.prototype.replaceAll = function(str1, str2, ignore)
{
return this.replace(new RegExp(str1.replace(/([\/\,\!\\\^\$\{\}\[\]\(\)\.\*\+\?\|\<\>\-\&])/g,"\\$&"),(ignore?"gi":"g")),(typeof(str2)=="string")?str2.replace(/\$/g,"$$$$"):str2);
}
这是非常快的,它将工作在所有这些条件 很多人都没有做到的:
"x".replaceAll("x", "xyz");
// xyz
"x".replaceAll("", "xyz");
// xyzxxyz
"aA".replaceAll("a", "b", true);
// bb
"Hello???".replaceAll("?", "!");
// Hello!!!
让我知道你是否可以打破它,或者你有更好的东西,但要确保它能通过这4个测试。
I tried a number of these suggestions after realizing that an implementation I had written of this probably close to 10 years ago actually didn't work completely (nasty production bug in an long-forgotten system, isn't that always the way?!)... what I noticed is that the ones I tried (I didn't try them all) had the same problem as mine, that is, they wouldn't replace EVERY occurrence, only the first, at least for my test case of getting "test....txt" down to "test.txt" by replacing ".." with "."... maybe I missed so regex situation? But I digress...
因此,我重写了我的实现如下。这是非常简单的,虽然我怀疑不是最快的,但我也不认为现代JS引擎的区别是重要的,除非你在一个紧密的循环中做这件事,但这总是任何事情的情况…
function replaceSubstring(inSource, inToReplace, inReplaceWith) {
var outString = inSource;
while (true) {
var idx = outString.indexOf(inToReplace);
if (idx == -1) {
break;
}
outString = outString.substring(0, idx) + inReplaceWith +
outString.substring(idx + inToReplace.length);
}
return outString;
}
希望这能帮助到别人!
仅从速度问题考虑,我相信上面链接中提供的区分大小写的示例将是目前为止最快的解决方案。
var token = "\r\n";
var newToken = " ";
var oldStr = "This is a test\r\nof the emergency broadcasting\r\nsystem.";
newStr = oldStr.split(token).join(newToken);
newStr是 “这是对紧急广播系统的一次测试。”
使用String对象的replace()方法。
正如所选答案中提到的,应该在正则表达式中使用/g标志,以便替换字符串中子字符串的所有实例。
var mystring = 'This is a string';
var newString = mystring.replace(/i/g, "a");
newString现在是'Thas as a strange '
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