反斜杠 - 正则表达式 - Javascript [英] Backslashes - Regular Expression - Javascript

问题描述

我想构建一个 JS 函数，将参数列表连接到有效路径(因为我无法确定路径的一部分是否带有斜杠)

I wanted to build a JS function concatting a list of arguments to a valid path (since I could not be sure whether a part of the path is given with or without slashes)

这是函数:

concatPath = function() {
    var path = "";
    for(var i = 0; i < arguments.length; i++)   {
        path += arguments[i].replace("(\|/)$|^(\|/)", "") + "/";
    }
    return path;
}

使用的 RegEx 匹配了 http://regexpal.com 上的所有开始和结束斜杠和反斜杠但是函数不能正常工作(RegEx 不匹配).此外，Chrome 声明

The used RegEx matched all beginning and ending slashes and backslashes on http://regexpal.com But the function does not work properly (RegEx does not match). Furthermore, Chrome states

语法错误:正则表达式无效:/()$|^()/:未终止组

当我只使用 RegEx 时

when I just use the RegEx

 (\)$|^(\)

但是，使用正则表达式

 (\)$|^(\)

工作正常.

是不是太晚了，还是我错过了一些特别的东西?

Is it too late or did I missed something special?

提前致谢！

狮子座

推荐答案

您应该使用正则表达式字面量 (/.../) 而不是字符串字面量 ('...' 或 "...") 在对 replace 的调用中.字符串对反斜杠有自己的解释，反斜杠在正则表达式构造函数得到破解之前就开始使用，因此您需要额外的引用级别.

You should use a regular expression literal (/.../) instead of a string literal ('...' or "...") in the call to replace. Strings have their own interpretation of backslashes that kicks in before the regular expression constructor gets a crack at it, so you need an extra level of quoting.

匹配一个反斜杠，正则表达式字面量:/\/

Match one backslash, regular expression literal: /\/

匹配一个反斜杠，字符串中的正则表达式:'\\'

Match one backslash, regular expression in a string: '\\'

但是在正则表达式文字中，您还必须在正斜杠前面放置反斜杠，因为正斜杠是整个事物的分隔符:

But in a regex literal, you also have to put backslashes in front of the forward slashes, since forward slashes are the delimiter for the whole thing:

path += arguments[i].replace(/(\|/)$|^(\|/)/, "") + "/";

或者，如果您出于某种原因习惯于使用字符串，这也应该有效:

Or, if you're married to the use of strings for some reason, this should also work:

path += arguments[i].replace("(\\|/)$|^(\\|/)", "") + "/";

作为旁注，当你的替代品是单个字符时，(x|y) 是矫枉过正；您可以只使用字符类 ([xy]).在这种情况下，您会得到:

As a side note, when your alternatives are single characters, (x|y) is overkillish; you can just use a character class ([xy]). In which case you get this:

path += arguments[i].replace(/[\/]$|^[\/]/, "") + "/";

path += arguments[i].replace("[\\/]$|^[\\/]", "") + "/";

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