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，将一个参数列表连接到一个有效的路径（因为我无法确定该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 http://regexpal.com
但该功能无法正常工作（RegEx不匹配）。
此外，Chrome状态

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

when I just use the RegEx

 (\\)$|^(\\)

但是，使用RegEx

However, using the RegEx

 (\\)$|^(\\)

工作正常。

是否太晚或者我错过了什么特别的东西？

Is it too late or did I missed something special?

提前致谢！

Leo

推荐答案

你应该使用正则表达式文字（ /.../ ）而不是字符串文字（'...'或...）调用替换。字符串有自己的反斜杠解释，在正则表达式构造函数得到破解之前就会启动，所以你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，定期expression literal： / \\ /

Match one backslash, regular expression literal: /\\/

在字符串中匹配一个反斜杠，正则表达式：'\\\\'

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

但是在正则表达式字面值中，你还必须在前面加上反斜杠正斜杠，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，如果您因某种原因与使用字符串结婚，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("(\\\\|/)$|^(\\\\|/)", "") + "/";

作为附注，当您的替代品是单个字符时，（x | y）过于苛刻;你可以使用字符类： [xy] 。在这种情况下你得到这个：

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

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

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

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