Backslashes - Regular Expression - Javascript

Question

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)

This is the function:

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

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

SyntaxError: Invalid regular expression: /()$|^()/: Unterminated group

when I just use the RegEx

 (\\)$|^(\\)

However, using the RegEx

 (\\)$|^(\\)

works fine.

Is it too late or did I missed something special?

Thanks in advance!

Leo

Answer 1

Try this... it's a little easier to follow with the [character classes]. to match a single \ with a javascript string you need \\\\, which may be what's going on.

new Regexp('^[\\\\/]|[\\\\/]$')

You can also try the /^[\\\/]|[\\\/]$/g notation.

s = 'c:\\folder\\'
console.log(s.replace(/^[\\\/]|[\\\/]$/g, ''))

Answer 2

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.

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

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

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(/(\\|\/)$|^(\\|\/)/, "") + "/";

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

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

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("[\\\\/]$|^[\\\\/]", "") + "/";