Is there a way to improve this Javascript Regex replace

Question

I am trying to clean-up a windows folder path using the following Javascript.




    function StandardizeFolderPath(inFolderPath) {
        var outFolderPath = inFolderPath.replace(/^\s+|\s+$/g, "");
        outFolderPath = outFolderPath.replace(/\\\s*/g, "\\");
        outFolderPath = outFolderPath.replace(/\s*\\/g, "\\");
        outFolderPath = outFolderPath.replace(/\\{2,}/, "\\");

        alert("^" + inFolderPath + "$           " + "^" + outFolderPath + "$");

        return outFolderPath;
    }

    function Test_StandardizeFolderPath() {
        StandardizeFolderPath("D:\\hel   xo  \\");
        StandardizeFolderPath("D:\\hello  \\        ");
        StandardizeFolderPath("D:\\hello  \\        \\");
        StandardizeFolderPath("D:\\hello  \\        mike \\");
        StandardizeFolderPath("  D:\\hello  \\        jack \\");
        StandardizeFolderPath("  D:\\hello Multiple Slashes \\\\");
    }

Each replace does specific parts:

Remove spaces from fron and back
Replace any "\ " with "\"
Replace any " \"
Replace multiple occurrences of "\" with a single.

It gets the job done, but I want to know if there is a better way (with explanation)

@dystroy 4th gets tested with this statement "D:\\hello \\ \\" — Pranav Shah
– Pranav Shah, Commented Sep 18, 2013 at 17:26
Partially only. There could be no space in between. But you know what can occur in the real cases, I was just pointing this omission relative to your rules. — Denys Séguret
– Denys Séguret, Commented Sep 18, 2013 at 17:29
@dystroy Thanks for pointing that out. Does the code that you provided below also solve that. — Pranav Shah
– Pranav Shah, Commented Sep 18, 2013 at 17:35

Denys Séguret · Accepted Answer · 2014-06-08 06:24:01Z

2

You could merge three of your replacements :

function StandardizeFolderPath(inFolderPath) {
    return inFolderPath.replace(/^\s+|\s+$/g, "").replace(/(\s*\\\s*)+/g, "\\");
}

Here's what /(\s*\\\s*)+/g means :

NODE                     EXPLANATION
--------------------------------------------------------------------------------
/                        start of regex literal
--------------------------------------------------------------------------------
  (                        group \1:
--------------------------------------------------------------------------------
    \s*                      whitespace (\n, \r, \t, \f, and " ") (0
                             or more times (matching the most amount
                             possible))
--------------------------------------------------------------------------------
    \\                       '\'
--------------------------------------------------------------------------------
    \s*                      whitespace (\n, \r, \t, \f, and " ") (0
                             or more times (matching the most amount
                             possible))
--------------------------------------------------------------------------------
  )+                       end of \1 . The + makes it work for one or
                           more occurences
--------------------------------------------------------------------------------
/                         end of regex literal
--------------------------------------------------------------------------------
g                         executes more than once

References :

edited Jun 8, 2014 at 6:24

answered Sep 18, 2013 at 17:16

Denys Séguret

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11 Comments

Denys Séguret Over a year ago

Well... not the first one that I've left untouched.

Peter Boughton Over a year ago

You're missing the {2,} from the fourth one.

Denys Séguret Over a year ago

@PeterBoughton No, I don't. There's the + for that. It also takes care of "\\\\" or "\\ \\".

Pranav Shah Over a year ago

@dystroy Thanks that does help. Could you please add a little explanation to your answer so that I can understand better. Significance of "/" at the beginning and "/g" at the end. Also how the "()+" works.

Denys Séguret Over a year ago

@PranavShah I detailed and added some links you should read to get a better understanding

|

6502 · Accepted Answer · 2013-09-18 19:43:11Z

0

A single regular expression

s.replace(/ *(\\)(\\? *)+|^ *| *$/g, "$1")

seems to do the trick.

The idea is that of a block made up with spaces followed by a backslash followed by a sequence of other backslashes or spaces you want to keep only the backslash.

The other cases are for removing initial and ending spaces.

edited Sep 18, 2013 at 19:43

answered Sep 18, 2013 at 17:39

6502

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4 Comments

Denys Séguret Over a year ago

I don't think it works for now. Try "\\ a". It can be fixed of course but I find this hard to maintain and check.

6502 Over a year ago

@dystroy: true, I don't see how to do everything with just one pass without look-behind zero-width assertions that are not supported everywhere. Added a second pass.

Denys Séguret Over a year ago

In fact I think that those assertions are supported in no browser.

6502 Over a year ago

@dystroy: while dining I got the idea that using a capturing group could make it... seems it works.

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