Rewrite Javascript path String

Question

In Javascript, I have a string for a path that looks like:

/xxx:Level1/yyy:Level2/xxx:Level3/ccc:Level4

The prefix may or may not be there for each level. I need to create a new string which eliminates the prefix on each folder level, something like:

/Level1/Level2/Level3/Level4

OK. I've done something like the following, but I think perhaps with regex it could be made more compact. How could I do that?

var aa = "/xxx:Level1/yyy:Level2/xxx:Level3/ccc:Level4"

var bb = aa.split("/").filter(String);
var reconstructed = "";

for( var index in bb )
{
 var dirNames = bb[index].split(":");
 if(dirNames.length==1) reconstructed += "/" + dirNames[0];
 else if(dirNames.length==2) reconstructed += "/" + dirNames[1];
 }

Wow! You guys are amazingly fast.

George Hernando
– George Hernando

2011-09-21 17:56:21 +00:00
Commented Sep 21, 2011 at 17:56 — George Hernando
– George Hernando, Commented Sep 21, 2011 at 17:56

jfriend00 · Accepted Answer · 2011-09-21 17:51:23Z

4

You can use regex like this:

var str = "/xxx:Level1/yyy:Level2/xxx:Level3/ccc:Level4";
var out = str.replace(/\/[^:\/]+:/g, "/");
alert(out);

This matches:

/
followed by one or more characters that is not a : or a /
followed by a :
and replaces all that with a / effectively eliminating the xxx:

Demo here: http://jsfiddle.net/jfriend00/hbUkz/

answered Sep 21, 2011 at 17:51

jfriend00

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Comments

SLaks · Accepted Answer · 2011-09-21 17:50:04Z

2

Like this:

var bb = aa.replace(/\/[a-z]+:/g, '/');

Change the [a-z] to include any characters that might appear in the prefix, or just use [^\/:].

answered Sep 21, 2011 at 17:50

SLaks

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Brendan · Accepted Answer · 2011-09-21 17:53:12Z

2

var a = "/xxx:Level1/yyy:Level2/xxx:Level3/ccc:Level4";
var b = a.replace(/\/[^\/]*:/g, "/");

answered Sep 21, 2011 at 17:53

Brendan

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Comments

Rob W · Accepted Answer · 2011-09-21 17:50:13Z

1

aa = aa.replace(/\/[^:\/]\:/g, "/");

This function will replace every occurence of "/xxx:" by "/" using a RE, where xxx: is a prefix.

answered Sep 21, 2011 at 17:50

Rob W

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