Grunt Task - Get Filename from path without extension

Question

Is it possible to get the filename without the extension from the src filepath.

As an example, let's say the src file is my-file.png - located at images/my-file.png.

In my task I have this at the moment:

var processName = options.processName || function (name) { return name; };
var filename = processName(filepath);

When I reference filename for output it returns:

images/my-file.png

I want to only return the actual filename, without the path and without the extension:

my-file.png

How can I achieve this?

Answer 1

Might be pretty old but if someone else finds this SO., in reply for @user3143218 's comment : slice(0, -4) will remove the last 4 characters from the name, so for the example my-file.png we will get my-file but for script.js we will get scrip . I suggest using a regex removing everything from the last dot.

Answer 2

You could use a regex like this:

var theFile = filename.match(/\/([^/]*)$/)[1];
var onlyName = theFile.substr(0, theFile.lastIndexOf('.')) || theFile;

That should give you my-file . The regex gives you the string after the last forward slash, and the next line removes everything after the last dot (and the dot).

Answer 3

Thanks to Andeersg's answer below I was able to pull this off. It might not be the best solution but it works. Final code is:

var processName = options.processName || function (name) { return name; };
var filename = processName(filepath);
var theFile = filename.match(/\/([^/]*)$/)[1];
var onlyName = theFile.slice(0, -4);

Now onlyName will return:

my-file