Get the current script file name

Just use the PHP magic constant __FILE__ to get the current filename.

But it seems you want the part without .php. So...

basename(__FILE__, '.php'); 

A more generic file extension remover would look like this...

function chopExtension($filename) {
    return pathinfo($filename, PATHINFO_FILENAME);
}

var_dump(chopExtension('bob.php')); // string(3) "bob"
var_dump(chopExtension('bob.i.have.dots.zip')); // string(15) "bob.i.have.dots"

Using standard string library functions is much quicker, as you'd expect.

function chopExtension($filename) {
    return substr($filename, 0, strrpos($filename, '.'));
}

You can investigate script collection at:

var scripts = document.getElementsByTagName("script");

For each element in the returned scripts array you can access its src attribute.

The currently executing include file will always be the last one in the scripts array. So you can access it at scripts[scripts.length-1].

Of course this will only work at time of initial code run and would not be useful for example within a function that is called after initial script is loaded, so if you need the value available later, you would need to save it to a variable.

// With help from huynhjl
// http://.com/questions/8129185#8131613

import scala.util.matching.Regex.MatchIterator

object ScriptName {
    val program = {
        val filenames = new RuntimeException("").getStackTrace.map { t => t.getFileName }
        val scala = filenames.indexOf("NativeMethodAccessorImpl.java")

        if (scala == -1)
            "<console>"
        else
            filenames(scala - 1)
    }

    def main(args: Array[String]) {
        val prog = program
        println("Program: " + prog)
    }
}

Rosetta Code