How to get the file-path of the currently executing javascript code

I'm trying to do something like a C #include "filename.c", or PHP include(dirname(__FILE__)."filename.php") but in javascript. I know I can do this if I can get the URL a js file was loaded from (e.g. the URL given in the src attribute of the tag). Is there any way for the javascript to know that?

Alternatively, is there any good way to load javascript dynamically from the same domain (without knowing the domain specifically)? For example, lets say we have two identical servers (QA and production) but they clearly have different URL domains. Is there a way to do something like include("myLib.js"); where myLib.js will load from the domain of the file loading it?

Sorry if thats worded a little confusingly.

Answers:

Answer

Within the script:

var scripts = document.getElementsByTagName("script"),
    src = scripts[scripts.length-1].src;

This works because the browser loads and executes scripts in order, so while your script is executing, the document it was included in is sure to have your script element as the last one on the page. This code of course must be 'global' to the script, so save src somewhere where you can use it later. Avoid leaking global variables by wrapping it in:

(function() { ... })();
Answer

All browsers except Internet Explorer (any version) have document.currentScript, which always works always (no matter how the file was included (async, bookmarklet etc)).

If you want to know the full URL of the JS file you're in right now:

var script = document.currentScript;
var fullUrl = script.src;

Tadaa.

Answer

I just made this little trick :

window.getRunningScript = () => {
    return () => {      
        return new Error().stack.match(/([^ \n])*([a-z]*:\/\/\/?)*?[a-z0-9\/\\]*\.js/ig)[0]
    }
}
console.log('%c Currently running script:', 'color: blue', getRunningScript()())

screenshot

? Works on: Chrome, Firefox, Edge, Opera

Enjoy !

Answer

I've more recently found a much cleaner approach to this, which can be executed at any time, rather than being forced to do it synchronously when the script loads.

Use stackinfo to get a stacktrace at a current location, and grab the info.file name off the top of the stack.

info = stackinfo()
console.log('This is the url of the script '+info[0].file)
Answer

I've coded a simple function which allows to get the absolute location of the current javascript file, by using a try/catch method.

// Get script file location
// doesn't work for older browsers

var getScriptLocation = function() {
    var fileName    = "fileName";
    var stack       = "stack";
    var stackTrace  = "stacktrace";
    var loc     = null;

    var matcher = function(stack, matchedLoc) { return loc = matchedLoc; };

    try { 

        // Invalid code
        0();

    }  catch (ex) {

        if(fileName in ex) { // Firefox
            loc = ex[fileName];
        } else if(stackTrace in ex) { // Opera
            ex[stackTrace].replace(/called from line \d+, column \d+ in (.*):/gm, matcher);
        } else if(stack in ex) { // WebKit, Blink and IE10
            ex[stack].replace(/at.*?\(?(\S+):\d+:\d+\)?$/g, matcher);
        }

        return loc;
    }

};

You can see it here.

Answer

Regardless of whether its a script, a html file (for a frame, for example), css file, image, whatever, if you dont specify a server/domain the path of the html doc will be the default, so you could do, for example,

<script type=text/javascript src='/dir/jsfile.js'></script>

or

<script type=text/javascript src='../../scripts/jsfile.js'></script>

If you don't provide the server/domain, the path will be relative to either the path of the page or script of the main document's path

Answer

I may be misunderstanding your question but it seems you should just be able to use a relative path as long as the production and development servers use the same path structure.

<script language="javascript" src="js/myLib.js" />
Answer

Refining upon the answers found here:

little trick

getCurrentScript and getCurrentScriptPath

I came up with the following:

//Thanks to https://stackoverflow.com/a/27369985/5175935
var getCurrentScript = function () {

    if ( document.currentScript && ( document.currentScript.src !== '' ) )
        return document.currentScript.src;
    var scripts = document.getElementsByTagName( 'script' ),
        str = scripts[scripts.length - 1].src;
    if ( str !== '' )
        return src;
    //Thanks to https://stackoverflow.com/a/42594856/5175935
    return new Error().stack.match(/(https?:[^:]*)/)[0];

};

//Thanks to https://stackoverflow.com/a/27369985/5175935
var getCurrentScriptPath = function () {
    var script = getCurrentScript(),
        path = script.substring( 0, script.lastIndexOf( '/' ) );
    return path;
};
Answer

The accepted answer here does not work if you have inline scripts in your document. To avoid this you can use the following to only target <script> tags with a [src] attribute.

/**
 * Current Script Path
 *
 * Get the dir path to the currently executing script file
 * which is always the last one in the scripts array with
 * an [src] attr
 */
var currentScriptPath = function () {

    var scripts = document.querySelectorAll( 'script[src]' );
    var currentScript = scripts[ scripts.length - 1 ].src;
    var currentScriptChunks = currentScript.split( '/' );
    var currentScriptFile = currentScriptChunks[ currentScriptChunks.length - 1 ];

    return currentScript.replace( currentScriptFile, '' );
}

This effectively captures the last external .js file, solving some issues I encountered with inline JS templates.

Answer

Refining upon the answers found here I came up with the following:

getCurrentScript.js

var getCurrentScript = function () {
  if (document.currentScript) {
    return document.currentScript.src;
  } else {
    var scripts = document.getElementsByTagName('script');
    return scripts[scripts.length-1].src;

  }
};

module.exports = getCurrentScript;

getCurrentScriptPath.js

var getCurrentScript = require('./getCurrentScript');

var getCurrentScriptPath = function () {
  var script = getCurrentScript();
  var path = script.substring(0, script.lastIndexOf('/'));
  return path;
};

module.exports = getCurrentScriptPath;

BTW: I'm using CommonJS module format and bundling with webpack.

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