How to get JavaScript caller function line number? How to get JavaScript caller source URL?

This works for me in chrome/QtWebView

function getErrorObject(){
    try { throw Error('') } catch(err) { return err; }
}

var err = getErrorObject();
var caller_line = err.stack.split("\n")[4];
var index = caller_line.indexOf("at ");
var clean = caller_line.slice(index+2, caller_line.length);

kangax's solution introduces unnecessary try..catch scope. If you need to access the line number of something in JavaScript (as long as you are using Firefox or Opera), just access (new Error).lineNumber.

I was surprised that most of these answers assumed that you wanted to handle an error rather than just output helpful debug traces for normal cases as well.

For example, I like using a console.log wrapper like this:

consoleLog = function(msg) {//See https://stackoverflow.com/a/27074218/470749
    var e = new Error();
    if (!e.stack)
        try {
            // IE requires the Error to actually be thrown or else the 
            // Error's 'stack' property is undefined.
            throw e;
        } catch (e) {
            if (!e.stack) {
                //return 0; // IE < 10, likely
            }
        }
    var stack = e.stack.toString().split(/\r\n|\n/);
    if (msg === '') {
        msg = '""';
    }
    console.log(msg, '          [' + stack[1] + ']');        
}

This ends up printing an output such as the following to my console:

1462567104174 [getAllPosts@http://me.com/helper.js:362:9]

See https://stackoverflow.com/a/27074218/ and also A proper wrapper for console.log with correct line number?