Getting stack trace for an error when running code from the console in Chrome

Question

I am calling a function from the console but when it throws an exception I do not receive a stack trace like I would if the code was executed normally.

Is there a way I can modify my command (perhaps with try/catch) to have it provide me this information?

to clarify:

page.js:

function otherStuff() { return ['a','b',undefined,'c'];
function doStuff() {
    var x = otherStuff();
    var z = parseInt(x[2]); // this will throw an error
}

console, after loading a html page that links page.js

> otherStuff();

I get no line number from the Error that is returned to me. When running it from the page (instead of the console) i would receive a line number and a stack trace.

what is the command? what version of chrome? (what is your mother's maiden name?) — Naftali
– Naftali, Commented Oct 18, 2011 at 19:40

millimoose · Accepted Answer · 2011-10-18 19:43:57Z

11

Although verbose, this will print the stack trace of an interactive error in the Chrome JS console:

try { 
    throw new Error(); 
} catch (e) { 
    console.error(e.stack); 
}

Unfortunately this won't work if a non-Error object is thrown.

answered Oct 18, 2011 at 19:43

millimoose

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3 Comments

Steven Lu Over a year ago

This works quite well. The data is formatted differently but all there (most important to me is line number). If a non Error object is thrown my stacktrace information will be lost to me, correct? Since we are retrieving it from the Error object in this case

millimoose Over a year ago

Yes, same for any custom error object that isn't based on JS's standard Error objects. I tried setting a breakpoint in the catch clause when throwing a string, trying to see if Chrome creates some sort of proper exception object internally one could use – no such luck I'm afraid.

Steven Lu Over a year ago

Just a note: MobileSafari does not appear to provide a "stack" property to its errors.

Naftali · Accepted Answer · 2011-10-18 20:08:31Z

2

You have an error in your code.

You are missing a closing brace:

function otherStuff() { return ['a','b',undefined,'c']; //} where am i?
function doStuff() {
    var x = otherStuff();
    var z = parseInt(x[2]); // this will throw an error
}

Side point:

parseInt(undefined) does not throw an error. case in point: http://jsfiddle.net/maniator/Zequj/2/

answered Oct 18, 2011 at 20:08

Naftali

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