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Creating a Wrapper for Firebug's Console [Tips & Tricks] | Posted at 12:20 PM

Thanks to Dimitar over at Frag on how to create a wrapper for Firebug's console, I was able to implement better debugging in my jQuery plugin. The one quip I had was that I needed to directly pass everything to the log function. The way Dimitar did it, it wrote a log line for each argument. I was passing formatted strings.

Here are the offending lines:

for (var i = 0, l = arguments.length; i < l; i++)
	console.log(arguments[i]);

And here's what you replace it with:

console.log.apply(this, arguments);

BAM! Works like a charm.

Thanks Dimitar!

Filed Under: Coding Tips and Tricks

Comments

hey - thanks for the tip. I should have read the firebug source / manual to see if there was a more appropriate method than looping through vars :)

updated and credited.

tried hours and hours to get this... thank you so much! really works like charm ;)

erm - you may want to update it as the apply breaks on Safari 4's console.log (probably not a normal function).

added a try block instead - check the original post :)

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About the Author

Kamran
Divide by Zero is Kamran Ayub's personal blog. Kamran owns and operates Intrepid Studios, a web design and development firm based in Minneapolis. Despite the above picture, Kamran is usually an intelligent person. Please forgive him.

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