On Tue, Jul 5, 2011 at 2:57 AM, Poetro wrote:
> 2011/7/5 David Marrs :
> > Why do you have a 2nd arg in your lambda called undefined?
> Because he might want to use the undefined value, and in case someone
> in the global scope created a variable named undefined, then it can
> cause trouble.
>
I
2011/7/5 David Marrs :
> Why do you have a 2nd arg in your lambda called undefined?
Because he might want to use the undefined value, and in case someone
in the global scope created a variable named undefined, then it can
cause trouble.
>
> Also, you will want to check for the existence of console.
Why do you have a 2nd arg in your lambda called undefined?
Also, you will want to check for the existence of console.log before you
call it, otherwise you will get a runtime error if firebug is switched off.
On 4 Jul 2011 20:30, Sam3k wrote:
Why does the following scrip
On Jul 4, 7:34 pm, "David Marrs" wrote:
[...]
> P.S. Please let me know if my MUA is a bit rubbish for mailing lists. I
> noticed the formatting of my email in your quotes was a bit ugly.
I'm using Google Groups, your replies are not well formatted at all:
http://groups.google.com/group/jsme
Doh, oh course it doesn't show in window as it is part of the Firebug
scope (not window) as you mentioned Poetro.
Thanks you two.
On Jul 4, 4:04 pm, Sam3k wrote:
> Oh wow, I was checking the window globals and this convenience
> function is not seen there. Hehe, learned something new.
>
> Thanks
Oh wow, I was checking the window globals and this convenience
function is not seen there. Hehe, learned something new.
Thanks so much Poetro.
On Jul 4, 3:58 pm, Poetro wrote:
> To be able to use it, you have to use the global context for it,
> namely window.debug each and every time, or you wil
To be able to use it, you have to use the global context for it,
namely window.debug each and every time, or you will be calling
Firebug's debug function, as that is the current scope.
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debug is a convinence function in firebug so you may not be able to
overwrite it, or get it to user your user defined function.
http://getfirebug.com/wiki/index.php/Command_Line_API#debug.28fn.29
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Try adding these four lines immediately before the debug.log() call inside
the wrapper function (the one that fails):
console.log( _debug );
console.log( window.debug );
console.log( debug );
console.log( debug.toString() );
That may lead to this interesting idea... What if you change these lines
Environment:
FireFox 5
Firebug 1.7.3
On Jul 4, 3:30 pm, Sam3k wrote:
> Why does the following script works within the application code but it
> doesn't when I type it in the firebug console. It throws the following
> error:
>
> TypeError: debug.log is not a function
>
> My goal with this scri
Why does the following script works within the application code but it
doesn't when I type it in the firebug console. It throws the following
error:
TypeError: debug.log is not a function
My goal with this script is to make an object from where I can log
things to the console, store javascript sp
On Fri, Jul 1, 2011 at 8:13 PM, Jason Mulligan wrote:
> yes i am. extjs/jquery are fantastic examples of this paradigm gone
> wrong. the syntax ends up on multiple lines as you try to figure out
> what's what, and no IDE can accurately parse the docblock/expression/
> literal to say Param1 is ...
jsdoc-toolkit [http://code.google.com/p/jsdoc-toolkit/] already has an
approach to mark up object parameters in the form of:
/**
* @param userInfo Information about the user.
* @param userInfo.name The name of the user.
* @param userInfo.email The email of the user.
*/
--
Poetro
--
To
On 4 Jul 2011 06:16, RobG wrote:
Choosing to use objects with defined property names instead of formal
parameters isn't a language feature, it's a design decision. That IDEs
don't support it likely means it is too difficult to do easily
I think IDEs already use docblocks to get param info. It s
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