On Oct 20, 11:35 pm, John J Barton <[email protected]>
wrote:
> On Oct 20, 7:01 pm, Greg <[email protected]> wrote:
>
> > Hello,
>
> > So I'm debugging some javascript on a site I'm working on, in one part
> > I get the current year with getFullYear(). The value seems to be
> > coming out correct, but firebug keeps telling me that g.getFullYear is
> > not a function in the console view. The only line I use getFullYear
> > is:
>
> > function getCurrentYear() {
> > today = new Date();
> > return today.getFullYear();
> > }
>
> > Any suggestions on what I'm missing?
>
> Yes, the source line that includes the string "g.getFullYear()".
> Also try Console > mini-menu options > Show stack trace.
>
> Also try with only one tab open in Firefox.
>
> jjb
>
> The error message should be linked to the source line;
>
> > I'm using Firefox 3.5.3 on Vista 64 Home. Any suggestions are welcome.
>
> > Thank you for your time
There is no source line in my code that includes "g.getFullYear", the
stack trace shows:
g.getFullYear is not a function
anonymous(Object date=2009-10-18 dateOnly=false setDateOnly=true)
core,med...ault.I.js (line 86)
anonymous()core,med...ault.I.js (line 109)
anonymous()core,med...ault.I.js (line 109)
anonymous(Object, listEvents(feedRoot), handleGDError(e), function(),
Object name=l, Object)core,med...ault.I.js (line 121)
anonymous(Object, listEvents(feedRoot), handleGDError(e), Object
name=h)core,med...ault.I.js (line 635)
loadCalendar("http://www.google.com/calendar/feeds/
[email protected]/public/full")
index.php?m=store (line 105)
index.php?m=store()
Now I've searched through every line of code in my files and have
found no reference to g.getFullYear, and only 1 reference to
getFullYear, which I posted earlier. I've tried with the site being
the only tab in firefox, but still same situation. And the message
does not take me to any source line, only opens/closes the stack trace
tree. Thank you for your suggestions but unless I have misinterpreted
them (which is possible) I don't believe they have helped so far.
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