Modify your class that it takes a Context as a constructor or method
parameter.   This is not unreasonable given that it's accessing
resources.

On Dec 3, 11:04 am, Kostya Vasilyev <[email protected]> wrote:
> Indeed - createPackageContext is not a static method.
>
> The compiler is correct :)
>
> Calling createPackageContext requires that you have a Context object. This
> can be a Context that corresponds to your package.
>
> -- Kostya
>
> 3 декабря 2011 г. 23:01 пользователь Matt Clark 
> <[email protected]>написал:
>
>
>
>
>
>
>
> > Still getting the error:
> > Cannot make a static reference to the non-static method
> > createPackageContext(String, int) from the type Context
>
> > On Dec 3, 1:52 pm, Kostya Vasilyev <[email protected]> wrote:
> > > createPackageContext takes a String as the first parameter, so it shoud
> > be
> > > createPackageContext("your.package.name.here"...). Note the double
> > quotes.
>
> > > Don't think you need the "ignore security" flag in this case, just to
> > load
> > > the resources.
>
> > > -- Kostya
>
> > > 3 декабря 2011 г. 22:45 пользователь Matt Clark <[email protected]
> > >написал:
>
> > > > I have no idea how to set that up, I found an example that would use:
>
> > > > Context r =
>
> > Context.createPackageContext(tinyClark.android.libraries.LanguagePack,Conte
> > xt.CONTEXT_IGNORE_SECURITY);
>
> > > > But i get an error that tinyClark.android.libraries.LanguagePack
> > > > cannot be resolved to a variable, and when i try to fix it it creates
> > > > a new class in tinyClark.android.*
>
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