Sorry, I didn't proof-read, correcting mistakes...

I'm implementing a new "import" feature into an app I have that's already
been published, designed to show your current list of contacts with a
check-mark next to them, letting you select which you would like to import
into the app. However, I'm looking to serve all platforms: 1.1, 1.5, 1.6,
2.0, which doesn't seem possible. It seems that I now have to choose whether
I want my app to be available to 2.0 or prior to 2.0, alienating all of my
prior customers.

The problem is the People class is deprecated in 2.0, and doesn't work. I
must use ContactsContract. However, I can't use ContactsContract unless I
include the 2.0 SDK which requires the user have a 2.0 phone. So it has to
be one, or the other. It doesn't seem like there is a way around it. Am I
missing something? Is there a method that will work across the board, or am
I forced to severely dwarf my potential market by choosing which SDK(s) I
want to support?

-Thanks,
Ryan Rowland

On Sun, Nov 15, 2009 at 7:27 PM, Ryan Rowland <[email protected]>wrote:

> I'm implementing a new "import" feature into an app I have that's already
> been published, designed to show your current list of contacts with a
> check-mark next to them, letting you select which you would like to import
> into the app. However, I'm looking to serve all platforms: 1.1, 1.5, 1.6,
> 2.0, which doesn't seem possible. It seems that I now have to choose whether
> I want my app to be available to 2.0 or prior to 2.0, alienating all of my
> prior customers.
>
> The problem is the People class is deprecated in 2.0, and doesn't work. I
> must use ContactsContract. However, I can't use ContactsContract unless I
> include the 2.0 SDK which requires the user have a 2.0 phone. So it has to
> be one, or the other. It doesn't seem like there is a way around it. Am I
> missing something? Is there a method that will work across the board, or am
> I forced to severely dwarf my potential market by choosing which SDKs I want
> to support?
>
> -Thanks,
> Ryan Rowland
>

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