You already asked this question and you received an answer... What more do
you want?  The problem is that nothing is calling your LaunchComponent()
method...

As you were told before, override the onCreate() method and call your
LaunchComponent() method from there.

Thanks,
Justin Anderson
MagouyaWare Developer
http://sites.google.com/site/magouyaware


On Fri, Jun 29, 2012 at 12:32 PM, Jaison Brooks <jaisonbro...@gmail.com>wrote:

> Ok, im feeling really stupid right now, for some reason i cannot figure
> out how to create just a signal application that when the user clicks the
> icon from the home screen, it opens a different application. For instance,
> you click on my app and it takes you to your email icon, or settings or
> something along those lines.
>
> Here is some code i got so far, but its not doing anything when i open the
> application, just showing blank white screen with the label of my
> application.
>
> JAVA File
>
> package com.xxx.xxx.xxx;
>
> import android.app.Activity;
> import android.content.Intent;
> import android.content.pm.PackageManager;
>
>
> public class MainActivity extends Activity {
>
>     public void  LaunchComponent (String packageName, String name){
>     Intent task = new
>             Intent(Intent.ACTION_MAIN);
>     PackageManager manager =
>             getPackageManager ();
>     task = manager.getLaunchIntentForPackage("com.xxx.xxx.xxx");
>     task.addCategory(Intent.CATEGORY_HOME);
>     startActivity(task);
>
>     }
> }
>
> i do NOT have a layout file, i didnt think i would need one, since i am
> not creating a UI at all.
> Any recommendations would help alot, i've tried a few links, and i just
> cannot get something to compile right.
>
> http://stackoverflow.com/questions/2780102/open-another-application-from-your-own-intent
>
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