Your other program must publish the Activity that is what it should
show when launched, and then you can use Context.startActivity() to
start it.
On Mar 23, 7:31 pm, Paul <[EMAIL PROTECTED]> wrote:
> Now I have a program named ATest, and i have a button in this
> program.
> I want to start another program CityListguide when i click the button
> in program ATest.
> I use the following exec() metohd, but it can't start the new program.
> The CityListguide program is stored in
> com.google.android.citylistguide package, and it is no relation with
> current program ATest.
>
> Can anyone give me some suggestion?
> Thanks.
>
> The source code as follow:
>
> OnClickListener btnClick = new OnClickListener() {
> @Override
> public void onClick(View arg0) {
> String cmd = "CityListguide";
> String[] envp = {"com.google.android.citylistguide"};
> try {
> Log.i("Runtime", "cmd = " + cmd);
> Process ps = Runtime.getRuntime().exec(cmd,
> envp, new File("/data/
> app"));
> //Process ps = Runtime.getRuntime().exec(cmd);
> ps.waitFor();
> Log.i("Runtime", "CityListguide is exected.");
> }catch(Exception e) {
> Log.i("Runtime", "e.toString() = " +
> e.toString());
> e.printStackTrace();
> }
>
> }
>
> };
>
> And the log is:
>
> INFO/Runtime(1286): cmd = CityListguide
> INFO/ProcessManager(1312): Setting environment:
> com.google.android.citylistguide
> ERROR/ProcessManager(1312): Error running CityListguide: No such file
> or directory
> INFO/Runtime(1286): CityListguide is exected.
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