到目前为止,我从在这里看到的示例中获得了这段代码:
public class testLayout extends Activity {
final int PICK_CONTACT = 0;
ImageView image = null;
/** Called when the activity is first created. */
@Override
public void onCreate(Bundle savedInstanceState) {
super.onCreate(savedInstanceState);
setContentView(R.layout.main);
image=(ImageView)findViewById(R.id.icon);
image.setOnClickListener(onChangePerson);
}
private View.OnClickListener onChangePerson=new View.OnClickListener() {
public void onClick(View v) {
Intent intent = new Intent(Intent.ACTION_PICK, People.CONTENT_URI);
startActivityForResult(intent, PICK_CONTACT);
}
};
@Override
public void onActivityResult(int reqCode, int resultCode, Intent data) {
super.onActivityResult(reqCode, resultCode, data);
switch (reqCode) {
case (PICK_CONTACT) :
if (resultCode == Activity.RESULT_OK) {
Uri contactData = data.getData();
Cursor c = managedQuery(contactData, null, null, null, null);
if (c.moveToFirst())
{
String name = c.getString(c.getColumnIndexOrThrow(People.NAME));
}
}
break;
}
}
}
这使我可以打开一个正确显示手机上联系人的活动,然后让我选择一个联系人。但是,每次我单击联系人时,程序都会崩溃。任何想法为什么会发生这种情况?谢谢