我在此处和其他网站上阅读了相互冲突的报告，这些报告说它是/不可能显示来自另一个活动的警报对话框。我正在尝试这样做：
public class Options extends Activity {
/** Include classes */
SharedPreferences sharedPrefs;
Preferences prefs;
Location loc;
LocationSQL locSQL;
NetworkConnection netConnect;
/** Declare buttons */
Button bLocation;
Button bRefresh;
public void onCreate(Bundle savedInstanceState) {
super.onCreate(savedInstanceState);
setContentView(R.layout.options);
sharedPrefs = PreferenceManager.getDefaultSharedPreferences(Options.this);
/** Set the location for the user */
setLocationPref();
/** Refresh selection */
bRefresh = (Button) findViewById(R.id.bRefresh);
bRefresh.setOnClickListener(new View.OnClickListener() {
public void onClick(View v) {
loc = new Location(getApplicationContext());
try{
loc.locationSelection();
}catch (Exception e){
Log.d("ERROR", "Catch " + e);
}
}
});
/** Location selection */
bLocation = (Button) findViewById(R.id.bLocation);
bLocation.setOnClickListener(new View.OnClickListener() {
public void onClick(View v) {
}
});
}
}
然后我想做一些处理，如果需要调用一个 AlertDialog 框。但我得到下面的错误。我知道如果我是正确的，我正在使用已被破坏的上下文？但我想不出如何解决这个问题？
public class Location {
private static final String Context = null;
NetworkConnection nc;
SharedPreferences getSharedPrefs;
SharedPreferences putSharedPrefs;
NetworkConnection netConnect;
Options opts;
Context context;
public Location (Context arg)
{
context = arg;
}
public void locationSelection(){
message();
}
public void message(){
AlertDialog.Builder alertDialogBuilder = new AlertDialog.Builder(context);
alertDialogBuilder.setTitle("Your Title");
AlertDialog alertDialog = alertDialogBuilder.create();
alertDialog.show();
}
}
捕获 android.view.WindowManager$BadTokenException: Unable to add window -- token null 不适用于应用程序。
非常感谢您的帮助 ：-）