我一直试图在用户试图退出活动时显示“您想退出吗?”类型的对话框。
但是我找不到合适的API钩子。onuserleavehint()最初看起来很有希望,但我找不到一种方法来阻止活动完成。
我一直试图在用户试图退出活动时显示“您想退出吗?”类型的对话框。
但是我找不到合适的API钩子。onuserleavehint()最初看起来很有希望,但我找不到一种方法来阻止活动完成。
当前回答
Just put this code in your first activity
@Override
public void onBackPressed() {
if (drawerLayout.isDrawerOpen(GravityCompat.END)) {
drawerLayout.closeDrawer(GravityCompat.END);
}
else {
// if your using fragment then you can do this way
int fragments = getSupportFragmentManager().getBackStackEntryCount();
if (fragments == 1) {
new AlertDialog.Builder(this)
.setMessage("Are you sure you want to exit?")
.setCancelable(false)
.setPositiveButton("Yes", new DialogInterface.OnClickListener() {
public void onClick(DialogInterface dialog, int id) {
finish();
}
})
.setNegativeButton("No", null)
.show();
} else {
if (getFragmentManager().getBackStackEntryCount() > 1) {
getFragmentManager().popBackStack();
} else {
super.onBackPressed();
}
}
}
}
其他回答
已修改@user919216代码..并使其与WebView兼容
@Override
public void onBackPressed() {
if (webview.canGoBack()) {
webview.goBack();
}
else
{
AlertDialog.Builder builder = new AlertDialog.Builder(this);
builder.setMessage("Are you sure you want to exit?")
.setCancelable(false)
.setPositiveButton("Yes", new DialogInterface.OnClickListener() {
public void onClick(DialogInterface dialog, int id) {
finish();
}
})
.setNegativeButton("No", new DialogInterface.OnClickListener() {
public void onClick(DialogInterface dialog, int id) {
dialog.cancel();
}
});
AlertDialog alert = builder.create();
alert.show();
}
}
在Android 2.0+中,这看起来像:
@Override
public void onBackPressed() {
new AlertDialog.Builder(this)
.setIcon(android.R.drawable.ic_dialog_alert)
.setTitle("Closing Activity")
.setMessage("Are you sure you want to close this activity?")
.setPositiveButton("Yes", new DialogInterface.OnClickListener()
{
@Override
public void onClick(DialogInterface dialog, int which) {
finish();
}
})
.setNegativeButton("No", null)
.show();
}
在早期版本中,它看起来像:
@Override
public boolean onKeyDown(int keyCode, KeyEvent event) {
//Handle the back button
if(keyCode == KeyEvent.KEYCODE_BACK) {
//Ask the user if they want to quit
new AlertDialog.Builder(this)
.setIcon(android.R.drawable.ic_dialog_alert)
.setTitle(R.string.quit)
.setMessage(R.string.really_quit)
.setPositiveButton(R.string.yes, new DialogInterface.OnClickListener() {
@Override
public void onClick(DialogInterface dialog, int which) {
//Stop the activity
YourClass.this.finish();
}
})
.setNegativeButton(R.string.no, null)
.show();
return true;
}
else {
return super.onKeyDown(keyCode, event);
}
}
我喜欢@GLee的方法,并像下面这样使用片段。
@Override
public void onBackPressed() {
if(isTaskRoot()) {
new ExitDialogFragment().show(getSupportFragmentManager(), null);
} else {
super.onBackPressed();
}
}
使用片段的对话框:
public class ExitDialogFragment extends DialogFragment {
@Override
public Dialog onCreateDialog(Bundle savedInstanceState) {
return new AlertDialog.Builder(getActivity())
.setTitle(R.string.exit_question)
.setPositiveButton(android.R.string.yes, new DialogInterface.OnClickListener() {
@Override
public void onClick(DialogInterface dialog, int which) {
getActivity().finish();
}
})
.setNegativeButton(android.R.string.no, new DialogInterface.OnClickListener() {
@Override
public void onClick(DialogInterface dialog, int which) {
getDialog().cancel();
}
})
.create();
}
}
我更喜欢双击后退按钮退出,而不是退出对话框。
在这个解决方案中,当第一次返回时,它会显示一个吐司,警告另一个回按将关闭应用程序。在这个例子中,不到4秒。
private Toast toast;
private long lastBackPressTime = 0;
@Override
public void onBackPressed() {
if (this.lastBackPressTime < System.currentTimeMillis() - 4000) {
toast = Toast.makeText(this, "Press back again to close this app", 4000);
toast.show();
this.lastBackPressTime = System.currentTimeMillis();
} else {
if (toast != null) {
toast.cancel();
}
super.onBackPressed();
}
}
令牌来自:http://www.androiduipatterns.com/2011/03/back-button-behavior.html
在中国,大多数App会通过“点击两次”确认退出:
boolean doubleBackToExitPressedOnce = false;
@Override
public void onBackPressed() {
if (doubleBackToExitPressedOnce) {
super.onBackPressed();
return;
}
this.doubleBackToExitPressedOnce = true;
Toast.makeText(this, "Please click BACK again to exit", Toast.LENGTH_SHORT).show();
new Handler().postDelayed(new Runnable() {
@Override
public void run() {
doubleBackToExitPressedOnce=false;
}
}, 2000);
}