无法驳回AlertDialog时pressing菜单键第二次

问题描述

我在这里呈现出 AlertDialog 通过pressing菜单键和显示弹出对话框，但当我试图 dissmiss 对话框由$ p $再次pssing菜单键第二次它似乎菜单键不触发。

这是我的code此

  @燮pressLint（{NewApi，NewApi，NewApi，NewApi}）
    @覆盖
 公共布尔的onkeydown（INT键code，KeyEvent的事件）{
  AlertDialog.Builder alertDialogBu​​ilder =新AlertDialog.Builder（背景）;    如果（键code == KeyEvent.KEY code_MENU＆放大器;＆安培; event.getAction（）== 0）{   如果（！打开）{
   showDialogItem（）;
   开放= TRUE;
   }   否则，如果（打开）{
   alertDialog.dismiss（
  开放= FALSE;
   }返回true;            }    返回super.onKeyDown（键code，事件）;    } 公共无效showDialogItem（）{
                     AlertDialog.Builder alertDialogBu​​ilder =新AlertDialog.Builder（背景）;
  alertDialogBu​​ilder .setItems（R.array.select_dialog_items，新DialogInterface.OnClickListener（）
     {
  公共无效的onClick（DialogInterface对话，诠释它）{
 。的String []项目= getResources（）getStringArray（R.array.select_dialog_items）;
 新AlertDialog.Builder（MainActivity.this）
 .setMessage（您选择：这+ +，+项目[它]）
                                            。显示（）;
                                }
                            }）;
   alertDialog = alertDialogBu​​ilder.create（）;   WindowManager.LayoutParams LP =新WindowManager.LayoutParams（）;
LP = alertDialog.getWindow（）的getAttributes（）。
 lp.gravity = Gravity.TOP | Gravity.LEFT;
   lp.x = -300;
lp.gravity = Gravity.BOTTOM;
   。alertDialog.getWindow（）setAttributes（LP）;    alertDialog.show（）;
    alertDialog.getWindow（）的setLayout（130，220）。    }
 

解决方案

 得到这个初始化公共静态INT X = 0;公共布尔的onkeydown（INT键code，KeyEvent的事件）{如果（键code == KeyEvent.KEY code_MENU＆放大器;＆安培; event.getAction（）== 0）
{AlertDialog.Builder建设者=新AlertDialog.Builder（本）
.setOnKeyListener（新DialogInterface.OnKeyListener（）
{
 @覆盖
 公共布尔安其（DialogInterface对话，诠释关键code，KeyEvent的事件）
{         如果（键code == KeyEvent.KEY code_MENU＆放大器;＆安培; event.getAction（）== 0）
{                      如果（X == 1）
                      {
                        alert.dismiss（）;
                        X = 0;
                      }
             返回true;
             }
             返回false;
             }
             }）;
如果（X == 0）
{
builder.setMessage（登录信息修改成功......）;
建设者.setCancelable（真）;
builder.setPositiveButton（OK，新DialogInterface.OnClickListener（）{
公共无效的onClick（DialogInterface对话，诠释的id）{
      System.exit（0）;
 }
     }）;
    警报= builder.create（）;
    alert.show（）;
    X = 1;
          }
          返回true;         }
 

Here I am showing a AlertDialog By pressing a menu key and its displaying the popup dialogs but when i am trying to dissmiss the dialog by pressing the menu key again 2nd time its seems the menu key doesn't triggered.

here is my code for this

     @SuppressLint({ "NewApi", "NewApi", "NewApi", "NewApi" })
    @Override
 public boolean onKeyDown(int keyCode, KeyEvent event) {
  AlertDialog.Builder alertDialogBuilder = new AlertDialog.Builder(context);

    if (keyCode == KeyEvent.KEYCODE_MENU && event.getAction() == 0 ) {

   if(!open){
   showDialogItem();
   open=true;


   }

   else if(open){
   alertDialog.dismiss(
  open =false;
   }return true;

            }

    return super.onKeyDown(keyCode, event);

    } 

 public void showDialogItem(){
                     AlertDialog.Builder alertDialogBuilder = new AlertDialog.Builder(context);
  alertDialogBuilder .setItems(R.array.select_dialog_items, new DialogInterface.OnClickListener()
     {
  public void onClick(DialogInterface dialog, int which) {


 String[] items = getResources().getStringArray(R.array.select_dialog_items);
 new AlertDialog.Builder(MainActivity.this)
 .setMessage("You selected: " + which + " , " + items[which])
                                            .show();
                                }
                            });


   alertDialog = alertDialogBuilder.create();

   WindowManager.LayoutParams lp = new WindowManager.LayoutParams();
lp = alertDialog.getWindow().getAttributes();
 lp.gravity = Gravity.TOP | Gravity.LEFT;
   lp.x=-300;
lp.gravity=Gravity.BOTTOM;
   alertDialog.getWindow().setAttributes(lp);

    alertDialog.show();
    alertDialog.getWindow().setLayout(130, 220);

    }

解决方案

Got this initialize 

public static int x=0;

public boolean onKeyDown(int keyCode, KeyEvent event) {

if (keyCode == KeyEvent.KEYCODE_MENU && event.getAction() == 0 ) 
{

AlertDialog.Builder builder = new AlertDialog.Builder(this)
.setOnKeyListener(new DialogInterface.OnKeyListener() 
{
 @Override
 public boolean onKey (DialogInterface dialog, int keyCode, KeyEvent event) 
{

         if (keyCode == KeyEvent.KEYCODE_MENU && event.getAction() == 0) 
{

                      if(x==1)
                      {
                        alert.dismiss();
                        x=0;
                      }
             return true;
             }
             return false;
             }
             });
if(x==0)
{
builder.setMessage("Login details Changed Successfully...");
builder .setCancelable(true);              
builder.setPositiveButton("OK", new DialogInterface.OnClickListener() {
public void onClick(DialogInterface dialog, int id) {
      System.exit(0);
 }
     });
    alert = builder.create();
    alert.show();
    x=1;
          }
          return true;

         }

这篇关于无法驳回AlertDialog时pressing菜单键第二次的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持！