2014-02-20 91 views
10

我想通过单击弹出窗口外或后退按钮来关闭弹出窗口,但是当单击后退按钮时我的应用程序退出的,而不是退出应用程序我想关闭弹出窗口。通过后退按钮关闭弹出窗口

这里是我的代码,

ivmainmenu.setOnClickListener(new OnClickListener() { 

     @Override 
     public void onClick(View v) { 
      // TODO Auto-generated method stub 

      LayoutInflater layoutInflater 
      = (LayoutInflater)getBaseContext() 
       .getSystemService(LAYOUT_INFLATER_SERVICE); 
      View popupView = layoutInflater.inflate(R.layout.popupwindow, null); 
      final PopupWindow popupWindow = new PopupWindow(popupView,LayoutParams.FILL_PARENT, 
        LayoutParams.WRAP_CONTENT); 
       popupWindow.showAsDropDown(ivmainmenu, 0,14); 
       popupView.setPadding(0, 0, 0, 10); 
       popupWindow.showAsDropDown(ivmainmenu); 

       popupWindow.setBackgroundDrawable(new BitmapDrawable()); 
       popupWindow.setOutsideTouchable(false); 

       TextView tvpopupwork = (TextView)popupView.findViewById(R.id.tvpopupwork); 
       TextView tvpopupabout = (TextView)popupView.findViewById(R.id.tvpopupabout); 
       TextView tvpopupservices = (TextView)popupView.findViewById(R.id.tvpopupservices); 
       TextView tvpopupcontact = (TextView)popupView.findViewById(R.id.tvpopupcontact); 

       Typeface typeFace2 = Typeface.createFromAsset(getAssets(),"fonts/arboriaboldregular.ttf"); 
       tvpopupwork.setTypeface(typeFace2); 
       tvpopupabout.setTypeface(typeFace2); 
       tvpopupservices.setTypeface(typeFace2); 
       tvpopupcontact.setTypeface(typeFace2); 

       tvpopupwork.setOnClickListener(new OnClickListener() { 

        @Override 
        public void onClick(View v) { 
         // TODO Auto-generated method stub 
         Intent intent = new Intent(Home.this,Ourwork.class); 
         intent.addFlags(Intent.FLAG_ACTIVITY_NO_ANIMATION); 
         startActivity(intent); 
        } 
       }); 

       tvpopupabout.setOnClickListener(new OnClickListener() { 

        @Override 
        public void onClick(View v) { 
         // TODO Auto-generated method stub 
         Intent intent = new Intent(Home.this,Aboutus.class); 
         intent.addFlags(Intent.FLAG_ACTIVITY_NO_ANIMATION); 
         startActivity(intent); 
        } 
       }); 

       tvpopupservices.setOnClickListener(new OnClickListener() { 

        @Override 
        public void onClick(View v) { 
         // TODO Auto-generated method stub 

         Intent intent = new Intent(Home.this,Services.class); 
         intent.addFlags(Intent.FLAG_ACTIVITY_NO_ANIMATION); 
         startActivity(intent); 
        } 
       }); 

       tvpopupcontact.setOnClickListener(new OnClickListener() { 

        @Override 
        public void onClick(View v) { 
         // TODO Auto-generated method stub 

         Intent intent = new Intent(Home.this,Contact.class); 
         intent.addFlags(Intent.FLAG_ACTIVITY_NO_ANIMATION); 
         startActivity(intent); 
        } 
       }); 
       ivmainmenu.setOnClickListener(new OnClickListener() { 

       @Override 
       public void onClick(View v) { 
        // TODO Auto-generated method stub 

        popupWindow.dismiss(); 
       } 
      } 
     }); 

它给我我想要的结果,但是当我关闭它不会再次打开,我想再次打开它的菜单所以我应该怎么办? 谢谢。

回答

14

更换

popupWindow.setOutsideTouchable(false); 

与此

popupWindow.setOutsideTouchable(true); 
popupWindow.setFocusable(true); 
+0

我不得不将背景设置为一个非空绘制,至少不上的是Android 4.4.2 –

+1

工作'popupWindow.setBackgroundDrawable(new ColorDrawable(Color.TRANSPARENT));' –

0

试试这样:实施onBackPressed()并添加

if(popup!=null) { 
    popup.dismiss(); 
    popup=null; 
} 

并设置PopWindow下面:

popup.setOutsideTouchable(true); 
+0

ACTION_OUTSIDE ???那有什么用? – akky777

+0

@ akky777当触摸在活动之外时,您将获得ACTION_OUTSIDE。但这只适用于我的要求 –

0

您可以在您的代码中覆盖onBackPressed()回调并检查弹出窗口是否已经显示(然后解除它),否则您会调用super来获得正常行为。

8

维护全球基准为PopUpWindow,并覆盖onBackPressed() ...

@Override 
public void onBackPressed() { 
    if (popupWindow != null && popupWindow.isShowing()) { 
     popupWindow.dismiss(); 
    } else { 
     super.onBackPressed(); 
    } 
} 

通过在同一Button解雇......

ivmainmenu.setOnClickListener(new View.OnClickListener() { 

     @Override 
     public void onClick(View v) { 
      if(popupWindow != null && popupWindow.isShowing()) { 
       popupWindow.dismiss(); 
       popupWindow = null; 
      } else { 
       // show pop up now 
      } 
     } 
    }); 
+0

是否有解决方案通过同一个按钮关闭弹出窗口? – akky777

+0

@ akky777看到编辑答案... –

+2

它不起作用 – akky777

1

试试这个..

使用PopupWindow popupWindow作为全球变量

使用popup.setOutsideTouchable(true);

@Override 
public void onBackPressed() { 
    if (popupWindow != null) { 
     if (popupWindow.isShowing()) { 
      popupWindow.dismiss(); 
     } 
    } else { 
     finish(); 
    } 
} 
+0

@GopalRao感谢您的编辑。 – Hariharan

+1

可能更安全地使用'if(popupWindow!= null && popupWindow.isShowing())' – tar

2

请写信onBackPressed(),并具有下列代码

if(popup!=null){ 
    //dismiss the popup 
    popup.dismiss(); 
    //make popup null again 
    popup=null; 
} 
+0

@Gopal Rao很多thx gopal ji +1对你 –

+0

不客气。无论如何!不要为编辑加注:-) ... –

3
popupWindow.setBackgroundDrawable (new BitmapDrawable()); 
popupWindow.setOutsideTouchable(true); 
;)