如何打开一个弹出窗口，点击列表查看项目

Question

我需要一个列表视图，在这个列表视图中点击每行应该打开一个弹出窗口。

在该弹出窗口中，我需要显示来自listview的项目。

如何打开项目的弹出窗口？

Answer 1

你必须尝试karan`s回答

AlertDialog.Builder builderSingle = new AlertDialog.Builder(DialogActivity.this); 
builderSingle.setIcon(R.drawable.ic_launcher); 
builderSingle.setTitle("Select One Name:-"); 

final ArrayAdapter<String> arrayAdapter = new ArrayAdapter<String>(
     DialogActivity.this, 
     android.R.layout.select_dialog_singlechoice); 
arrayAdapter.add("Hardik"); 
arrayAdapter.add("Archit"); 
arrayAdapter.add("Jignesh"); 
arrayAdapter.add("Umang"); 
arrayAdapter.add("Gatti"); 

builderSingle.setNegativeButton(
     "cancel", 
     new DialogInterface.OnClickListener() { 
      @Override 
      public void onClick(DialogInterface dialog, int which) { 
       dialog.dismiss(); 
      } 
     }); 

builderSingle.setAdapter(
     arrayAdapter, 
     new DialogInterface.OnClickListener() { 
      @Override 
      public void onClick(DialogInterface dialog, int which) { 
       String strName = arrayAdapter.getItem(which); 
       AlertDialog.Builder builderInner = new AlertDialog.Builder(
         DialogActivity.this); 
       builderInner.setMessage(strName); 
       builderInner.setTitle("Your Selected Item is"); 
       builderInner.setPositiveButton(
         "Ok", 
         new DialogInterface.OnClickListener() { 
          @Override 
          public void onClick(
            DialogInterface dialog, 
            int which) { 
           dialog.dismiss(); 
          } 
         }); 
       builderInner.show(); 
      } 
     }); 
builderSingle.show(); 

Answer 2

使用

PopupMenu popupMenu = new PopupMenu(context, view); 
popupMenu.inflate(R.menu.menu_popup); 
popupMenu.setOnMenuItemClickListener(itemClickListener); 
popupMenu.show(); 

view是视图点击，将被用作锚点。