8

我有这个代码,这很完美。只有我想在没有 xml 文件(actions.xml)的情况下使这个动态。我怎么做?

public void showPopup(View v) {
    PopupMenu popup = new PopupMenu(this, v);
    MenuInflater inflater = popup.getMenuInflater();
    inflater.inflate(R.menu.actions, popup.getMenu());
    popup.show();
}
4

2 回答 2

6

使用popup.getMenu()然后使用add.

于 2012-03-05T20:16:11.553 回答
3

在 xml 文件中删除未使用的项目(只是为了实现菜单主题)。所以,它会像:

<menu xmlns:android="http://schemas.android.com/apk/res/android"
xmlns:app="http://schemas.android.com/apk/res-auto"
xmlns:tools="http://schemas.android.com/tools"
android:theme="@style/AppTheme" />

然后用于getMenu添加新的菜单项,如下所示:

Button btn1= (Button) findViewById(R.id.btn_test);
PopupMenu popup = new PopupMenu(yourFormName.this, btn1);  
                    //Inflating the Popup using xml file  
                 popup.getMenu().add("Menu1 Label");
                 popup.getMenu().add("Menu2 Label");
                 popup.getMenuInflater().inflate(R.menu.YourXMLFileName, popup.getMenu());  


                    //registering popup with OnMenuItemClickListener  
                    popup.setOnMenuItemClickListener(new PopupMenu.OnMenuItemClickListener() {  
                     public boolean onMenuItemClick(MenuItem item) {  
                       //---your menu item action goes here ....
                      Toast.makeText(DisplayTransactions.this,"You Clicked : " + item.getTitle(),Toast.LENGTH_SHORT).show();  
                      return true;  
                     }  
                    });  
                    popup.show();//showing popup menu  
于 2016-04-11T09:24:55.030 回答