android - 在android中填充没有XML文件的弹出菜单

Question

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();
}

Answer 1

使用popup.getMenu()然后使用add.

Answer 2

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