App开发中的弹出式菜单

1.首先需要在根目录下新建menu才能创建menu的资源文件

2.然后创建main.xml文件来作为弹出式菜单的填充内容

<?xml version="1.0" encoding="utf-8"?>
<menu xmlns:android="http://schemas.android.com/apk/res/android">
    <item android:id="@+id/copy"
        android:orderInCategory="100"
        android:title="复制"/>
    <item android:id="@+id/del"
        android:orderInCategory="200"
        android:title="删除"/>
</menu>

3.绑定一个按钮并添加按钮的单击事件

btn = (Button)findViewById(R.id.btn);

public void showPopupMenu(View view){
        //弹出式菜单
        PopupMenu popupMenu = new PopupMenu(this,view);
        //获取菜单填充器
        MenuInflater inflater = popupMenu.getMenuInflater();
        //填充菜单到弹出式菜单中
        inflater.inflate(R.menu.main,popupMenu.getMenu());

        popupMenu.setOnMenuItemClickListener(this);
        popupMenu.show();
    }

4.需要继承按钮菜单的监听事件并重写按钮菜单点击的方法

public boolean onMenuItemClick(MenuItem item) {
        switch (item.getItemId()){
            case R.id.copy:
                Toast.makeText(this,"复制",Toast.LENGTH_SHORT).show();
            case R.id.del:
                Toast.makeText(this,"删除",Toast.LENGTH_SHORT).show();
               default:
                   break;
        }
        return false;
    }