如何使用菜单项中的开关案例启动不同的活动？

Question

 public void onPopup(View view)
{
    final PopupMenu menu=new PopupMenu(this,view);
    menu.getMenuInflater().inflate(R.menu.menu1,menu.getMenu());


    menu.setOnMenuItemClickListener(new PopupMenu.OnMenuItemClickListener()
    {
        public boolean onMenuItemClick(MenuItem item)
        {

            Toast toast=Toast.makeText(MainActivity.this,
                    item.getTitle()+"Selected",Toast.LENGTH_SHORT);
            //Intent intent2 = new Intent(MainActivity.this, YourSpotActivity.class);
            //startActivity(intent2);

            //startActivity(new Intent(MainActivity.this,YourSpotActivity.class));

    toast.show();
    return true;
        }
    });
    menu.show();

}

When i click any one of the list item then it will start another activity. How can i do that by modify an above code. explain me please. I have use four car model in the menu. when i choose any one of that car then it will go to particular activity.

Answer 1

You can use switch statement as below inside onMenuItemClick :

 switch (item.getItemId()) {
        case R.id.menuitem1:
            //calling intent ( activity1 )
        case R.id.menuitem2:
          //calling intent ( activity 2)
         default:
          //default intent
   }

Answer 2

You need to use switch as below

  switch (item.getItemId()) {
    case R.id.menuitem1:
    Toast.makeText(getApplicationContext(), "StartActiviy 1", Toast.LENGTH_SHORT).show();
        // start activity 1
       return true;
    case R.id.menuitem2:
    Toast.makeText(getApplicationContext(), "StartActiviy 2", Toast.LENGTH_SHORT).show();
       // start activity 2
       return true;
     default:
      //default intent
       return true;
     }

http://developer.android.com/reference/android/widget/PopupMenu.html