Question

我正在尝试重新设计我用Java编写的基于文本的终端rpg，作为Android应用程序。我在游戏的一小部分上工作，玩家可以从一个房间走到另一个房间。

代码停止并提示用户在终端应用程序中输入，现在它只是一直运行到最后并且不会接受任何输入。

public class MainActivity extends AppCompatActivity {

//rooms
Room r1 = new Room("The Great Hall", "A long wide room draped here and there in curtains.");
Room r2 = new Room("The Side Study", "a small cluttered room.");
Room r3 = new Room("The Gardens", "a misty path between beds, once carefully tended now overgrown and disturbed by the encroaching weeds.");
Room r4 = new Room("A Long Passage", "a narrow hallway lined with brooding portraits. Set into the walls to your left and right are blank grey doors, three to a side. They are locked.");
Room currentRoom;

//room maps
HashMap<String, Room> mapR1 = new HashMap<>();
HashMap<String, Room> mapR2 = new HashMap<>();
HashMap<String, Room> mapR3 = new HashMap<>();
HashMap<String, Room> mapR4 = new HashMap<>();

//user
String userDir;
String endgame;

@Override
protected void onCreate(Bundle savedInstanceState) {
    super.onCreate(savedInstanceState);
    setContentView(R.layout.activity_main);

    //begin setup
    //

    //map for great hall
    mapR1.put("north", r4);
    mapR1.put("east", r2);
    mapR1.put("west", r3);
    r1.setRoomMap(mapR1);

    //map for side study
    mapR2.put("west", r1);
    r2.setRoomMap(mapR2);

    //map for gardens
    mapR3.put("east", r1);
    r3.setRoomMap(mapR3);

    //map for long passage
    mapR4.put("south", r1);
    r4.setRoomMap(mapR4);

    //init
    currentRoom = r1;
    endgame = "non";

    //views
    final TextView textView = (TextView) findViewById(R.id.youmum);
    final EditText editText = (EditText) findViewById(R.id.editText);

    //
    //end setup

    //the actual game play
    //

    assert textView != null;
    textView.setText("You are in the " + currentRoom.getRoomTitle() + "; " + currentRoom.getRoomDescription());
    textView.append("\nThere are exits to the, ");

    for (String ky : currentRoom.getRoomMap().keySet()) {
        textView.append(ky + " ");
    }

    textView.append("\nChoose a direction");

    assert editText != null;
    editText.setOnEditorActionListener(new TextView.OnEditorActionListener() {
        @Override
        public boolean onEditorAction(TextView v, int actionId, KeyEvent event) {
            boolean handled = false;
            if (actionId == EditorInfo.IME_ACTION_SEND) {
                sendMessage();
                handled = true;
            }
            return handled;
        }

        private void sendMessage() {
            userDir = editText.getText().toString().toLowerCase();
            if (userDir.contains("go") || userDir.contains("walk") || userDir.contains("move")) {
                boolean notDir = false;
                for (String kee : currentRoom.getRoomMap().keySet()) {
                    if (userDir.contains(kee)) {
                        currentRoom = currentRoom.getExitRoom(kee);
                        notDir = false;
                        break;
                    } else {
                        notDir = true;
                    }
                }
                if (notDir) {
                    textView.setText("You cannot go that way.\n");
                }
            }
        }

    });

    textView.setText("You are in the " + currentRoom.getRoomTitle() + "; " + currentRoom.getRoomDescription());
    textView.append("\nThere are exits to the, ");

    for (String ky : currentRoom.getRoomMap().keySet()) {
        textView.append(ky + " ");
    }

    textView.append("\nChoose a direction");

  }
 }

我对Android很新（这实际上是我的第一次尝试）所以我假设我只是遗漏了一些明显的东西。

Answer 1

底部的代码（显示有关您搬入的新房间的文本）需要在ActionListener中移动，以便每次事件触发时都会运行，而不是只是一次初始化。您可以在更新currentRoom的现有代码之后将其放在sendMessage（）中。

您可能已经这样做了，但为了触发IME_ACTION_SEND事件，定义EditText的XML应包括：

android:imeOptions="actionSend"

您可能还想添加一些代码来清除EditText内容，因为当用户点击＆＃34;发送＆＃34;后，它将保留在那里。

等待用户输入？

1 个答案: