java - Android从其他线程获取EditText内容

Question

在我的问题中，我有我的 MainActivity ，带有 TextView和EditText ，以及一个方法，该方法应将消息发送到 TextView并收到 EditText 的内容。问题是从 EditText 接收文本，并使该方法等待用户输入

我将尝试发布我的裁剪代码，以便您了解我想要实现的目标。

public class MainActivity extends Activity {

public class MonitorObject{
}
final MonitorObject mSync = new MonitorObject();
private TextView mConsoleOut;
private EditText mInputLine;
public String inputString;
public synchronized String getInputString(String value){this.inputString = value; return inputString;}

IOHandler IOhandler;

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

    mConsoleOut = (TextView) findViewById(R.id.consoleOut);
    mInputLine = (EditText) findViewById(R.id.actionInField);


    EditText editText = (EditText) findViewById(R.id.actionInField);
    editText.setOnEditorActionListener(new TextView.OnEditorActionListener() {
        @Override
        public boolean onEditorAction(TextView v, int actionId, KeyEvent event) {
            synchronized (mSync){
                mSync.notify();
            }
            return true;
        }
    });
    GameCycle gameCycle = new GameCycle();
    Thread gameloop = new Thread(gameCycle);
    gameloop.start();
}

public class GameCycle implements Runnable{
    public void run(){
        game();
    }
}

public void game(){
findViewById(R.id.actionInField);
    Integer time=0;
    PlayerClass p1 = new PlayerClass(1,4);
    TileClass tile = new TileClass.Forest();
    Integer gamestate=0;


    while(gamestate==0){
        time++;
        tile.initialize_tilesettings(time, p1);
        tile.passed=true;
        List actionResponse=new ArrayList(Arrays.asList("repeat", "type", "returnval"));

        while(gamestate==0 && !actionResponse.get(1).equals("tileChange")){

            actionResponse=Arrays.asList("repeat", "type", "returnval");
            Boolean tileActions = true;

            while(!actionResponse.get(0).equals("continue")){
                //actions to be repeated

                String exe=null;

                tileActions=true;

                List<String> params = new ArrayList<>();

                <<<<<print>>>>>("\nWhat will you do?");

                synchronized (mSync){
                    try {
                        mSync.wait();
                    } catch (InterruptedException e) {
                        e.printStackTrace();
                    }
                }

                <<<<<getinput>>>>>String inp=returnInputString();
                String[] splitIn=inp.split(" ");


                for(String text:splitIn){
                    params.add(text.toLowerCase());
                }

                for(ActionClass action:p1.getActions()){
                    if(action.method.contains(params.get(0))){
                        exe=action.method;
                    }
                }

                if(exe==null){
                    <<<<<print>>>>>("Not a valid action");

                    actionResponse=Arrays.asList("repeat","","");
                }
                else{
                    try{
                        actionResponse=p1.excecFunctionByTag(exe, Arrays.asList(p1, tile, params));
                        if(actionResponse==null){
                            throw new RuntimeException();
                        }
                    }
                    catch (Exception e){
                        for(ActionClass action:p1.getActions()){
                            if(exe.equals(action.method)){
                                String actionEnt=action.actionEntry;
                                <<<<<print>>>>>("\nInvalid parameters, try: "+actionEnt);
                                actionResponse=Arrays.asList("repeat","","");
                            }
                        }
                    }
                    return;
                }


            }
        }
        if(gamestate==1){
            <<<<<print>>>>>('You died! Game over!\n\n<-<-< New Game >->->\n\n')
            game();
        }
        else if(gamestate==2){
            <<<<<print>>>>>("You have defeated the boss! Behind the fallen enemy you can see a path. You follow it and find a small village. You are safe!\n\n<-<-< New Game >->->\n\n")
            game();
        }
    }
}

我已经用 <<<<< >>>>> 标记了所需的输入/输出方法.

请怜悯我的java技能，因为这是我用java编写的第一个大型项目，整个结构都是从python转换而来的。

再次，非常感谢所有帮助。

Answer 1

问题很简单，通常线程不会通知用户界面或主线程。为了解决这个问题，android 有 AsyncTask。您可以将 AsyncTask 用作:

1. 在 onPreExecute() 方法中启动 TextView 和 EditText。
2. 发送或调用您的方法以在 doInBackground() 方法中发送您想要的任何文本。
3. 然后您将在 onPostExecute() 方法中获得结果。在这里您可以将 Text() 设置为您的 EditText。

注意:AsyncTask 是一个联系主 ui 线程的线程，它有一个 onPostExecute() 方法。

为了更好地理解 AsyncTask，请查看此链接: http://www.androidhive.info/2012/01/android-json-parsing-tutorial/