Question

这样的问题（ID 1606993）已经发布了，但并没有真正回答我的问题，所以这里有：

在我正在创建的Android应用中，用户必须从键盘输入数字才能继续使用该应用。当他们输入他们的号码时，他们可以按返回键或主按钮。问题是，它只能工作一次。

这是我的听力方法：

public boolean onKeyDown(int key, KeyEvent event)//Listens for key events
{
    if(key==KeyEvent.KEYCODE_0)
        add(0);
    else if(key==KeyEvent.KEYCODE_1)
        add(1);
    else if(key==KeyEvent.KEYCODE_2)
        add(2);
    else if(key==KeyEvent.KEYCODE_3)
        add(3);
    else if(key==KeyEvent.KEYCODE_4)
        add(4);
    else if(key==KeyEvent.KEYCODE_5)
        add(5);
    else if(key==KeyEvent.KEYCODE_6)
        add(6);
    else if(key==KeyEvent.KEYCODE_7)
        add(7);
    else if(key==KeyEvent.KEYCODE_8)
        add(8);
    else if(key==KeyEvent.KEYCODE_9)
        add(9);
    else if(key==KeyEvent.KEYCODE_ENTER)
        setNum();
    else if(key==KeyEvent.KEYCODE_DPAD_CENTER)
        setNum();
    return true;
}

add（int numb）方法将num添加到占据屏幕的文本字段，不会造成任何问题。

这是setNum（）：

protected void setNum()//Sets the number and tells the Runner it is set.
{
    if(ray.size()==1)
        num=ray.get(0);
    else if(ray.size()==0)
        num=0;
    else
        num=(ray.get(0)*10)+ray.get(1);
    ray=new ArrayList<Integer>();//or ray.clear(), I've tried both
    ready=true;
}

ArryayList ray是存储数字的地方。计算工作正常。 Int num是代码使用的数字。布尔准备好了，因为在runner类中有一个while循环等待ready为true以继续代码。

while(!a.ready)
{
    for(int x=0;x<100;x++);
}

有什么想法吗？

编辑：以下是Runner中名为的方法：

while(!go)
    {
        addText("It is White's move");
        addText("Possible pieces to move");
        for(int x=0;x<b.getWhite().getPiecesWithMoves(b).size();x++)//Loop to print White's pieces that can move
        {
            addText(""+(x+1)+") "+b.getWhite().getPiecesWithMoves(b).get(x));
        }   
        got=false;
        p=0;
        while(!got)//Loop to enter number
        {
            addText("Input the number of the piece you want to move");
            while(!a.ready)
            {
                for(int x=0;x<100;x++);
            }
            p=a.num-1;
            if(p<b.getWhite().getPiecesWithMoves(b).size()&&p>=0)//Checks to make sure that p is valid
                got=true;
            a.num=0;
            a.ready=false;
        }
        gl=b.getWhite().getPiecesWithMoves(b).get(p).getLocation();//Makes a location that is where the piece currently is
        addText("Possible moves");
        for(int x=0;x<b.getPiece(gl).getMoves(b).size();x++)//Loop to print where the piece can go
        {
            addText(""+(x+1)+") "+b.genL(b.getPiece(gl).getMoves(b).get(x)));
        }
        got=false;//reset
        while(!got)//Loop to enter number
        {
            addText("Input the number of one of these moves.");
            addText("If you wish to change, enter 0.");
            while(!a.ready)
            {
                for(int x=0;x<100;x++);
            }
            p=a.num-1;
            if(p==-1)
                got=true;
            else if(p<b.getPiece(gl).getMoves(b).size()&&p>=0)//Checks to make sure that p is valid
            {
                got=true;
                go=true;
            }
            a.num=0;
            a.ready=false;
        }
    }
    gk=b.getPiece(gl).getMoves(b).get(p);//The location that the piece is going to
    b.move(gk, b.getPiece(gl),gk.isTaken(),gk.isCastle());//Moves the piece

Answer 1

public class MyView {
  int num;
  int currentState = 1; //much, much more pretty if you'd use Enum, 
                        //but I'll keep it simple now.

  public boolean onKeyDown(int key, KeyEvent event) {
    //no changes here
  }

  protected void setNum(){
    if(ray.size()==1)
        num=ray.get(0);
    else if(ray.size()==0)
        num=0;
    else
        num=(ray.get(0)*10)+ray.get(1);
    ray=new ArrayList<Integer>();
    if (currentState == 1) //(TODO:should really use Enum here....)
       part1();
    else if (currentState == 2)
       part2();
    else if (...
  }

  void part1() {
    addText("It is White's move");
    addText("Possible pieces to move");
    for(int x=0;x<b.getWhite().getPiecesWithMoves(b).size();x++)
    {
      addText(""+(x+1)+") "+b.getWhite().getPiecesWithMoves(b).get(x));
    }

    addText("Input the number of the piece you want to move");
    currentState = 2;//now we do nothing, we wait untill user finishes input
  }
  void part2()
  {//will be called after part1 finishes and a number has been entered
    p=a.num-1;
    if(p<b.getWhite().getPiecesWithMoves(b).size()&&p>=0) {
      gl=b.getWhite().getPiecesWithMoves(b).get(p).getLocation();
      addText("Possible moves");
      //and more code
      //Need input? stop and set next entry point
      currentState = 3;//continue to part 3
    }
    else
    {
      addText("Input the number of the piece you want to move");
      currentState = 2;//oops, stay at part2, we didn't get a good number!
    }
  }

}

当用户必须输入时，想法是什么都不做。此时没有代码运行，例如在part1（）完成之后。当用户完成输入并按下某个键时，您必须确保执行正确的代码，然后调用part2。然后在需要再次输入时停止运行代码。

祝你好运！

输入仅在Android中运行一次

1 个答案: