如何以特定方式重复此倒数计时器...? [英] How can I repeat this countdown timer in a specific way...?
问题描述
在游戏中，输出一个随机数(loadG1)并显示4秒钟.一旦四秒钟结束，它就会消失，用户必须输入其值才能获得分数.一旦用户按下Enter键，他们的输入框就会消失，程序必须等待，直到更长的CountDownTimer(当前为18秒)结束才可以看到他们的分数.到目前为止，他们只能得分1分.我想发生的是，一旦输入答案(无论是否正确)，便要重复4秒倒数计时器的内容.虽然，我希望在它们的倒数计时器再次启动时也输出loadG1.因此，他们输入答案，新的随机负载G1会显示另外4秒钟，然后消失，然后输入另一个答案，依此类推，直到更长的计时器结束.我将如何去做呢?我知道循环不起作用，就像我尝试过的那样，它只是使我的应用崩溃了.如果有人可以帮助我做到这一点，并向我展示如何实现代码，我将不胜感激. 注意:我是学习Java和Android开发的新手. 预先非常感谢.
In the game a random number (loadG1) outputted and is shown for 4 seconds. Once the four seconds are up, it disappears and the user must input its value to gain a point. Once the user presses enter, their input box disappears and the program must wait until the longer CountDownTimer (currently 18 seconds) has ended to see their score. So far, they can only score one point. What I want to happen is for the contents of the 4 second countdowntimer to be repeated once they enter an answer (whether correct or not). Although, I want loadG1 to also be outputted when they countdown timer starts again. So they enter an answer, a new random loadG1 is shown for another 4 secs then disappears, they enter another answer, etc... until the longer timer ends. How would I go about doing this? I'm aware that loops don't work, as I tried and it just crashed my app. I'd be very grateful if someone could help me do this, and show me how to implement the code. NOTE: I am new to learning Java and Android dev. Many thanks in advance.
代码如下:
public class game1 extends AppCompatActivity {
@Override
protected void onCreate(Bundle savedInstanceState) {
super.onCreate(savedInstanceState);
setContentView(R.layout.activity_game1);
final Button loseStarter1;
loseStarter1 = (Button) findViewById(R.id.Starter1);
loseStarter1.setOnClickListener(new View.OnClickListener() {
@Override
public void onClick(View v) {
loseStarter1.setVisibility(View.GONE);
final int[] score = {0};
Random generateG1 = new Random();
final int loadG1 = generateG1.nextInt(1000000)+10000;
final TextView number = (TextView) findViewById(R.id.number);
number.setText(" "+loadG1);
new CountDownTimer(18000, 1000) {
@Override
public void onTick (long millisUntilFinished) {
}
public void onFinish() {
TextView result = (TextView) findViewById(R.id.outcome);
result.setText("Score: "+ score[0]);
TextView prompt = (TextView) findViewById(R.id.prompt);
prompt.setVisibility(View.GONE);
}
}.start();
new CountDownTimer(4000, 1000) {
@Override
public void onTick(long millisUntilFinished) {
}
@Override
public void onFinish() {
number.setVisibility(View.GONE);
final TextView prompt = (TextView) findViewById(R.id.prompt);
prompt.setText(" Enter the number");
final EditText input = (EditText) findViewById(R.id.enterAnswer);
input.setVisibility(View.VISIBLE);
input.setOnKeyListener(new View.OnKeyListener() {
@Override
public boolean onKey(View v, int keyCode, KeyEvent event) {
if (event.getAction() == KeyEvent.ACTION_DOWN) {
switch (keyCode) {
case KeyEvent.KEYCODE_ENTER:
Editable answer = input.getText();
int finalAnswer = Integer.parseInt(String.valueOf(answer));
int finalLoadG1 = Integer.parseInt(String.valueOf(loadG1));
input.setVisibility(View.GONE);
prompt.setVisibility(View.GONE);
if (finalAnswer == finalLoadG1) {
score[0]++;
}
return true;
default:
}
}
return false;
}
});
}
}.start();
};
});
}
}
推荐答案
已解决
我在分数[0] ++之后加上了以下内容:
I put the following after the score[0]++:
number.setVisibility(View.VISIBLE);
final int loadG1 = generateG1.nextInt(1000000) + 10000;
number.setText(" " + loadG1);
input.getText().clear();
start();
这样，将进行所有必要的更改，并且计时器将重新开始.我还修改了提示 TextView的可见性，以确保仅当用户能够输入答案时才显示它.花了我一段时间，但问题已解决.
This way, all necessary changes are made and the timer begins anew. I also modified the visibility of the prompt TextView to ensure it only appear when the user is able to input an answer. Took me a while but this has been resolved.
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