【发布时间】:2015-05-12 20:20:56
【问题描述】:
所以我正在尝试制作一个应用程序,其中有一个ImageView,它显示图像列表中的随机图像。还有两个按钮,根据图像,需要按下正确的按钮。这会一直持续下去,直到你得到一个不正确的答案为止。
我在 Async 方法中使用了在 ImageView 中随机显示图像,并在我的按钮上设置条件。但是,当我运行应用程序时,该条件仅适用于显示的第一张图像。之后,无论显示什么图像,按钮条件都会像显示第一张图像一样工作。
这是代码
public class Game extends ActionBarActivity {
static TextView timeDisplay;
int[] cardGallery = {R.drawable.tile0, R.drawable.tile1, R.drawable.tile2, R.drawable.tile3, R.drawable.tile4, R.drawable.tile5, R.drawable.tile6, R.drawable.tile7, R.drawable.tile8, R.drawable.tile9};
int score = 0;
int imageId = (int) (Math.random() * cardGallery.length);
@Override
protected void onCreate(Bundle savedInstanceState) {
super.onCreate(savedInstanceState);
setContentView(R.layout.activity_game);
ImageView cardImageView = (ImageView) findViewById(R.id.cardImage);
ImageButton bigButton = (ImageButton) findViewById(R.id.upButton);
ImageButton smallButton = (ImageButton)findViewById(R.id.downButton);
final TextView scoreDisplay = (TextView)findViewById(R.id.scoreMessage);
timeDisplay = (TextView) findViewById(R.id.timerMessage);
cardImageView.setImageResource(cardGallery[imageId]);
scoreDisplay.setText("Current score: " + score);
.
.
.
bigButton.setOnClickListener(new View.OnClickListener() {
@Override
public void onClick(View v) {
/* may need to implement switch and cases for each button
switch (cardGallery[imageId]){
case R.drawable.tile0:
GameTimer.onFinish();
}*/
if (imageId == 0) {
GameTimer.cancel();
GameTimer.onFinish();
}
else if (imageId == 1) {
GameTimer.cancel();
GameTimer.onFinish();
}
else if (imageId == 2) {
GameTimer.cancel();
GameTimer.onFinish();
}
else if (imageId == 3) {
GameTimer.cancel();
GameTimer.onFinish();
}
else if (imageId == 4) {
GameTimer.cancel();
GameTimer.onFinish();
}
else if (imageId == 5) {
GameTimer.start();
scoreDisplay.setText("Current score: " + ++score);
new CardAsyncTask().execute();
}
else if (imageId == 6) {
GameTimer.start();
scoreDisplay.setText("Current score: " + ++score);
new CardAsyncTask().execute();
}
else if (imageId == 7) {
GameTimer.start();
scoreDisplay.setText("Current score: " + ++score);
new CardAsyncTask().execute();
}
else if (imageId == 8) {
GameTimer.start();
scoreDisplay.setText("Current score: " + ++score);
new CardAsyncTask().execute();
}
else if (imageId == 9) {
GameTimer.start();
scoreDisplay.setText("Current score: " + ++score);
new CardAsyncTask().execute();
}
else {
GameTimer.cancel();
GameTimer.onFinish();
}
}});
smallButton.setOnClickListener(new View.OnClickListener() {
@Override
public void onClick(View v) {
if (imageId == 0) {
GameTimer.start();
scoreDisplay.setText("Current score: " + ++score);
new CardAsyncTask().execute();
}
else if (imageId == 1) {
GameTimer.start();
scoreDisplay.setText("Current score: " + ++score);
new CardAsyncTask().execute();
}
else if (imageId == 2) {
GameTimer.start();
scoreDisplay.setText("Current score: " + ++score);
new CardAsyncTask().execute();
}
else if (imageId == 3) {
GameTimer.start();
scoreDisplay.setText("Current score: " + ++score);
new CardAsyncTask().execute();
}
else if (imageId == 4) {
GameTimer.start();
scoreDisplay.setText("Current score: " + ++score);
new CardAsyncTask().execute();
}
else if (imageId == 5) {
GameTimer.cancel();
GameTimer.onFinish();
}
else if (imageId == 6) {
GameTimer.cancel();
GameTimer.onFinish();
}
else if (imageId == 7) {
GameTimer.cancel();
GameTimer.onFinish();
}
else if (imageId == 8) {
GameTimer.cancel();
GameTimer.onFinish();
}
else if (imageId == 9) {
GameTimer.cancel();
GameTimer.onFinish();
}
else {
GameTimer.cancel();
GameTimer.onFinish();
}
}
});
}
class CardAsyncTask extends AsyncTask<Integer, Void, Integer> {
@Override
protected Integer doInBackground(Integer... params) {
int imageId = (int) (Math.random() * cardGallery.length);
return imageId;
}
@Override
protected void onPostExecute(Integer imageId) {
ImageView cardImageView = (ImageView) findViewById(R.id.cardImage);
cardImageView.invalidate();
cardImageView.setImageResource(cardGallery[imageId]);
}
}
}
我认为这是因为我已将新旧 imageId 值声明为相同,但如果我将它们更改为 imageId 和 newimageId,这将改变我的代码中我要求比较值的条件到 imageId。 谢谢
【问题讨论】:
-
发布您的代码,以便我们弄清楚。
-
刚刚添加。如果它看起来凌乱且写得不好,请道歉。我正在尝试掌握一些东西。
-
@hofmeister 抱歉,下次会记住的
标签: java android button android-asynctask android-studio