-1
我有一个方法,启动一个if语句onClick以显示列表数组中的问题,删除该项目,然后增加一个int以便再次显示另一个问题onClick。该变量在onCreate之前定义。当我第一次点击按钮时,一个随机项从数组中拉出并显示出来。如果我再次点击它,它什么也不做。什么?将变量传递到其他方法
爪哇
public class Careunderfirequiz extends AppCompatActivity{
int questionsAnswered = 0;
List<String> questions = new ArrayList<>();
@Override
protected void onCreate(Bundle savedInstanceState) {
super.onCreate(savedInstanceState);
getSupportActionBar().hide();
setContentView(R.layout.myLayout);
questions.add("1");
questions.add("2");
questions.add("3");
questions.add("4");
questions.add("5");
questions.add("6");
questions.add("7");
questions.add("8");
questions.add("9");
questions.add("10");
}
public void nextQuestion(View view){
TextView questionsTextView = (TextView)findViewById(R.id.questions);
Button next = (Button)findViewById(R.id.next);
if(questionsAnswered == 0){
Random randomGenerator = new Random();
String randQuestion = questions.get(randomGenerator.nextInt(questions.size()));
questionsTextView.setText(randQuestion);
questions.remove(randQuestion);
questionsAnswered++;
}
if(questionsAnswered == 1){
Random randomGenerator = new Random();
String randQuestion = questions.get(randomGenerator.nextInt(questions.size()));
questionsTextView.setText(randQuestion);
questions.remove(randQuestion);
questionsAnswered++;
}
}
}
的'如果(questionAnwerered == 0)'和'如果(questionAnswered == 1)'似乎没有必要,因为你的代码始终是相同的反正。只要注意questionAnswered == 10,因为只有10个问题。另外'next'变量没有被使用,因此无论如何都不需要找到它,因为按钮被传递给方法,因为'view'变量总是(假设你设置了android:onclick =“nextQuestion”布局中的按钮。 – Ridcully