java - Android 应用程序崩溃，按钮或条件不起作用？

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我在这方面几乎是新手，所以我可以使用一些帮助来了解为什么我的应用程序不断崩溃？我尝试了很多方法，这是迄今为止最好的方法。

public class MainActivity extends Activity {

@Override
protected void onCreate(Bundle savedInstanceState) {
    super.onCreate(savedInstanceState);
    setContentView(R.layout.activity_main);
    final TextView lastAnswer = (TextView) findViewById(R.id.textView5);
    final TextView answerLabel = (TextView) findViewById(R.id.textView1);
    final TextView answerLabel1 = (TextView) findViewById(R.id.textView2);
    final TextView answerLabel2 = (TextView) findViewById(R.id.textView3);
    final EditText enteredNumber = (EditText) findViewById(R.id.editText1);
    Button getAnswerButton = (Button) findViewById(R.id.button1);

    getAnswerButton.setOnClickListener(new View.OnClickListener() { 

            @Override   
        public void onClick(View arg0) {
                 Random randomGenerator = new Random();
                  //rand gene
                     int randomNumber2 = randomGenerator.nextInt(99);
                      int randomNumber = randomGenerator.nextInt(99);

            String number = "";
            String number2 = "";
            String jtext = " Times ";
            number2= Integer.toString(randomNumber2);
            number= Integer.toString(randomNumber);
            answerLabel.setText(jtext);
            answerLabel1.setText(number2);
            answerLabel2.setText(number);

            String content = enteredNumber.getText().toString();

When i comment the conditional the app works fine, but thats not all that i want I suppose the problem is in the conditional but i can't seem to find it. no errors what so ever on eclipse.

            if (content != null){
                int anInt=Integer.parseInt(content);
                int result= randomNumber2 *randomNumber;

                    if(anInt == result ){

                      Context context =getApplicationContext();
                            CharSequence text = "That's correct!!";
                            int duration = Toast.LENGTH_SHORT;
                            Toast toast = Toast.makeText(context, text, duration);
                            toast.show();

                    } else{ 

                    Context context = getApplicationContext();
                    CharSequence text = "That's not correct";
                    int duration = Toast.LENGTH_SHORT;
                    Toast toast = Toast.makeText(context, text, duration);
                    toast.show(); 
                    }
                    lastAnswer.setText(content);
            }

            }
        });}

所以我在这里想做的(注意菜鸟代码)只是生成 2 个随机数，显示它们，然后让用户输入结果，比较它，通知是否正确，然后重新生成..我已经已经为此工作好几天了

感谢任何帮助，甚至是遵循此代码的新方法。谢谢。

Answer 1

您可能应该用 try/catch 包围您的解析。用户输入被解析回数字可能并不总是有效 - 特别是对于 EditTexts。改变

int anInt=Integer.parseInt(content);

原来如此

  int anInt= -1;
  try{
   anInt = Integer.parseInt(content);
  }
  catch (NumberFormatException e)
  {
    //Show toast or dialog
  }

为了让生活更轻松，您应该将 EditText 设置为 accept in only numbers .