Question

在比较EditText的输入整数时遇到问题。我无法找到它的错误。请帮我。这是下面的代码。

    protected void onCreate(Bundle savedInstanceState) {
        // TODO Auto-generated method stub
        super.onCreate(savedInstanceState);
        setContentView(R.layout.problem2);
        textIn = (EditText) findViewById(R.id.probText);
        Button ans3 = (Button) findViewById(R.id.answer3);



        ans3.setOnClickListener(new View.OnClickListener() {

            public void onClick(View v) {
                // TODO Auto-generated method stub
                String probString = textIn.getText().toString();
                Integer probInt = Integer.parseInt(probString);
                Integer prob = 31;
                if (probInt.equals(prob)) {
                    Toast toast = Toast.makeText(answer3.this,"CORRECT!",Toast.LENGTH_SHORT);
                    toast.setGravity(Gravity.CENTER_VERTICAL, 0, 0);
                    toast.show();
                    startActivity(new Intent("com.sample.androidsample.ANSWER4")    );

                } else {
                    Toast toast = Toast.makeText(answer3.this,"Wrong answer! Try again.",Toast.LENGTH_SHORT);
                    toast.setGravity(Gravity.CENTER_VERTICAL, 0, 0);
                    toast.show();
                }
            }
        });
}

Answer 1

试试这个：

Integer probInt = Integer.parseInt(probString);
            Integer prob = 31;
            //changed from equals() to ==
            if (probInt == prob)) {
                Toast toast = Toast.makeText(answer3.this,"CORRECT!",Toast.LENGTH_SHORT);
                toast.setGravity(Gravity.CENTER_VERTICAL, 0, 0);
                toast.show();
                startActivity(new Intent("com.sample.androidsample.ANSWER4")    );

            } else {
                Toast toast = Toast.makeText(answer3.this,"Wrong answer! Try again.",Toast.LENGTH_SHORT);
                toast.setGravity(Gravity.CENTER_VERTICAL, 0, 0);
                toast.show();
            }

根据我的理解equals是字符串。

Answer 2

您可以使用Integer probInt = Integer.parseInt(probString);代替Integer probInt = Integer.valueOf(probString);，但我不确定这是否是问题。而且probInt.equals(prob)代替probInt.equalIgnoreCase(prob)，您可以使用{{1}}。

Answer 3

我认为这是错误的。

Integer prob = 31;

这里Integer是一个类，所以你必须像下面那样实例化它。

Integer prob = new Integer(31);

或你可以使用

int prob =31;

Answer 4

而不是

 Integer probInt = Integer.parseInt(probString);
 Integer prob = 31;
 if (probInt.equals(prob)) {
 } else {
 }

使用

int probInt = (int) Integer.parseInt(probString);
int prob = 31;
if (probInt == prob) {
} else {
}

Answer 5

尝试这件事，它会起作用

Integer probInt = Integer.parseInt(probString);
        Integer prob = 31;
        //changed from equals() to ==
        if (probInt.intValue() == prob.intValue()) {
            Toast toast = Toast.makeText(answer3.this,"CORRECT!",Toast.LENGTH_SHORT);
            toast.setGravity(Gravity.CENTER_VERTICAL, 0, 0);
            toast.show();
            startActivity(new Intent("com.sample.androidsample.ANSWER4")    );

        } else {
            Toast toast = Toast.makeText(answer3.this,"Wrong answer! Try again.",Toast.LENGTH_SHORT);
            toast.setGravity(Gravity.CENTER_VERTICAL, 0, 0);
            toast.show();
        }

Answer 6

以下是清理输入的示例。为简洁起见省略了异常处理。

private Pattern patternNum = Pattern.compile("^(\\d{1,5})$", 
                               Pattern.CASE_INSENSITIVE | Pattern.UNICODE_CASE);

// Wrap it in a try/catch for PatternSyntaxException!
private boolean validateNum(String inputNum){
    return patternNum.matcher(inputNum);
}

然后假设validateNum例程返回true，意味着它匹配至少5位数，然后这样说：

if (validateNum(probString)){
   int probInt = Integer.parseInt(probString);
   if (probInt == prob){
      // Success
   }else{
      // Failure
   }
}else{
   // Whoops! Bad input caught!
}

EditText整数比较

6 个答案: