android - 检查输入的值是否为数字

Question

我试过这个片段，但它不起作用

try 
    {
    Integer.parseInt(enteredID.getText().toString());
    Log.i("enteredID value", "enterdID is numeric!!!!!!!!!!!^^^");
    flag=1;
} catch (NumberFormatException e) {
    flag=-1;
    Log.i("enteredID value", "enterdID isn't numeric!!!!!!!!!!!^^^");
}

注意它可以接受用户名或 id 来检查值，我不希望它只接受数字！！

score 22 · Accepted Answer

如果布尔值为真，则为数字，否则为字符串值

boolean digitsOnly = TextUtils.isDigitsOnly(editText.getText());

或例如

String text = edt_email.getText().toString();
            boolean digitsOnly = TextUtils.isDigitsOnly(text);
            if (digitsOnly) {
                 if (text.length() == 0) {
                    Toast.makeText(getApplicationContext(), "field can't be empty.", Toast.LENGTH_LONG).show();
                 } else {
                   Toast.makeText(getApplicationContext(), "field is int value", Toast.LENGTH_LONG).show();
                 }
            }else {
                    Toast.makeText(getApplicationContext(), "Field is string value", Toast.LENGTH_LONG).show();
                }
            }

score 21 · Accepted Answer

仅将此表达式用于验证编号

String regexStr = "^[0-9]*$";

if(et_number.getText().toString().trim().matches(regexStr))
{
    //write code here for success
}
else{
    // write code for failure
}

score 7 · Accepted Answer

设置 EditText 属性 inputType = number 它总是将数字作为输入

android:inputType="number"

score 5 · Accepted Answer

试试这个正则表达式：

String regex = "-?\\d+(\\.\\d+)?";

if (enteredID.getText().toString().matches(regex))  {
     Log.i("enteredID value", "enterdID is numeric!!!!!!!!!!!^^^");
     flag=1;
 } else {
     flag=-1;
     Log.i("enteredID value", "enterdID isn't numeric!!!!!!!!!!!^^^");
 }

score 5 · Accepted Answer

为此使用TextUtils类函数

TextUtils.isDigitsOnly("gifg");

如果只有数字，它将返回 true，如果字符串中存在任何字符，则返回 false

score 2 · Accepted Answer

使用这个方法

 boolean isNumber(String string) {
    try {
      int amount = Integer.parseInt(string);
      return true;
    } catch (Exception e) {
      return false;
    }
  }

像这样：

if (isNumber(input)) {
  // string is int
}

score 0 · Accepted Answer

也许是这样的：

    String text = enteredID.getText().toString();

    if(text.matches("\\w+")){
      //--words--
    }else if (text.matches("\\d+")){
      //--numeric--
    }else {
      //-- something else --
    }

您可以更改正则表达式以匹配复杂格式。

score 0 · Accepted Answer

你真的不应该以这种方式使用异常。当您期望像这样抛出异常时，您应该找到另一种处理方法。

尝试使用：http: //developer.android.com/reference/android/widget/TextView.html#setRawInputType%28int%29

这样的事情应该这样做：

editText.setInputType(InputType.TYPE_CLASS_NUMBER | InputType.TYPE_NUMBER_FLAG_DECIMAL | InputType.TYPE_NUMBER_FLAG_SIGNED);

score 0 · Accepted Answer

0

在活动中只需添加 android:inputType="number"

于 2019-05-20T10:00:32.990 回答

score 0 · Accepted Answer

String text = editText123.getText().toString();
try {
   int num = Integer.parseInt(text);
   Log.i("",num+" is a number");
} catch (NumberFormatException e) {
   Log.i("",text+" is not a number");
}

score 0 · Accepted Answer

我使用这个函数来检查小数点后两位的三位整数和浮点值。如果您不想限制位数，请从正则表达式中删除 {x,x}。

private boolean isNumeric(String string) {
     if(Pattern.matches("\\d{1,3}((\\.\\d{1,2})?)", string))
          return true;
     else
          return false;
}

score 0 · Accepted Answer

/**
 * If @param digit is null or not any of the numbers - @return false.
 */
static boolean isNumber(String digit) {
    try {
        Double.parseDouble(digit);
    } catch (NullPointerException | NumberFormatException ignored) {
        return false;
    }
    return true;
}

score 0 · Accepted Answer

Pattern ptrn = Pattern.compile(regexStr);

if (!ptrn.matcher(et_number.getText().toString().trim()).matches()) {

    //write code here for success

}else{

    //write code here for success

}

android - 检查输入的值是否为数字

13 回答 13

Related

Reference