在Java中检查有效的double

Question

I want to check if a String contains a Double not an Integer . I am working in this way;

 private boolean isDouble(String str) {
        try {
            Double.parseDouble(str);
            return true;
        }
        catch(NumberFormatException e) {
            return false;
        }

    } 

For checking it, i simply passed;

isDouble("123");

But it is not working, giving true in both conditions ("123" ,"123.99"). What is wrong here?

Answer 1

If you want to check that it is a number that does not fit into in Integer, you may round the double. Eg exploit the fact that round(1.2) != 1.2 , but round(1) == 1 .

private boolean isDouble(String str) {
    try {
        // check if it can be parsed as any double
        double x = Double.parseDouble(str);
        // check if the double can be converted without loss to an int
        if (x == (int) x)
            // if yes, this is an int, thus return false
            return false;
        // otherwise, this cannot be converted to an int (e.g. "1.2")
        return true;
        // short version: return x != (int) x;
    }
    catch(NumberFormatException e) {
        return false;
    }

} 

Answer 2

You could use Scanner(String) and use the hasNextDouble() method. From its javadoc:

Returns true if the next token in this scanner's input can be interpreted as a double value using the nextDouble() method. For example:

if(source.contains(".")){
    Scanner scanner = new Scanner(source);
    boolean isDouble = scanner.hasNextDouble();
    return isDouble;
}
return false;

Answer 3

You can also always start by parsing to double , and then test if the double is an int or not.

private void main() {

    String str = "123";

    Double value = parseDouble(str);
    boolean isInt = isInt(value);
}

private void isInt(Double value) {
    if(value != null) {
        return (value == (int) value) ? true : false;
    }
    return false;
} 

private double parseToDouble(String str) {
    Double value = null;
    try {
        value = Double.parseDouble(str);
    }
    catch(NumberFormatException e) {
        // Do something
    }
    return value;
}

Answer 4

The problem is due to the fact that 1.00 is 1 and it's a double . This means you can't simply parse a double and pretend that the code detect itself if it's an int or not. For this you should add a check, I think the easiest is the following:

private boolean isDouble(String str) {
  try {
    double myDouble = Double.parseDouble(str); 
    myDouble -= (int)myDouble; //this way you are making the (for example) 10.3 = 0.3

    return myDouble != (double)0.00; //this way you check if the result is not zero. if it's zero it was an integer, elseway it was a double
  }
  catch(NumberFormatException e) {
    return false;
  }
} 

I did it without editor, so tell me if something is wrong.

Hope this helps

Answer 5

Check the simple code

private boolean isDecimalPresent(d){
 try {
    return d%1!=0;
}
catch(NumberFormatException e) {
    return false;
}