Java - checking if parseInt throws exception

Question

I\'m wondering how to do something only if Integer.parseInt(whatever) doesn\'t fail.

More specifically I have a jTextArea of user specified values seperated by line

Answer 1

It would be something like this.

String text = textArea.getText();
Scanner reader = new Scanner(text).useDelimiter("\n");
while(reader.hasNext())
    String line = reader.next();

    try{
        Integer.parseInt(line);
        //it worked
    }
    catch(NumberFormatException e){
       //it failed
    }
}

Answer 2

public static boolean isParsable(String input) {
    try {
        Integer.parseInt(input);
        return true;
    } catch (final NumberFormatException e) {
        return false;
    }
}

Answer 3

parseInt will throw NumberFormatException if it cannot parse the integer. So doing this will answer your question

try{
Integer.parseInt(....)
}catch(NumberFormatException e){
//couldn't parse
}

Answer 4

instead of trying & catching expressions.. its better to run regex on the string to ensure that it is a valid number..

Answer 5

You can use a scanner instead of try-catch:

Scanner scanner = new Scanner(line).useDelimiter("\n");
if(scanner.hasNextInt()){
    System.out.println("yes, it's an int");
}

Answer 6

Check if it is integer parseable

public boolean isInteger(String string) {
    try {
        Integer.valueOf(string);
        return true;
    } catch (NumberFormatException e) {
        return false;
    }
}

or use Scanner

Scanner scanner = new Scanner("Test string: 12.3 dog 12345 cat 1.2E-3");

while (scanner.hasNext()) {
    if (scanner.hasNextDouble()) {
        Double doubleValue = scanner.nextDouble();
    } else {
        String stringValue = scanner.next();
    }
}

or use Regular Expression like

private static Pattern doublePattern = Pattern.compile("-?\\d+(\\.\\d*)?");

public boolean isDouble(String string) {
    return doublePattern.matcher(string).matches();
}