Java - checking if parseInt throws exception

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
}

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
    }
}

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

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();
}