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();
}