Whenever I am trying to lookup on the internet to see if an int value which is accepted from a userfield has some characters like A-z or any special characters, all the sites happen to presume that the input is in the form of String. Function
boolean numberValidation(int number){
if number has characters mentioned above, return false
else return true
}
In that case, the compiler will complain about the parameter passed.
If the parameter was a String and you want to detect if the String is a valid integer, you can parse it controlling exceptions. The following code is a suggestion:
boolean numberValidation(String number){
try{
Integer.parseInt(number);
return true;
}catch(NumberFormatException e){
return false;
}
}
Regards,
If you're looking for hexadecimal numbers with any digit > 9, you can do it as such:
// hex representation of -889275714
int i = 0xcafebabe;
// just 1
int j = 1;
// represents both as hexadecimal String representations
String is = Integer.toHexString(i);
String ij = Integer.toHexString(j);
// pattern to check for letters
Pattern alpha = Pattern.compile("[a-z]", Pattern.CASE_INSENSITIVE);
System.out.printf("\"%s\" contains characters, \"%s\" does not.%n", is, ij);
Output
"cafebabe" contains characters, "1" does not.
Note
For Strings representing numbers, as shown in other answers, use Integer.parseInt and catch a NumberFormatException.
Try to parse it and if exception occurs return false else return true
boolean numberValidation(String supposedNumber) {
boolean flag = false;
int x;
try {
x = Integer.parseInt(supposedNumber);
flag = true;
} catch(NumberFormatException ex) {
flag = false;
}
return flag;
}
numberhas typeint, it can't contain alphanumeric characters.