I'm taking user input from System.in
using a java.util.Scanner
. I need to validate the input for things like:
What's the best way to do this?
Scanner.hasNextXXX
methodsjava.util.Scanner
has many hasNextXXX
methods that can be used to validate input. Here's a brief overview of all of them:
hasNext()
- does it have any token at all?hasNextLine()
- does it have another line of input?hasNextInt()
- does it have a token that can be parsed into an int
?hasNextDouble()
, hasNextFloat()
, hasNextByte()
, hasNextShort()
, hasNextLong()
, and hasNextBoolean()
hasNextBigInteger()
and hasNextBigDecimal()
hasNext(String pattern)
hasNext(Pattern pattern)
is the Pattern.compile
overloadScanner
is capable of more, enabled by the fact that it's regex-based. One important feature is useDelimiter(String pattern)
, which lets you define what pattern separates your tokens. There are also find
and skip
methods that ignores delimiters.
The following discussion will keep the regex as simple as possible, so the focus remains on Scanner
.
Here's a simple example of using hasNextInt()
to validate positive int
from the input.
Scanner sc = new Scanner(System.in);
int number;
do {
System.out.println("Please enter a positive number!");
while (!sc.hasNextInt()) {
System.out.println("That's not a number!");
sc.next(); // this is important!
}
number = sc.nextInt();
} while (number <= 0);
System.out.println("Thank you! Got " + number);
Here's an example session:
Please enter a positive number!
five
That's not a number!
-3
Please enter a positive number!
5
Thank you! Got 5
Note how much easier Scanner.hasNextInt()
is to use compared to the more verbose try/catch
Integer.parseInt
/NumberFormatException
combo. By contract, a Scanner
guarantees that if it hasNextInt()
, then nextInt()
will peacefully give you that int
, and will not throw any NumberFormatException
/InputMismatchException
/NoSuchElementException
.
hasNextXXX
on the same tokenNote that the snippet above contains a sc.next()
statement to advance the Scanner
until it hasNextInt()
. It's important to realize that none of the hasNextXXX
methods advance the Scanner
past any input! You will find that if you omit this line from the snippet, then it'd go into an infinite loop on an invalid input!
This has two consequences:
hasNextXXX
test, then you need to advance the Scanner
one way or another (e.g. next()
, nextLine()
, skip
, etc).hasNextXXX
test fails, you can still test if it perhaps hasNextYYY
!Here's an example of performing multiple hasNextXXX
tests.
Scanner sc = new Scanner(System.in);
while (!sc.hasNext("exit")) {
System.out.println(
sc.hasNextInt() ? "(int) " + sc.nextInt() :
sc.hasNextLong() ? "(long) " + sc.nextLong() :
sc.hasNextDouble() ? "(double) " + sc.nextDouble() :
sc.hasNextBoolean() ? "(boolean) " + sc.nextBoolean() :
"(String) " + sc.next()
);
}
Here's an example session:
5
(int) 5
false
(boolean) false
blah
(String) blah
1.1
(double) 1.1
100000000000
(long) 100000000000
exit
Note that the order of the tests matters. If a Scanner
hasNextInt()
, then it also hasNextLong()
, but it's not necessarily true
the other way around. More often than not you'd want to do the more specific test before the more general test.
Scanner
has many advanced features supported by regular expressions. Here's an example of using it to validate vowels.
Scanner sc = new Scanner(System.in);
System.out.println("Please enter a vowel, lowercase!");
while (!sc.hasNext("[aeiou]")) {
System.out.println("That's not a vowel!");
sc.next();
}
String vowel = sc.next();
System.out.println("Thank you! Got " + vowel);
Here's an example session:
Please enter a vowel, lowercase!
5
That's not a vowel!
z
That's not a vowel!
e
Thank you! Got e
In regex, as a Java string literal, the pattern "[aeiou]"
is what is called a "character class"; it matches any of the letters a
, e
, i
, o
, u
. Note that it's trivial to make the above test case-insensitive: just provide such regex pattern to the Scanner
.
hasNext(String pattern)
- Returns true
if the next token matches the pattern constructed from the specified string.java.util.regex.Pattern
Scanner
at onceSometimes you need to scan line-by-line, with multiple tokens on a line. The easiest way to accomplish this is to use two Scanner
, where the second Scanner
takes the nextLine()
from the first Scanner
as input. Here's an example:
Scanner sc = new Scanner(System.in);
System.out.println("Give me a bunch of numbers in a line (or 'exit')");
while (!sc.hasNext("exit")) {
Scanner lineSc = new Scanner(sc.nextLine());
int sum = 0;
while (lineSc.hasNextInt()) {
sum += lineSc.nextInt();
}
System.out.println("Sum is " + sum);
}
Here's an example session:
Give me a bunch of numbers in a line (or 'exit')
3 4 5
Sum is 12
10 100 a million dollar
Sum is 110
wait what?
Sum is 0
exit
In addition to Scanner(String)
constructor, there's also Scanner(java.io.File)
among others.
Scanner
provides a rich set of features, such as hasNextXXX
methods for validation.hasNextXXX/nextXXX
in combination means that a Scanner
will NEVER throw an InputMismatchException
/NoSuchElementException
.hasNextXXX
does not advance the Scanner
past any input.Scanner
if necessary. Two simple Scanner
is often better than one overly complex Scanner
.Scanner
method that takes a String pattern
argument is regex-based.
String
into a literal pattern is to Pattern.quote
it.