Logo Questions Linux Laravel Mysql Ubuntu Git Menu
 

Validating input using java.util.Scanner [duplicate]

I'm taking user input from System.in using a java.util.Scanner. I need to validate the input for things like:

  • It must be a non-negative number
  • It must be an alphabetical letter
  • ... etc

What's the best way to do this?

like image 751
bhavna raghuvanshi Avatar asked Jun 17 '10 06:06

bhavna raghuvanshi


People also ask

How do I validate input when using scanner?

To validate input the Scanner class provides some hasNextXXX() method that can be use to validate input. For example if we want to check whether the input is a valid integer we can use the hasNextInt() method.

How do you validate data in Java?

In the Java programming language, the most natural way of doing data validation seems to be the following: try to build an object. if no problem is found, then just use the object. if one or more problems are found, then ensure the caller has enough information to tell the user about the issues.


2 Answers

Overview of Scanner.hasNextXXX methods

java.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?
  • For Java primitives
    • hasNextInt() - does it have a token that can be parsed into an int?
    • Also available are hasNextDouble(), hasNextFloat(), hasNextByte(), hasNextShort(), hasNextLong(), and hasNextBoolean()
    • As bonus, there's also hasNextBigInteger() and hasNextBigDecimal()
    • The integral types also has overloads to specify radix (for e.g. hexadecimal)
  • Regular expression-based
    • hasNext(String pattern)
    • hasNext(Pattern pattern) is the Pattern.compile overload

Scanner 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.


Example 1: Validating positive ints

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.

Related questions

  • How to use Scanner to accept only valid int as input
  • How do I keep a scanner from throwing exceptions when the wrong type is entered? (java)

Example 2: Multiple hasNextXXX on the same token

Note 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:

  • If you need to skip the "garbage" input that fails your hasNextXXX test, then you need to advance the Scanner one way or another (e.g. next(), nextLine(), skip, etc).
  • If one 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.


Example 3 : Validating vowels

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.

API links

  • hasNext(String pattern) - Returns true if the next token matches the pattern constructed from the specified string.
  • java.util.regex.Pattern

Related questions

  • Reading a single char in Java

References

  • Java Tutorials/Essential Classes/Regular Expressions
  • regular-expressions.info/Character Classes

Example 4: Using two Scanner at once

Sometimes 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.


Summary

  • Scanner provides a rich set of features, such as hasNextXXX methods for validation.
  • Proper usage of hasNextXXX/nextXXX in combination means that a Scanner will NEVER throw an InputMismatchException/NoSuchElementException.
  • Always remember that hasNextXXX does not advance the Scanner past any input.
  • Don't be shy to create multiple Scanner if necessary. Two simple Scanner is often better than one overly complex Scanner.
  • Finally, even if you don't have any plans to use the advanced regex features, do keep in mind which methods are regex-based and which aren't. Any Scanner method that takes a String pattern argument is regex-based.
    • Tip: an easy way to turn any String into a literal pattern is to Pattern.quote it.
like image 74
polygenelubricants Avatar answered Oct 06 '22 23:10

polygenelubricants


Here's a minimalist way to do it.

System.out.print("Please enter an integer: ");
while(!scan.hasNextInt()) scan.next();
int demoInt = scan.nextInt();
like image 20
dsharhon Avatar answered Oct 07 '22 00:10

dsharhon