Validating input using java.util.Scanner [duplicate]

Answer 1

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.

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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)

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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.

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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

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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.

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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.

Answer 2

For checking Strings for letters you can use regular expressions for example:

someString.matches("[A-F]");

For checking numbers and stopping the program crashing, I have a quite simple class you can find below where you can define the range of values you want. Here

    public int readInt(String prompt, int min, int max)
    {
    Scanner scan = new Scanner(System.in);

    int number = 0;

    //Run once and loop until the input is within the specified range.
    do 
    {
        //Print users message.
        System.out.printf("\n%s > ", prompt);

        //Prevent string input crashing the program.
        while (!scan.hasNextInt()) 
        {
            System.out.printf("Input doesn't match specifications. Try again.");
            System.out.printf("\n%s > ", prompt);
            scan.next(); 
        }

        //Set the number.
        number = scan.nextInt();

        //If the number is outside range print an error message.
        if (number < min || number > max)
            System.out.printf("Input doesn't match specifications. Try again.");

    } while (number < min || number > max);

    return number;
}

Answer 3

One idea:

try {
    int i = Integer.parseInt(myString);
    if (i < 0) {
        // Error, negative input
    }
} catch (NumberFormatException e) {
    // Error, not a number.
}

There is also, in commons-lang library the CharUtils class that provides the methods isAsciiNumeric() to check that a character is a number, and isAsciiAlpha() to check that the character is a letter...

Answer 4

what i have tried is that first i took the integer input and checked that whether its is negative or not if its negative then again take the input

Scanner s=new Scanner(System.in);

    int a=s.nextInt();
    while(a<0)
    {
    System.out.println("please provide non negative integer input ");
    a=s.nextInt();
    }
    System.out.println("the non negative integer input is "+a);

Here, you need to take the character input first and check whether user gave character or not if not than again take the character input

    char ch = s.findInLine(".").charAt(0);
    while(!Charcter.isLetter(ch))
    {
    System.out.println("please provide a character input ");
    ch=s.findInLine(".").charAt(0);
    }
    System.out.println("the character  input is "+ch);

Answer 5

If you are parsing string data from the console or similar, the best way is to use regular expressions. Read more on that here: http://java.sun.com/developer/technicalArticles/releases/1.4regex/

Otherwise, to parse an int from a string, try Integer.parseInt(string). If the string is not a number, you will get an exception. Otherise you can then perform your checks on that value to make sure it is not negative.

String input;
int number;
try
{
    number = Integer.parseInt(input);
    if(number > 0)
    {
        System.out.println("You positive number is " + number);
    }
} catch (NumberFormatException ex)
{
     System.out.println("That is not a positive number!");
}

To get a character-only string, you would probably be better of looping over each character checking for digits, using for instance Character.isLetter(char).

String input
for(int i = 0; i<input.length(); i++)
{
   if(!Character.isLetter(input.charAt(i)))
   {
      System.out.println("This string does not contain only letters!");
      break;
   }
}

Good luck!

Answer 6

Here's a minimalist way to do it.

System.out.print("Please enter an integer: ");
while(!scan.hasNextInt()) scan.next();
int demoInt = scan.nextInt();