Data type recognition

Question

I want to create a Java program where a user will input something and the output will show what type of data the input is..? For example:

Input: 25
Output: integer
Input: ABC
Output: string
Input: 12.7
Output: float/double.

Please help as I am clueless on how to work this out

Possible duplicate of [Checking input type...how?](http://stackoverflow.com/questions/5333110/checking-input-type-how) — Jack Deeth, Dec 08 '16 at 18:23
[This previous answer](http://stackoverflow.com/a/40062753/2449857) shows exactly how to do this. — Jack Deeth, Dec 08 '16 at 18:23

GhostCat · Answer 1 · 2016-12-08T18:19:10.480

A simple approach could go like this; starting with some input string X.

If X can be parsed as Integer --> it is an int/Integer
Then you try Long
Then Float
Then Double
If nothing worked, you probably have a string there

( with "parsing" I mean using methods such as Integer.parseInt() ... you pass in X; and when that method doesn't throw an exception on you, you know that X is an Integer/int )

But: such a detection very much depends on your definition of valid inputs; and potential mappings. As there zillions of ways of interpreting a string. It might not be a number; but given a correct format string ... it could be timestamp.

So the very first step: clarify your requirements! Understand the potential input formats you have to support; then think about their "mapping"; and a potential check to identify that type.

ItamarG3 · Answer 2 · 2016-12-08T18:21:41.307

1

You can get a string and try to parse it as other types:

Scanner s = new Scanner(System.in);//allows user input
input = s.nextLine();//reads user input
try{//java will try to execute the code but will go to the catch block if there's an exception.
    int inputInt = Integer.parseInt(input);//try to convert input to int
catch(Exception e){
    e.printStackTrace();//this tell you exactly what went wrong. If you get here, then the input isn't an integer.
}
//same with double

edited Dec 08 '16 at 18:21

answered Dec 08 '16 at 18:16

ItamarG3

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Now a hint for you: never ever suggest to a newbie to go with an **empty** catch block. That is super-bad practice! – GhostCat Dec 08 '16 at 18:20
@GhostCat Yes, I was thinking how exactly to explain what **should** go there – ItamarG3 Dec 08 '16 at 18:20

Ébe Isaac · Accepted Answer · 2017-02-09T03:42:46.903

This should work for your purpose:

import java.util.Scanner;

public class DataType 
{
   public static void main(String[] args)
   {
      Scanner in = new Scanner(System.in);
      if(in.hasNextByte(2))
         System.out.println("Byte");
      else if(in.hasNextInt())
         System.out.println("Integer");
      else if(in.hasNextFloat())
         System.out.println("Float");
      else if(in.hasNextBoolean())
         System.out.println("Boolean");
      else if(in.hasNext())
         System.out.println("String");
   }
}

Note that the order of if...else statements is very important here because of the following set relations with respect to patterns:

All byte patterns can be integers
All integer patterns can be floats
All float patterns can be Strings
All booleans can be Strings

There are quite a lot of hasNext..() methods in the Scanner class, such as BigInteger, short, and so on. You may refer the Scanner class documentation for further details.

Data type recognition

3 Answers3