How can I check if an input is a integer or String, etc.. in JAVA?

Question

I am wondering how I can check if a user's input is a certain primitive type (I mean integer, String, etc... I think it's called primitive type?). I want a user to input something, then I check if it's a String or not, and do a certain action accordingly. (JAVA) I have seen some codes like this:

if (input == (String)input) { (RANDOM STUFF HERE) }

or something like

if input.equals((String) input)

And they don't work. I want to know how I can Check for only a String? (EDITED OUT) I was wondering if there was a function that can do that? Thanks for the answers

EDIT: With the help of everyone I have created my fixed code that does what I want it to:

package files;
import java.util.*;
public class CheckforSomething {
static Scanner inputofUser = new Scanner(System.in);    
static Object userInput;
static Object classofInput;
public static void main(String[] args){
    try{
    System.out.print("Enter an integer, only an integer: ");
    userInput = inputofUser.nextInt();

    classofInput = userInput.getClass();
    System.out.println(classofInput);
    } catch(InputMismatchException e) {
        System.out.println("Not an integer, crashing down");

    }

 }



 }

No need for answers anymore, thanks!

What is input? Where did it come from, and how? ie. If it is a String, then it will always be a String.. and the casts make no sense. — user2864740
– user2864740, Commented Sep 13, 2014 at 4:55

Rustam · Accepted Answer · 2014-09-13 05:52:45Z

5

Use instanceof to check type and typecast according to your type:

 public class A {

    public static void main(String[]s){

        show(5);
        show("hello");
    }
    public static void show(Object obj){
        if(obj instanceof Integer){
            System.out.println((Integer)obj);
        }else if(obj instanceof String){
            System.out.println((String)obj);
        }
    }
}

answered Sep 13, 2014 at 5:52

Rustam

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2 Comments

user4017662 Over a year ago

This works in that context but not with a Scanner. :(

roottraveller Over a year ago

what about if I take input from stdio? I need help in this case.

user2575725 · Accepted Answer · 2014-09-13 05:00:38Z

3

You may try this with Regex:

String input = "34";
if(input.matches("^\\d+(\\.\\d+)?")) {
  //okay
} else {
  // not okay !
}

Here,

^\\d+ says that input starts with a digit 0-9,

()? may/or may not occur

\\. allows one period in input

answered Sep 13, 2014 at 5:00

user2575725

Comments

T. Short · Accepted Answer · 2019-12-27 16:36:19Z

1

Scanner input = new Scanner (System.in);

if (input.hasNextInt()) System.out.println("This input is of type Integer.");
else if (input.hasNextFloat()) System.out.println("This input is of type Float.");
else if (input.hasNextLine()) System.out.println("This input is of type string."); 
else if (input.hasNextDouble()) System.out.println("This input is of type Double."); 
else if (input.hasNextBoolean()) System.out.println("This input is of type Boolean.");  

  else if (input.hasNextLong())
    System.out.println("This input is of type Long.");

edited Dec 27, 2019 at 16:36

T. Short

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answered Dec 27, 2019 at 15:52

Ajay Gupta

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Comments

Petar_D · Accepted Answer · 2020-05-30 10:26:29Z

Hate to bring this up after 6 years but I found another possible solution.

Currently attending a coding bootcamp and had to solve a similar problem. We introduce booleans and change their values depending on the result of the try/catch blocks. We then check the booleans using simple if statements. You can omit the prints and input your code instead. Here's what it looks like:

import java.io.IOException;
import java.util.Scanner;

public class DataTypeFinder {
    public static void main(String[] args) throws IOException {
        Scanner scan = new Scanner(System.in);
        String input = "";

        while (true) { //so we can check multiple inputs (optional)
            input = scan.nextLine();

            if ("END".equals(input)) { //a way to exit the loop
                break;
            }

            boolean isInt = true; //introduce boolean to check if input is of type Integer
            try { // surround with try/catch
                int integer = Integer.parseInt(input); //boolean is still true if it works
            } catch (NumberFormatException e) {
                isInt = false; //changed to false if it doesn't
            }

            boolean isDouble = true; //same story
            try {
                double dbl = Double.parseDouble(input);
            } catch (NumberFormatException e) {
                isDouble = false;
            }

            if (isInt) {
                System.out.printf("%s is integer type%n", input);
            } else if (isDouble) {
                System.out.printf("%s is floating point type%n", input);
            } else if (input.length() == 1) { //this could be useless depending on your case
                System.out.printf("%s is character type%n", input);
            } else if ("true".equals(input.toLowerCase()) || "false".equals(input.toLowerCase())) {
                System.out.printf("%s is boolean type%n", input);
            } else {
                System.out.printf("%s is string type%n", input);
            }
        }
    }
}

Pingolin · Accepted Answer · 2019-08-20 13:27:16Z

0

class Main{
    public static void main(String args[]){
        String str;
        Scanner sc=new Scanner(System.in);
        int n,
        boolean flag=false;
        while(!flag){
            try{
                str=sc.nextLine();
                n=Integer.parseInt(str);
                flag=true;
            }
            catch(NumberFormatException e){
                System.out.println("enter an no");
                str=sc.nextLine();
            }
        }
    }
}

edited Aug 20, 2019 at 13:27

Pingolin

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answered Aug 20, 2019 at 13:07

G Bulli Swamy Reddy

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Comments

subham soni · Accepted Answer · 2014-09-13 05:00:37Z

-1

Is this ok?

class Test
{
    public static void main(String args[])
    {
        java.util.Scanner in = new java.util.Scanner(System.in);
        String x = in.nextLine();
        System.out.println("\n The type of the variable is : "+x.getClass());
    }
}

Output:

subham@subham-SVE15125CNB:~/Desktop$ javac Test.java 
subham@subham-SVE15125CNB:~/Desktop$ java Test
hello

 The type of the variable is : java.lang.String

answered Sep 13, 2014 at 5:00

subham soni

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Comments

Govind · Accepted Answer · 2014-09-13 05:18:59Z

-1

But Zechariax wanted an answer with out using try catch

You can achieve this using NumberForamtter and ParsePosition. Check out this solution

import java.text.NumberFormat;
import java.text.ParsePosition;


    public class TypeChecker {
        public static void main(String[] args) {
        String temp = "a"; // "1"
        NumberFormat numberFormatter = NumberFormat.getInstance();
        ParsePosition parsePosition = new ParsePosition(0);
        numberFormatter.parse(temp, parsePosition);
        if(temp.length() == parsePosition.getIndex()) {
            System.out.println("It is a number");
        } else {
            System.out.println("It is a not number");
        }
    }
}

answered Sep 13, 2014 at 5:18

Govind

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3 Comments

user4017662 Over a year ago

This looks really good, but I'm not that advanced (Sorry) and changed my mind of the whole Try catch it looks easier than this complicated code. But thanks anyways :)

Govind Over a year ago

I agree Zechariax. Try catch is a simple solution. But if you don't have that option you could use this at anytime.

user4017662 Over a year ago

Ok, also is there a way to use if(randomVar = (int)) {} or something like that? That would be the most useful.

Asaf Nevo · Accepted Answer · 2014-09-13 04:45:47Z

-2

Try instanceof function with Integer instead of int.. each primitive also have a class

answered Sep 13, 2014 at 4:45

Asaf Nevo

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1 Comment

user4017662 Over a year ago

Can you give more description or a code? I don't know exactly how to use it, that will help thanks

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