Find out whether user have entered string, integer or double values

Question

Scanner sc = new Scanner(System.in);
System.out.println("Enter whatever you want");
String st = sc.nextLine();
try {
    String d = String.valueOf(st);
    if (d == (String) d) {
        System.out.println((String) d);
    } else {
        System.out.println(d);
    }
} catch(Exception e) {
    System.out.println("integer");
}

When I try to execute this, it keeps printing "if" part even for integer and double.

Since `d` is already a `String`, casting it to a `String` is redundant; then `d == d` is obviously always true; but checking `d == d` is redundant, since you do the same thing in both the true and false branches. And `String.valueOf(st)` is also redundant, since `st` is already a `String`, so `st.toString() == st`. — Andy Turner, Aug 02 '16 at 07:06
To know what user has given as input you can use `instanceof`or apparently use sc.hasNextDouble() or hasNextInt() — Imran, Aug 02 '16 at 07:10
this is also a useful link you can refer http://stackoverflow.com/questions/1102891/how-to-check-if-a-string-is-numeric-in-java — Santhucool, Aug 02 '16 at 07:14
Possible duplicate of [Validating input using java.util.Scanner](http://stackoverflow.com/questions/3059333/validating-input-using-java-util-scanner) — Tom, Aug 02 '16 at 07:23

Liviu Stirb · Answer 1 · 2016-08-02T07:14:48.793

In order to use your code: Try to parse to integer first. If this is successful it means you have an int. If this doesn't work try parsing to a double, again if this works it means you have a double otherwise you have a string.

Use Integer.valueOf and Double.valueOf:

    System.out.println("Enter whatever you want");
    String st = sc.nextLine();
    try {
        Integer d = Integer.valueOf(st);

        System.out.println("Integer: " + d);
    } catch (NumberFormatException e) {
        try {
            Double d = Double.valueOf(st);
            System.out.println("Double: " + d);
        }catch (NumberFormatException nf) {
            System.out.println("String: " + st);
        }
    }

But I wouldn't build it this way. A better option is to use sc.hasNextInt and sc.hasNextDouble

may be @Mureinik's answer is more relavent – Imran Aug 02 '16 at 07:14 — Imran, Aug 02 '16 at 07:14

score 1 · Answer 2 · answered Aug 02 '16 at 07:10

Any input can be evaluated as a string, even if it is "2.9". Instead, you could use Scanner's hasXYZ methods:

if (sc.hasNextInt()) {
    System.out.println("Integer");
} else if (sc.hasNextDouble()) {
    System.out.println("Double");
} else {
    System.out.println("String");
}

score 0 · Answer 3 · answered Aug 02 '16 at 07:18

You can use this program to output the type:

import java.util.Scanner;

public class MyTest {
public static void main(String args[]) throws Exception {
    Scanner sc = new Scanner(System.in);

    System.out.println("Enter whatever you want");

    String st = sc.nextLine();
    try {
        Integer.parseInt(st);
        System.out.println("Integer");
    } catch (NumberFormatException nfe) {

        try {
            Double.parseDouble(st);
            System.out.println("Double");
        } catch (NumberFormatException nfe2) {
            System.out.println("String");
        }

    }
    sc.close();
}
}

score -1 · Answer 4 · answered Aug 02 '16 at 07:09

-1

try this:

String d= String.valueOf(st);
try{
 //If double or integer comes here
  double val = Double.parseDouble(d);
}catch(Exception e){
  //If String comes here
}

answered Aug 02 '16 at 07:09

Santhucool

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Find out whether user have entered string, integer or double values

4 Answers4