Issues reading string input in Java with scanner class

Question

I'm pretty much a complete newbie when it comes to Java. I've dabbled a bit in python and VB.net, and that's about it.

I'm trying to write a program in java that literally reads the user's input and displays it back to them with this code:

import java.util.Scanner;

public class InputTesting
{   
    public static void main( String[] args )
    {
        Scanner input = new Scanner ( System.in );

        String str1;

        System.out.println("Input string: ");

        str1 = input.nextString();

        System.out.println( str1 );
    }
}

And I get the error:

InputTesting.java:13: error: cannot find symbol
        str1 = input.nextString();
                    ^
  symbol:   method nextString()
  location: variable input of type Scanner
1 error

Can someone tell what why it's not compiling? Thanks!

Learning to read compile errors will help you immensely. This is literally telling you that scanner has no method nextString. At that point, if you shoot on over to the java API for scanner you can look for the method you need. — Cruncher
– Cruncher, Commented Aug 30, 2013 at 13:43

christopher · Accepted Answer · 2013-08-30 13:38:07Z

11

input.nextString();

There is no method called nextString in the Scanner class. That's why you're getting the error cannot find symbol.

Try

input.nextLine(); // If you're expecting the user to hit enter when done.
input.next(); // Just another option.

answered Aug 30, 2013 at 13:38

christopher

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Comments

Rakesh KR · Accepted Answer · 2013-08-30 13:42:33Z

0

import java.util.Scanner;

public class InputTesting
{   
    public static void main( String[] args )
    {
        Scanner input = new Scanner ( System.in );

         String str1;

        System.out.println("Input string: ");

            str1 = input.next();

        System.out.println( str1 );
    }
}

answered Aug 30, 2013 at 13:42

Rakesh KR

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Comments

Prashant Bhate · Accepted Answer · 2013-08-30 13:59:36Z

You may use one of input.nextXXX() as detailed in Scanner javadoc. to resolve compiler error

in your case

you may use input.nextLine(); get entire line of text.

By Default whitespace delimiter used by a scanner. However If you need to scan sequence of inputs separated by a delimiter useDelimiter can be used to set regex delim

public static void main(String[] args) {
    try (Scanner input = new Scanner(System.in);) {
        input.useDelimiter(";");
        while (input.hasNext()) {
            String next = input.next();
            if ("DONE".equals(next)) {
                break;
            }
            System.out.println("Token:" + next);
        }
    }
}

Input

1a;2b;3c;4d;5e;DONE;

Output

Token:1a
Token:2b
Token:3c
Token:4d
Token:5e

Just be cautious while scanning decimal and signed numbers as scanning is Locale Dependent

Krzysztof Wolny · Accepted Answer · 2015-11-13 09:49:07Z

0

You need to rather use method nextLine(). This methods prints entire line. By using next() it will print only one word in your sentence.

edited Nov 13, 2015 at 9:49

Krzysztof Wolny

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answered Nov 13, 2015 at 9:27

user5558192

1

Comments

Rubyyat Abir · Accepted Answer · 2022-12-25 17:08:21Z

0

// That's how i managed solution for input a string

import java.util.Scanner; public class PalindromeWord {

public static void main(String[] args) {
    
    //System.out.println("Enter a string: ");
    Scanner input = new Scanner(System.in);
    


     String S = input.next(); 
     String reverseStr = "";
    
    int strLength = S.length();

    for (int i = (strLength - 1); i >=0; --i) {
      reverseStr = reverseStr + S.charAt(i);
    }

    if (S.equals(reverseStr)) {
      System.out.println("Yes");
    }
    else {
      System.out.println("No");
    }

}

}

edited Dec 25, 2022 at 17:08

answered Dec 22, 2022 at 5:36

Rubyyat Abir

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1 Comment

Gauthier Tassin Over a year ago

Please, format your answer properly with your code block

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