Java scanner input mismatch when using space in input

Question

I am using the scanner in java and am trying to enter a space in my input for option 2 (removing a user from my hashmap) but when I add a space in my answer I get an InputMismatchException. while researching I came across this thread Scanner Class InputMismatchException and Warnings that says to use this line of code to solve the issue: .useDelimiter(System.getProperty("line.separator")); i have added this and now my option 2 goes into a never-ending loop of me inputting data. Here is my code:

public class Test {

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

        AddressBook ad1 = new AddressBook();
        String firstName="";
        String lastName="";
        String key="";
        int choice=0;
      do{
        System.out.println("********************************************************************************");
        System.out.println("Welcome to the Address book. Please pick from the options below.\n");
        System.out.println("1.Add user \n2.Remove user \n3.Edit user \n4.List Contact \n5.Sort contacts \n6.Exit");

          System.out.print("Please enter a choice: ");
         choice = scan.nextInt();

        if(choice==1){
            //Add user
            System.out.print("Please enter firstname: ");
            firstName=scan.next();
            System.out.print("Please enter lastname: ");
            lastName=scan.next();
            Address address = new Address();
            key = lastName.concat(firstName);
            Person person = new Person(firstName,lastName);
            ad1.addContact(key,person);
            System.out.println("key: " + key);
        }

        else if(choice==2){
            //Remove user
            System.out.println("Please enter name of user to remove: ");
            scan.useDelimiter(System.getProperty("line.separator"));
            key=scan.next();
            System.out.println("name:" + key);
            ad1.removeContact(key);  
        }

        else if(choice==3){
            //Edit user
        }

        else if(choice==4){
            //List contact
            ad1.listAllContacts();

        }

       else if(choice==5){
            //Sort contacts
        }
      }while(choice!=6);
    }
}

The reason why I need to use a space is to remove a user from my hashmap I need to enter their full name as the key is a concatenation of their last and firstname, any help will be appreciated

You should discard previous line by using `scan.nextLine();` instead of `useDelimiter()` — Yoshikage Kira, Sep 17 '19 at 14:41
Do it before print line and don't save return value anywhere. — Yoshikage Kira, Sep 17 '19 at 14:51
Thank you, it works fine now, would I have to do that every time I need to scan a full line then? — Charlie Ansell, Sep 17 '19 at 15:00
You can do it after `choice = scan.nextInt();` then you don't have to do it in every if-else — Yoshikage Kira, Sep 17 '19 at 15:02

score 0 · Accepted Answer · answered Sep 17 '19 at 15:48

nextInt() behaves similar to next() that is when it read a line it places the cursor behind it.

Example: You give 6 as input

6 
 ^(scanner's cursor)

So next time when you call nextLine(). It will return the whole line after that cursor which is empty in this case.

To fix this issue you need to call an extra nextLine() so that the Scanner closes the previous line it was reading and move on to the next line.

You could do this

System.out.print("Please enter a choice: ");
choice = scan.nextInt(); // Reads the int
scan.nextLine(); // Discards the line

And in choice 2 since you want full name of user you could just use nextLine() to get whole line along with space.

//Remove user
System.out.println("Please enter full name of user to remove: ");
key=scan.nextLine();
System.out.println("name:" + key);
ad1.removeContact(key);

Or you could do something similar to what you did in choice 1

System.out.print("Please enter firstname: ");
firstName=scan.next();
System.out.print("Please enter lastname: ");
lastName=scan.next();
key = lastName.concat(firstName);
System.out.println("name:" + key);
ad1.removeContact(key);  
scan.nextLine(); // This is will make sure that in you next loop `nextInt()` won't give an input mismatch exception

Java scanner input mismatch when using space in input

1 Answers1