DrJava Error message

KimboForte

Hi there,

I am a studying java on a beginners SDLC course and currently working on an assignment.

I need to create a user name and password but I keep coming up with a "The local variable user_name is never read" error message.

What's wrong? here's what I've got so far.


// Begin Program
import java.util.Scanner;
// Define class
public class groceryOnline

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

String user_name;
System.out.println("Enter; ");
first_name = user_input.next();
}
}

Umekichi

Hi there,
you have declared a String for user_name, yet you used first_name to store the input.
Did you mean to do:

String user_name;
System.out.println("Enter; ");
user_name = user_input.next();

KimboForte

Just tried that and it didnt work,

import java.util.Scanner;
// Define class
public class groceryOnline

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

String user_name;
System.out.println("Enter: ");
user_name = user_input.next();
}
}

Umekichi

Just fiddled with it there the problem is you aren't using user_name. It just sits there waiting to be used.
if you put in a System.out.println(user_name); statement after user_name = user_input.next(); you'll see what i mean.

onemorechance

A simple text scanner which can parse primitive types and strings using regular expressions.

A Scanner breaks its input into tokens using a delimiter pattern, which by default matches whitespace. The resulting tokens may then be converted into values of different types using the various next methods.

For example, this code allows a user to read a number from System.in:

Scanner sc = new Scanner(System.in);
int i = sc.nextInt();

As another example, this code allows long types to be assigned from entries in a file myNumbers:

Scanner sc = new Scanner(new File("myNumbers"));
while (sc.hasNextLong()) {
long aLong = sc.nextLong();
}
The scanner can also use delimiters other than whitespace. This example reads several items in from a string:

String input = "1 fish 2 fish red fish blue fish";
Scanner s = new Scanner(input).useDelimiter("\\s*fish\\s*");
System.out.println(s.nextInt());
System.out.println(s.nextInt());
System.out.println(s.next());
System.out.println(s.next());
s.close();
prints the following output:

1
2
red
blue
The same output can be generated with this code, which uses a regular expression to parse all four tokens at once:

String input = "1 fish 2 fish red fish blue fish";
Scanner s = new Scanner(input);
s.findInLine("(\\d+) fish (\\d+) fish (\\w+) fish (\\w+)");
MatchResult result = s.match();
for (int i=1; i<=result.groupCount(); i++)
System.out.println(result.group(i);
s.close();
The default whitespace delimiter used by a scanner is as recognized by Character.isWhitespace.

A scanning operation may block waiting for input.

Java docs

onemorechance

Don't forget to close your scanner.

user_input.close();

System.out.println("user name: " + user_name);

Also in Java class names should start with upper case, always do this.

KimboForte

Finally :- 0 it works! great thanks so much!

mad turnip

user_input.next();

I think should be

user_input.nextLine();

although according to the API next() should work.

onemorechance

mad turnip wrote: »

user_input.next();

I think should be

user_input.nextLine();

although according to the API next() should work.

It's because the scanner is not closed; it blocks still accepting input.

KimboForte

What if I don't want to system to display the username i just entered? It should go like this:

Enter Username:
Enter Password:

Welcome/invalid log in.

Obviously I need to set a value to the username string.

Umekichi

KimboForte wrote: »

What if I don't want to system to display the username i just entered? It should go like this:

Enter Username:
Enter Password:

Welcome/invalid log in.

Obviously I need to set a value to the username string.

Maybe try using an if statement?

if(username.equals(what you want username to be ) && password.equals(what you want pass to be)){
System.out.println("Welcome!");
}else{
System.out.println("Invalid log-in");
}

KimboForte

Whatever I put in the brackets for username, the error message being returned is that it can't be resolved to a variable

Umekichi

This is what I did:

import java.util.Scanner;
public class groceryOnline {
public static void main(String[] args) {
Scanner user_input = new Scanner(System.in);

String password;//password declared
String user_name;//username declared
System.out.println("Enter: ");
user_name = user_input.nextLine();

System.out.println("Password: ");
password = user_input.nextLine();

if(user_name.equals("hi") && password.equals("ok")){
System.out.println("Welcome!");
}else{
System.out.println("Invalid log-in");
}
user_input.close();
}
}

So what I did was, took info from user for Username and password. Then I ran an if statement that compared the user_name and password to ones I specified and if both were equal it printed "Welcome!" and if they both weren't equal then it prints, "Invalid log-in".