Reading String in java

Question

I've been working on this code for school to create a course, users, students, and teachers but I ran into a snag while creating a course.

What I want to do is (in the terminal) read an input stream such as "CSC 110 Into to java" and enter all of it into a string. I am using Scanner for input. and have tried hasNext() and nextLine() besides next(). I thought about writing a function to use a char array get the entire string and then convert the char array into a string.

First I wanted to know if there was an easier solution?

Edit: Sorry I wasn't clear. Without going into all of my code.

current Code:

        System.out.print("Enter the Course Title: ");
        title = keyboard.nextLine();
        System.out.print("Enter a Brief Course Description: ");
        desc = keyboard.nextLine();

        if (t != null)
        {   
            do
            {
                System.out.println("\nPick a teacher\n");

                for (int j = 0; j < t.size(); j++)
                {
                    nTeacher = t.get(j);
                    s += "\t(" + (j+1) + ") " + nTeacher + "\n";
                }

                System.out.print(s + "\nEnter the Teacher ID: ");
                int tId = keyboard.nextInt(); // My error

Current error:

Enter the Course Title: CSC 110 Into to java

Enter a Breif Course Description: 
Pick a teacher

Enter the Teacher ID: 
Exception in thread "main" java.util.InputMismatchException
    at java.util.Scanner.throwFor(Scanner.java:857)
    at java.util.Scanner.next(Scanner.java:1478)
    at java.util.Scanner.nextInt(Scanner.java:2108)
    at java.util.Scanner.nextInt(Scanner.java:2067)
    at Prototype.classCreator(Prototype.java:530)
    at Prototype.mainMenu(Prototype.java:181)
    at Prototype.login(Prototype.java:121)
    at Prototype.main(Prototype.java:48)

When I use nextLine()

Create Class
Enter the courseID: 27222
Enter the Course Title: Enter a Brief Course Description:

Intended result:

Create Class
Enter the courseID: 27299
Enter the Course Title: CSC 110 Into to Java
Enter a Brief Course Description: Introduction to programming
Pick a teacher
(1) ID: 1111    Name: Bob
Enter the Teacher ID: 1
Class Created

I understand that when I have three strings "CSC 110 Intro" it get to the input for nextInt() but I don't understand why nextLine() is writing my println's like this.

Sorry new at SO, virtually. I fixed that :)

Jeremy
– Jeremy

2011-04-04 06:56:39 +00:00
Commented Apr 4, 2011 at 6:56 — Jeremy
– Jeremy, Commented Apr 4, 2011 at 6:56

donnyton · Accepted Answer · 2011-04-04 20:53:48Z

6

Well, you haven't told us exactly what the error is, so I'll just give some sample Scanner code and let you work with that:

//create the Scanner
Scanner terminalInput = new Scanner(System.in);

//read input
String s = terminalInput.nextLine();

That should work; Scanner is one of the simplest classes to use. Perhaps you accidentally used

new Scanner(); //no input source!

rather than

new Scanner(System.in); //links to terminal

EDIT:

I've surmised from your output that you are calling

keyboard.nextInt();

and later

keyboard.nextLine();

This is what is messing up your program. The nextInt() method leaves the "\n" endline symbol and is picked up immediately by nextLine(), skipping over the next input. What you want to do is use nextLine for everything, and parse it later:

String nextIntString = keyboard.nextLine(); //get the number as a single line
int nextInt = Integer.parseInt(nextIntString); //convert the string to an int

This is by far the easiest way to avoid problems--don't mix your "next" methods. Use only nextLine() and then parse ints/separate words afterwards.

edited Apr 4, 2011 at 20:53

answered Apr 4, 2011 at 4:59

donnyton

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4 Comments

Jeremy Over a year ago

Could you check my edit. I've tried nextLine() and it ran into a problem.

donnyton Over a year ago

Your edit fails to provide the code you used when it ran into the nextLine() problem...

Jeremy Over a year ago

I have the output for when I use nextLine() I'm confused on what I'm missing to help you understand.

donnyton Over a year ago

What I mean is, you need to give us EXACTLY the code you are using with getLine(). Giving us your old code and error isn't really helpful, because there could be a dozen different issues. Show us exactly how you are using nextLine() and we will surely have a solution faster.

Cory G. · Accepted Answer · 2011-04-04 04:59:31Z

4

Scanner scan = new Scanner(System.in);
String str_input = scan.nextLine();

Would that work for you?

answered Apr 4, 2011 at 4:59

Cory G.

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3 Comments

Jeremy Over a year ago

Could you check my edit. I've tried nextLine() and it ran into a problem

Cory G. Over a year ago

Are you entering an integer there? InputMismatchException sounds like it did not receive an integer. However, I would probably change it to not use nextInt(), and instead use nextLine() throughout. nextInt() and the others allow users to enter many ints on the same line (e.g. "25 18 76"), which could create problems for your interface if you later expect full lines. While fixable, it would probably be easier to get the string using nextLine(), and then use Integer.parseInt().

Jeremy Over a year ago

I'm entering an integer after the String input. When I use the nextLine() it prints the next prompt.. checkout the example of the output.

Shankar Raju · Accepted Answer · 2011-04-04 05:08:59Z

3

You can use the new Scanner(System.in) as mentioned by a couple of people, or you could also try this:

BufferedReader br = new BufferedReader(new InputStreamReader(System.in));

String s = br.readLine();

answered Apr 4, 2011 at 5:08

Shankar Raju

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Comments

chantheman · Accepted Answer · 2011-04-04 05:42:19Z

0

BufferedReader bufferedReader = new BufferedReader(new FileReader(fileName));

That would be how you would read it in from a file btw.

I think BufferedReader is better than scanner simply because it does what it says, it buffers.

But yea scanner works too.

Try making a config file instead of reading it off the command line.

Properties configFile = new Properties();
configFile.load(new FileInputStream("conf.properties"));

myString = configFile.getProperty("MY_STRING");

The config file would be named conf.properties and then it would have in it:

MY_STRING = blahblahbah

answered Apr 4, 2011 at 5:42

chantheman

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