User string input. Counting characters

Question

My code works except I have to make it pass some Junit test. It passes all but one. It passes when the character enters nothing, enters upper case, lower case, or a mix of the two, and it works when the enter Hello World!

public static void main(String[] args)  {
    Scanner input = new Scanner(System.in);
    int[] textArray = new int[26];

    System.out.print("Enter text: ");
    readText(input, textArray);
}

public static void readText(Scanner input, int[]text){
    char letter = 0;

    if (input.hasNext() == false) {
        System.out.println();
    }
    else {
        while (input.hasNext()) {
            String a = input.nextLine();

            for (int i = 0; i < a.length(); i++) {
                letter = a.charAt(i);

                if (letter >= 'A' && letter <= 'Z') {
                    text[letter-65] = (text[letter-65]) + 1;
                }
                else if (letter >= 'a' && letter <= 'z') {
                    text[(letter - 32) - 65] = (text[(letter - 32) - 65]) + 1;
                }
                else if (letter == ' ') {
                    System.out.print("");
                }
                else {
                    System.out.print("");
                }
            }
        }
    }

    String[] alphabet = {"A","B","C","D","E","F","G","H","I","J","K","L","M","N","O","P","Q","R","S","T","U","V","W","X","Y","Z"};

    for (int y = 0; y < text.length; y++) {
        if (text[y] > 0) {
            System.out.println(alphabet[y] + ": " + text[y]);
        }
    }
}

The Junit tests this input: 1 2 3%n! ? >%n:) !!%n

The expected output is

Enter text: 
    //empty line here

But instead the output from my code is

Enter text: //with no line after

I'm not sure how to get the extra line after without ruining my other junit tests. I tried one way and it worked but then my Hello World didn't work properly.

And example of when its working:
When I hit run the console will say

Enter text:

I have a Scanner input so the user will enter some words and it will look like this in console

Enter text: sOme wordS

Then it will count the number of times each letter was used and print that to console like this

Enter text: sOme wordS
D: 1
E: 1
M: 1
O: 2
R: 1
S: 2
W: 1

If I don't enter anything when asked and just hit the enter key the output is

Enter text:
//empty line here

But when I enter

Enter text: 1 2 3
? ! >
:) !!

The output doesn't add an extra line at the end.

Use System.out.println("Enter text: ") instead of System.out.print("Enter text: "). — JB Nizet
– JB Nizet, Commented Nov 11, 2016 at 17:03
System.out.println("Enter text: " + "\n"); You want a new line after "Enter text". If you also want to tab it, you can use System.out.println("Enter text: " + "\n\t"); — Munib
– Munib, Commented Nov 11, 2016 at 17:08
@return0 Can't do that because user input has to be on the same line as "Enter text: " — Jay58t
– Jay58t, Commented Nov 11, 2016 at 17:11
@JaynineReturns I thought you wanted the expected output with a new line... Not sure what you mean by user input. — Munib
– Munib, Commented Nov 11, 2016 at 17:13
Where does the user input come in at this line? Before the Enter text? @JaynineReturns Enter text: //empty line here — Munib
– Munib, Commented Nov 11, 2016 at 17:21

dammina · Accepted Answer · 2016-11-11 17:50:11Z

1

Try using System.out.println("Enter text: "); instead of System.out.print("Enter text: ");

Update: After getting clarified your requirement I propose to use the following code.

public static void readText(Scanner input, int[]text){
        char letter = 0;

        while (input.hasNext()) {
            String a = input.nextLine();

            for (int i = 0; i < a.length(); i++) {
                letter = a.charAt(i);

                if (letter >= 'A' && letter <= 'Z') {
                    text[letter-65] = (text[letter-65]) + 1;
                }
                else if (letter >= 'a' && letter <= 'z') {
                    text[(letter - 32) - 65] = (text[(letter - 32) - 65]) + 1;
                }
                else if (letter == ' ') {
                    System.out.print("");
                }
                else {
                    System.out.print("");
                }
            }
        }

        String[] alphabet = {"A","B","C","D","E","F","G","H","I","J","K","L","M","N","O","P","Q","R","S","T","U","V","W","X","Y","Z"};

        boolean emptyOutput = true;
        for (int y = 0; y < text.length; y++) {
            if (text[y] > 0) {
                System.out.println(alphabet[y] + ": " + text[y]);
                emptyOutput = false;
            }
        }
        if (emptyOutput) {
            System.out.println();
        }
    }

edited Nov 11, 2016 at 17:50

answered Nov 11, 2016 at 17:08

dammina

6552 gold badges8 silver badges23 bronze badges

Sign up to request clarification or add additional context in comments.

6 Comments

Jay58t Over a year ago

I tired that already. I cannot do that because then when the user enters text it will be on the line under "Enter text: " when its supposed to be on the same line.

dammina Over a year ago

why don't you add a System.out.println(); after the line that you read input String a = input.nextLine();

Jay58t Over a year ago

I edited the post. The last sections shows what the output should be like

Jay58t Over a year ago

The only thing you added is the code under where I declared my alphabet array correct? I'm just trying to understand how that wouldn't add an extra line for when I do enter words. It worked though thank you very much! I'm just trying understand how it did is all haha.

Jay58t Over a year ago

Oh I see now! That took me a second but I understand now. Yeah that was my problem I wasn't understanding how I can check if the amount of letters used is 0. But now I get it. That was a simple fix wow I don't know why I was blanking. Thank you so much!

|

Collectives™ on Stack Overflow

User string input. Counting characters

1 Answer 1

6 Comments

Your Answer

Hot Network Questions

Collectives™ on Stack Overflow

1 Answer 1

6 Comments

Your Answer

Sign up or log in

Post as a guest

Related