Develop Reference

JAVA: verify user input for one character

this is my first time asking a question. If I'm breaking any rules let me know please :)
I want to verify that the user only types in only one character and store in a variable I have already declared initially. As well, loop back the question for user to type in again if they did not do what they are asked for
Here is a what I have done so far
import java.util.Scanner;
public class arraytesting {
public static void main(String[] args) {
Scanner myKeyboard = new Scanner(System.in);
int user_Choice;
int rowAndcolumns;
char[][] user_Array;
char user_Char;
do {
System.out.print("Enter your choice (1 to 9): ");
user_Choice = myKeyboard.nextInt();
if (user_Choice < 1 || user_Choice > 9)
System.out.println("Illegal choice, please try again.");
} while (user_Choice < 1 || user_Choice > 9);
switch (user_Choice) {
case 1:
do {
System.out.print("\nHow many rows and columns (min 4 & max 20)? ");
rowAndcolumns = myKeyboard.nextInt();
if (rowAndcolumns < 1 || rowAndcolumns > 9)
System.out.println("Illegal choice, please try again.");
} while (rowAndcolumns < 4 || rowAndcolumns > 20);
do {
System.out.print("Which character do you want to fill your square with? (only one character)");
user_Char = myKeyboard.next().charAt(0);
if () // error message for user if they did not type correctly, Idk what to put in the
System.out.println("Illegal choice, please try again.");// boolean for it to compare
System.out.print(user_Char);
} while (); // using do-while loop to loop back question, if they don't type in correctly, i
// would only like for user to type in only one character
break;
}
}
}
I know I can put both of them in one do-while loop, but I want to focus on getting the boolean to check for user input.
edit: I would only like the user to enter only one single character
ex. '#' or 'a'
whereas "##" or "i am typing something that is not one character" is wrong
inside the spaces of if and while are how I want it to be verified

There is no need to do any check for "only 1 character entered". That makes no sense. You can't predict the future, so you cannot know if a user will enter more characters after 1 character has been entered. You will either just take the first character entered and work with it and ignore any potential additional characters - or you have to wait for more than 1 character, essentially breaking the program for users who do the right thing (enter only one character), just to be able to give them an error message when they finally do the wrong thing (enter another character).
That being said, this code:
user_Char = myKeyboard.next().charAt(0);
will actually wait for several characters to be entered until some kind of delimiter (per default some whitespace character, e.g. newline) is entered. That's exactly what you do not want.
You want to get one character from input, and one only. You don't have to care about more characters being entered after that:
user_Char = myKeyboard.next(".").charAt(0);
This tells myKeyboard to return the next String that matches the regex ".", which is any character, and only 1 character.
If you want to validate the entered character, e.g. only alphanumeric characters allowed, you can update your if and while to something like this:
if (!Pattern.matches("[a-zA-Z0-9]", new String(user_Char)))
or even better, use the String returned by myKeyboard.next("."):
String user_String = myKeyboard.next(".");
user_Char = user_String.charAt(0);
if (!Pattern.matches("[a-zA-Z0-9]", user_String))
or you could directly tell myKeyboard to only allow valid characters and skip the entire do/if/while error handling:
user_Char = myKeyboard.next("[a-zA-Z0-9]").charAt(0);
Edit
One thing your code doesn't handle right now is invalid inputs, e.g. letters when you call nextInt. This will actually throw a java.util.InputMismatchException, and you might want to wrap your nextInt() and next(...) calls in try-catch blocks to handle these exceptions.

Please check the code below, based on the discussion with Max, I used the .length() method to check the lenght of the string that the user typed.
You can check the type of the character to avoid the runtime exception in the first if statement using some methods in Character class that you use to check if the input is digit/letter or not ?
Character.isDigit(char)
Character.isLetter(char)
Character.isLetterOrDigit(char)
I also changed some variable names, Java is following the camel case style and class name has to be capitalized. I also refactored some code to check the range of the numbers to git rid of repeating same code on and on, check the method betweenExclusive
package stackoverflow.q2;
import java.util.Scanner;
public class Question2 {
public static void main(String[] args) {
Scanner myKeyboard = new Scanner(System.in);
int userChoice;
int rowAndcolumns;
char[][] user_Array;
char userChar;
do {
System.out.print("Enter your choice (1 to 9): ");
userChoice = myKeyboard.nextInt();
if ( !betweenExclusive(userChoice, 1,9) )
System.out.println("Illegal choice, please try again.");
} while (!betweenExclusive(userChoice, 1,9));
switch (userChoice) {
case 1:
do {
System.out.print("\nHow many rows and columns (min 4 & max 20)? ");
rowAndcolumns = myKeyboard.nextInt();
if (!betweenExclusive(rowAndcolumns ,1 , 9))
System.out.println("Illegal choice, please try again.");
} while (!betweenExclusive(rowAndcolumns ,4 , 20));
String input;
while (true){
System.out.print("Which character do you want to fill your square with? (only one character)");
input = myKeyboard.next();
// error message for user if they did not type correctly, Idk what to put in the
// boolean for it to compare
if ( input.length()>1){
System.out.print("Illegal character, try again please !!! ");
}else{
userChar = input.charAt(0);
System.out.print(userChar);
break;
}
} // using do-while loop to loop back question, if they don't type in correctly, i
// would only like for user to type in only one character
break;
}
}
public static boolean betweenExclusive(int x, int min, int max)
{
return x>=min && x<=max;
}
}

Check String is in a range of numbers Java

Trying to design a simple lottery program. Everything works except checking if the numbers entered are between 1 to 59.
Exercise says the numbers must be stored in a String variable.
so
if(num<0 || num>59) //wont work for me
Tried making another variable
int numConverted = Integer.parseInt(num)
We haven't covered converting String to int in class though so I don't think this is what expected. Got confused trying that way anyway so probably this is wrong.
Here is the code I have currently.
{
Scanner scan = new Scanner(System.in);
String num=""; //num variable is empty untill user inputs numbers
for(int i =0; i<6; i++)
{
System.out.println("Enter your number between 1-59");
num = num +" "+ scan.nextLine();
}
System.out.println("Ticket printed £2. Your numbers are " + num);
}

In your posted code it's obvious that you want the User to supply 6 specific numerical values. These values are appended to the String variable named num (space delimited). You need to obviously do a few things here:
1) Make sure the value supplied by the user is indeed a numerical value;
2) Make sure the numerical values supplied fall within the minimum and maximum scope of the lottery itself (which you have stated is: 1 to 59);
3) Make sure the number entered by the User hasn't been supplied already.
You've been tasked to store the entered values into a String data type variable and that is all fine but at some point you want to carry out value comparisons to make sure that all the entered values actually play within the limits of the lottery.
When the User completes his/her entries, you end up with a space delimited string held in the num string variable. You now need to make sure that these values entered are indeed....numbers from 1 to 59 and none contain alpha characters.
In my opinion (and this is only because you need to store entered values into a String variable), it's best to use your String variable to gather User input, then test the input to make sure it is indeed a string representation of an actual integer number. Once this is established then we test to make sure if falls within the value min/max limits (1-59). Now we need to test to make sure the number entered hasn't already been entered before for this ticket.
Of course with each test described above, if one fails then the User should be prompted to re-enter a proper value. You can do this by utilizing a while loop. Plenty examples of this in StackOverflow but here's a quick example:
Scanner scan = new Scanner(System.in);
String ticketNumbers = "";
for(int i = 0; i < 6; i++) {
Boolean isOK = false;
while (!isOK) {
System.out.println("\nPlease enter your desired 6 ticket numbers:\n"
+ "(from 1 to 59 only)");
String num = scan.nextLine();
//Is the string entered an actual integer number?
//We use the String.matches() method for this with
//a regular expression.
if(!num.matches("\\d+")) {
System.out.println("You must supply a numerical value! "
+ "Try Again...");
continue;
}
if (ticketNumbers.contains(num + " ")) {
System.out.println("The number you supplied has already been chosen!"
+ " Try Again...");
continue;
}
if (Integer.parseInt(num) >= 1 && Integer.parseInt(num) <= 59) {
ticketNumbers+= num + " ";
isOK = true;
}
else {
System.out.println("The number you supply must be from "
+ "1 to 59! Try Again...");
}
}
}
System.out.println("Ticket printed £2. Your numbers are " + ticketNumbers);

How about -
if(Integer.parseInt(num) < 0 || Integer.parseInt(num) > 59)
This should work, place it after the input.
If it works, please mark this as correct, I need the rep!!!

Easy way would be add available numbers (suppose it wont grow more than 60. You can use a loop to add to this as well)
String numbers[] = {"1","2","3", "..."};
Then inside the loop
Arrays.asList(numbers).contains(num);
You can remove prefixing zero in order avoid conflicts with values like '02'
Here everything is String related.

If you don't want to explicitly convert to int, you could use a regular expression.
if (num.matches("[1-5]?[0-9]")) {
...
This checks whether the String consists of (1) maybe a digit from 1 to 5, followed by (2) definitely a digit from 0 to 9. That'll match any number in the range 0-59.
If you've got a whole series of numbers separated by spaces, you could expand this to cover a whole series like this.
if (num.matches("([1-5]?[0-9]\\s+)*[1-5]?[0-9]")) {
This matches any number of repetitions (including zero) of "a number followed by spaces", followed by a single repetition without a space. The "\\s" means "any whitespace character", the "+" after it means "one or more of what precedes", and the "*" means "zero more of what precedes" - which in this case is the term in parentheses.

Oh I see what you are trying to do
This is what you want
Scanner scan = new Scanner(System.in);
String allNums = "";
for(int i =0; i<6; i++)
{
System.out.println("Enter your number between 1-59");
int num = scan.nextInt();//Take the number in as an int
if(num >0 && num < 59)//Check if it is in range
{
allNums += num + " ";//if it is add it to a string
}
else
{
System.out.println("Number not in range");
i--;//go back one iteration if its not in range
}
}
System.out.println("Ticket printed £2. Your numbers are " + allNums);

Palindrome code not working well in Java

This is a assignment I'm doing and it seems I can't get it to work properly.
The question is below.
A palindrome is a word or phrase that reads the same forward and
backward, ignoring blanks and considering uppercase and lowercase
versions of the same letter to be equal.for example,the following are
palindromes:
warts n straw
radar
able was I ere I saw Elba
xyzczyx
Write a program that will accept a sequence of characters terminated
by a period and will decide whether the string--without the
period---is a palindrome.You may assume that the input contains only
letters and blanks and is at most 80 characters long.Include a loop
that allows the user to check additional strings until she or he
requests that the program end.
Hint: Define a static method called isPalindrome that begins as
follows:
Precondition: The array a contains letters and blanks in
positions a[0] through a[used - 1]. Returns true if the string is a
palindrome and false otherwise.
public static boolean isPalindrome(char[] a, int used)
Your program should read the input characters into an array whose base
type is char and then call the preceding method. The int variable used
keeps track of how much of the array is used, as described in the
section entitled "Partially Filled Arrays."
This is my class code:
public class Palindrome_class
{
// instance variable
char[] characterArray;
//constructor
//#param data is a string of characters
public Palindrome_class(String data)
{
characterArray = data.toUpperCase().toCharArray();
}
//#return true if the word is a palindrome, otherwise returns false.
public boolean isPalindrome(char[] a, int used)
{
int i = 0, j = used - 1;
while (i < j)
{
if(characterArray[i] == characterArray[j])
{
i++;
j--;
}
else
{
return false;
}
}
return true;
}
}
This is my main code:
import java.util.Scanner;
public class palindromeTest
{
public static void main(String[] args)
{
int used = 0;
char[] chars = new char[80];
Scanner inputWord = new Scanner(System.in);
Scanner reply = new Scanner(System.in);
System.out.println("Enter a string characters, terminated by a period.");
String data;
String cq;
Palindrome_class word;
do
{
//input word from user.
data = inputWord.nextLine();
word = new Palindrome_class(data);
//check for palindrome.
if(word.isPalindrome(chars, used))
System.out.println(data + " is a palindrome.");
else
System.out.println(data + " is not a palindrome.");
//request to continue or quit.
System.out.println("Continue or Quit?");
cq = reply.nextLine();
}
while (cq.equalsIgnoreCase("continue"));
System.exit(0);
}
}
This is the results:
Enter a string characters, terminated by a period.
radar.
radar. is a palindrome.
Continue or Quit?
continue
use
use is a palindrome.
Continue or Quit?
continue
use.
use. is a palindrome.
Continue or Quit?
continue
apple.
apple. is a palindrome.
Continue or Quit?
Quit
Please tell me where I'm making a mistake.

You are checking whether a String is a palindrome with this call :
if(word.isPalindrome(chars, used))
However, used is 0, so your method always returns true.
You are also ignoring the instructions of your assignment. You are not doing anything with the chars array, you are not removing the period that's supposed to be at the end of the input String, your isPalindrome method is not static, etc...

U did a very little mistake.. U are sending "used" variable as 0 each time. ideally it should be length of a word.
please check it. use
used = data.length();
before sending it to the check method

Need help splitting a string into two separate integers for processing

I am working on some data structures in java and I am a little stuck on how to split this string into two integers. Basically the user will enter a string like '1200:10'. I used indexOf to check if there is a : present, but now I need to take the number before the colon and set it to val and set the other number to rad. I think I should be using the substring or parseInt methods, but am unsure. The code below can also be viewed at http://pastebin.com/pJH76QBb
import java.util.Scanner; // Needed for accepting input
public class ProjectOneAndreD
{
public static void main(String[] args)
{
String input1;
char coln = ':';
int val=0, rad=0, answer=0, check1=0;
Scanner keyboard = new Scanner(System.in); //creates new scanner class
do
{
System.out.println("****************************************************");
System.out.println(" This is Project 1. Enjoy! "); //title
System.out.println("****************************************************\n\n");
System.out.println("Enter a number, : and then the radix, followed by the Enter key.");
System.out.println("INPUT EXAMPLE: 160:2 {ENTER} "); //example
System.out.print("INPUT: "); //prompts user input.
input1 = keyboard.nextLine(); //assigns input to string input1
check1=input1.indexOf(coln);
if(check1==-1)
{
System.out.println("I think you forgot the ':'.");
}
else
{
System.out.println("found ':'");
}
}while(check1==-1);
}
}

Substring would work, but I would recommend looking into String.split.
The split command will make an array of Strings, which you can then use parseInt to get the integer value of.
String.split takes a regex string, so you may not want to just throw in any string in it.
Try something like this:
"Your|String".split("\\|");, where | is the character that splits the two portions of the string.
The two backslashes will tell Java you want that exact character, not the regex interpretation of |. This only really matters for some characters, but it's safer.
Source: http://www.rgagnon.com/javadetails/java-0438.html
Hopefully this gets you started.

make this
if(check1==-1)
{
System.out.println("I think you forgot the ':'.");
}
else
{
String numbers [] = input1.split(":"); //if the user enter 1123:2342 this method
//will
// return array of String which contains two elements numbers[0] = "1123" and numbers[1]="2342"
System.out.print("first number = "+ numbers[0]);
System.out.print("Second number = "+ numbers[1]);
}

You knew where : is occurs using indexOf. Let's say string length is n and the : occurred at index i. Then ask for substring(int beginIndex, int endIndex) from 0 to i-1 and i+1 to n-1. Even simpler is to use String::split

How to display the middle digit in java

I am trying to get java to display the middle digit of a 1-4 digit integer, and if the integer has an even number of digits i would get it to display that there is no middle digit.
I can get the program to take get the integer from the user but am pretty clueless as how to get it to pick out the middle digit or differentiate between different lengths of integer.
thanks

Hint: Convert the integer to a String.
Java int to String - Integer.toString(i) vs new Integer(i).toString()
You may also find the methods String.length and String.charAt useful.

This sounds like it might be homework, so I will stay away from giving an exact answer. This is much more about how to display characters than it is about integers. Think about how you might do this if the questions was to display the middle letter of a word.

Something like this?
import java.util.Scanner;
public class Main {
public static void main(String args[]) {
Scanner s = new Scanner(System.in);
System.out.print("Enter an integer: ");
while (!s.hasNextInt()) {
s.next();
System.out.println("Please enter an integer.");
}
String intStr = "" + s.nextInt();
int len = intStr.length();
if (len % 2 == 0)
System.out.println("Integer has even number of digits.");
else
System.out.println("Middle digit: " + intStr.charAt(len / 2));
}
}
Uhm, I realized I might just have done someones homework... :-/ I usually try to avoid it. I'll blame the OP for not being clear in this case