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;
}
}

I want to know how to find a special character in a string

I am programming in Java in Eclipse, I want to ask users to enter their specific ID, Which starts with an uppercase G and has 8 digit numbers. like G34466567. if the user enters an invalid ID it will be an error. how can i separate a valid ID from others?

You can use regex. This pattern checks if the first character is a capital G and there are 8 digits following:
([G]{1})([0-9]{8})$
As you see there are two expressions which are separated by the (). The first says "only one character and this one has to be a capital G". And the second one says, there have to be 8 digits following and the digits can be from 0 to 9.
Every condition contains two "parts". The first with the [] defines which chars are allowed. The pattern inside the {} show how many times. The $ says that the max length is 9 and that there can't be more chars.
So you can read a condition like that:
([which chars are allowed]{How many chars are allowed})
^------------------------\/---------------------------^
One condition
And in Java you use it like that:
String test= "G12345678";
boolean isValid = Pattern.matches("([G]{1})([0-9]{8})$", test);
As you see that matches method takes two parameters. The first parameter is a regex and the second parameter is the string to check. If the string matches the pattern, it returns true.

Create an ArrayList. Ask the user to input the ID, check if it is already there in the list, ignore, otherwise add that ID to the list.
EDIT: For ensuring that the rest 8 characters of the String ID are digits, you can use the regex "\\d+". \d is for digits and + is for one or more digits.
Scanner sc = new Scanner(System.in);
ArrayList<String> IDS = new ArrayList();
char more = 'y';
String ID;
String regex = "\\d+";
while (more == 'y') {
System.out.println("Pleaes enter you ID.");
ID = sc.next();
if (IDS.contains(ID)) {
System.out.println("This ID is already added.");
} else if (ID.length() == 9 && ID.charAt(0) == 'G' && ID.substring(1).matches(regex)) {
IDS.add(ID);
System.out.println("Added");
} else {
System.out.println("Invalid ID");
}
System.out.println("Do you want to add more? y/n");
more = sc.next().charAt(0);
}

Assuming that you save the id as a string, you can check the first letter and then check if the rest is a number.
Ex.
String example = "G12345678";
String firstLetter = example.substring(0, 1); //this will give you the first letter
String number = example.substring(1, 9); //this will give you the number
To check that number is a number you could do the following instead of checking every character:
try {
foo = Integer.parseInt(number);
} catch (NumberFormatException e) {
//Will Throw exception!
//do something! anything to handle the exception.
// this is not a number
}

regular expression in java gui for text box

I am trying to perform user input on textbox to check whether they had enter any number or special character in the textbox for their username. I check several resource at online also not able to find out what is my problem inside my code. it's showing me four error message
1) error: 'else' without 'if'
else if(tf3.getText().isEmpty())
2.) error: ';' expected
if(!(Pattern.matches("^[a-zA-Z]$")),tf3.getText())
3.) error: variable declaration not allowed here
if(!(Pattern.matches("^[a-zA-Z]$")),tf3.getText())
4.) error: ';' expected
if(!(Pattern.matches("^[a-zA-Z]$")),tf3.getText())
Here is my full program for my whole program
if(e.getSource()== btn2)
{
if(!(Pattern.matches("^[a-zA-Z]$")),tf3.getText())
{
JOptionPane.showMessageDialog(null,"Please enter a valid name");
}
else if(tf3.getText().isEmpty())
{
JOptionPane.showMessageDialog(null,"Check your name");
}
else if(tf2.getText().isEmpty())
{
JOptionPane.showMessageDialog(null,"Check your id");
}
else if(cmb1.getSelectedIndex()== 0)
{
JOptionPane.showMessageDialog(null,"Check your year");
}
else if(cmb2.getSelectedIndex()== 0)
{
JOptionPane.showMessageDialog(null,"Check your major");
}
else if(cmb3.getSelectedIndex() == 0)
{
JOptionPane.showMessageDialog(null,"Check your selection");
}
else
{
String name = tf3.getText();
String id = tf2.getText();
String job = String.valueOf(cmb1.getSelectedIndex());
String country = String.valueOf(cmb2.getSelectedIndex());
String software = String.valueOf(cmb3.getSelectedIndex());
runCC(software,id,name,job,country);
}
}

Parentheses:
if(!(Pattern.matches("^[a-zA-Z]$")),tf3.getText())
should look like:
if(!Pattern.matches("^[a-zA-Z]$", tf3.getText()))
At least if i got what you meant.

Change
if(!(Pattern.matches("^[a-zA-Z]$")),tf3.getText())
to
if (! Pattern.matches("^[a-zA-Z]$", tf3.getText()))

Ok, since I'm unable to understand what your comments under my answer mean, (not only because i'm drunk!), I'll provide a 'step by step' whats-goin-on:
Pattern class provides a static method: Pattern#matches
It takes two arguments:
First - a string with a regular expression you want to test your string with
Second - a string to test that regex with.
So, whenever you are calling method:
Pattern.matches(regex, string)
You're asking, whether:
"string" matches regular expression "regex"
So in your code you want to say:
"Ok, i want to know, if the text inside tx3 field contains only characters, that are between a-z or between A-Z.
So your regex is: "any sequence of characters, that are between a-z or between A-Z.
That corresponds to
^[a-zA-Z]+$
Now, you call the method that will check, whether your strings matches that regex:
Pattern.matches("%[a-zA-Z]+$", tx3.getText())
That method returns boolean: true, if it matches, or false otherwise.
I don't think anyone will be able to put this more straight-forward.

How to check if there are double letters in a 4 digit code in Java

The above question might seems vague but it's actually a very simple idea which i can't seem to figure out.
It basically is a 4 digit letter code containing letters from A to F for example: ABDF, BAAF, DBAF etc.
Now I'm trying to do some post input-handling where it must become impossible to enter a letter that is already in the code cause it has to be a unique 4 digit code with no repeating letter. I've been trying to make it work but none of my code seems to work so i'm back to scratch asking you guys for help :)
I hope this is somewhat clear otherwise i'll be happy to clear it up.
Thanks in advance.

Kind of a pseudocode but it would work.
String uniquePass="";
while(uniquePass.length<4){
String userInput=getUserInputChar()
if(uniquePass.contains(userInput))
rejectInputAndNotifyUser
else
uniquePass=uniquePass+userInput
}

public static boolean hasDuplicateChars(String string) {
Set<Character> chars = new HashSet<Character>();
for (char c : string.toCharArray()) {
if (!chars.add(c)) return false;
}
return true;
}
Set is a collection that contains no duplicate elements. We will use add method which returns true if this set did not already contain the specified element.
hasDuplicateChars functions iterates over characters in the input string using toCharArray function and for loop; each character is added to the chars set which is initially empty. If add method returns false it means that we have already encountered same character before. So we return false from our function.
Otherwise input is valid and method returns true.

using this function you'll be able to see if the string contains unique characters
public static boolean checkForUnique(String str){
boolean containsUnique = false;
for(char c : str.toCharArray()){
if(str.indexOf(c) == str.lastIndexOf(c)){
containsUnique = true;
} else {
containsUnique = false;
}
}
return containsUnique;
}
Update:
This will be ran everytime a user enters a character and if it fails, this would mean there is a duplicate. You have the choice of discarding that input or showing an error.

If you're validating the complete input, you can lean on the set semantics, and a few tricks
String s = "ABAF";
int count = new HashSet<>(Arrays.asList(s.split(""))).size();
if (count - 1 == 4) {
System.out.println("All ok");
} else {
System.out.println("Repeated letters");
}
the split("") will split the string to a an array like {"","A", "B", "A", "F"}.
The new HashSet<>(Arrays.asList(s.split(""))) will create a Set with String elements, and as the Set will bounce back the elements already contained, the size of the set for e.g. "AAAF" will be 3 (it'll contain the "", "A" and "F"). This way you can use the size of the set to figure out if all letters of a String are unique
If its while typing you'll than the solution depends on the input method, but you can have something like (pseudo stuff)
if (pass.contains(letter)) {
breakAndNotifyUser();
} else {
pass+=letter;
}

Java - validating a subject Code

I have a subject code e.g: ABC123 that is a string
I need to ensure that it is of length 6, the first 3 characters are letters and the last 3 are numbers.
I would like to try and do it all in an if statement? I can work the length but cannot figure out the numeric and letter part of things. e.g:
public void isValidCode(String subjectCode2){
str = subjectCode2;
if (str.length() == 6 && """"NEED TO TEST OTHERS HERE??""" ) {
System.out.println("The code is valid");
}
else {
System.out.println("The code is not valid");
}

You can always use Regular Expressions, and the matches() method of the String class.
if (str.matches("[a-zA-Z]{3}[0-9]{3}")) {
// Validation succeeded
}
else {
// Validation failed
}

To test that the first three letters are letters, you could use a loop. Similarly, use a loop for testing that the last three digits are numbers. You might find the functions in the Character class helpful.

I would change the method signature so that it is not a void method but rather declared to return a boolean. Then you could have several if statements that if false returns false. At the bottom, return true if it passes all tests.
public boolean isValidCode(String code) {
if (code.length() != 6) {
return false;
}
// here check if first 3 chars are letters
// here check if last 3 chars are numbers
return true;
}
Then the calling code can do a println if desired.
String test = "ABC123";
if (isValidCode(test)) {
System.out.println("The code is valid");
} else {
System.out.println("The code is not valid");
}

If by "letter", you mean to include letters in alphabets other than English, you'll need this.
if (str.matches("\\p{L}{3}\\d{3}")) {
Here, \p{L} matches any character that Unicode considers to be a letter, in any language.