I'm currently doing a project in my computer science class and we are suppose to validate each character of a variable to see if it is legal or not. If it starts with a number it's illegal. If it starts with a special character it's legal but bad style. If it has a space it is again illegal. I'll post my current code now:
import java.util.Scanner;
public class classOfValidation {
public static void main(String[] args) {
Scanner scan = new Scanner(System.in);
String theVariable = null;
System.out.println("This program checks the validity of variables");
System.out.println("Please enter a variable (or press 'q' to quit");
theVariable = scan.nextLine();
do {
System.out.println("The variable is illegal");
theVariable = scan.nextLine();
} while (theVariable.startsWith("[0123456789]"));
do {
System.out.println("The variable is illegal");
theVariable = scan.nextLine();
} while (theVariable.contains("[ ]"));
do {
System.out.println("The variable is legal, but has bad style");
theVariable = scan.nextLine();
} while (theVariable.startsWith("[!##$%^&*]"));
}
}
If you couldn't already tell i'm new to programming and as confused as i possibly could be. If you have any advice or anything else you need me to explain then please leave a comment. Thanks everyone
You can use the single regex to validate your input via String#matches() method. But as for the example you've provided, you should use while loop, but not do-while, because in do while case, you are always running it's body once befor condition checked. So, you better do it like:
theVariable = scan.nextLine();
while (theVariable.startsWith("[0123456789]")) {
System.out.println("The variable is illegal");
theVariable = scan.nextLine();
}
while (theVariable.contains("[ ]")) {
System.out.println("The variable is illegal");
theVariable = scan.nextLine();
}
while (theVariable.startsWith("[!##$%^&*]")) {
System.out.println("The variable is legal, but has bad style");
theVariable = scan.nextLine();
}
The second, in your solution, you are using String.startsWith() method and passing into it some regex. Take a look at javadoc for this method. It's said there:
Tests if this string starts with the specified prefix.
That means, that this method doesn't support regexes, but simply checks whether the string starts with the passed string. So, your conditions seems, never to become true. I don't think, someone will input the [0123456789] or [!##$%^&*].
One more, any conditions are checked once, but after that user can modify the input and the previewsly passed condition will not be checked again. Seems, it's better to run into infinite loop with continue and break in some conditions, like:
//infinit loop, until user enter the `q` or the input is correct
while (true) {
//read the input
theVariable = scan.nextLine();
//chtck, whether is `quit` command entered
if ("q".equals(theVariable)) {
break;
}
//if string starts with digit or contains some whitespaces
//then print alert and let the user to
//modify the input in a new iteration
if (theVariable.matches("^\d+.*|.*\s+.*")) {
System.out.println("The variable is illegal");
continue;
}
//if string contains some special characters print alert
//and let the user to modify the input in a new iteration
if (theVariable.matches("^[!##$%^&*].*")) {
System.out.println("The variable is legal, but has bad style");
continue;
}
//if all the conditions checked, then break the loop
break;
}
I think the best way if you are use regex.
Here is an answer how to do that.
Related
I want to run a loop and continue to accept a string input until "x" is pressed. I'm trying following do-while loop, but it doesn't get in the loop, it runs only once.
Scanner sc = new Scanner(System.in);
String name;
do {
System.out.print("Enter name or 'x' to quit: ");
name = sc.next();
if (name.equals("x")) {
break;
}
System.out.printf("The name is %s", name);
} while (name.equals("x"));
You need to remove if condition so that it doesn't break out of the loop early and modify the break condition, e.g.:
Scanner sc = new Scanner(System.in);
String name;
do {
System.out.print("Enter name or 'x' to quit: ");
name = sc.next();
System.out.printf("The name is %s", name);
} while (!name.equals("x"));
It continues looping when "x" is pressed. So you did the exact opposite.
Add an exclamation mark to negate your while condition:
while (!name.equals("x"))
I might be missing something, but it seems like your code contradicts itself.
You read user input and then check if it equals to x and breaks the loop in that case. If it does not, you continue until the while statement, where x gets checked again for the same condition.
So in summary, you break the loop for the same condition you have for it to run.
You might want to try, changing your condition in the while statement to
while (!name.equals("x"))
I am a java beginner, and in this particular problem I practiced making a program that converts any given string to lowercase string. Is there a a better way to achieve this goal in java (in terms of design)?
Also, how does the "else" (after "else if") catches or waits for me to make an input. Somehow that part does not make sense to me, even though I achieved what I wanted. How is the value of "ans" from input transferred to the entire loop and used until the loop is closed?
After many attempts and failures, I used a separate method for the conversion part. My second question is a bit too complicated to be researched.
import static java.lang.System.out;
import java.util.Scanner;
public class MyClass {
public static Scanner s = new Scanner(System.in);
public static String ans;
public static void main(String args[]) {
Conversion();
do {
ans = new String(s.nextLine());
if (ans.equalsIgnoreCase("Y")) {
Conversion();
} else if (ans.equalsIgnoreCase("N")) {
out.println("Thank you for using this program!");
break;
} else {
out.println("Invalid entry!");
out.println("Would you like to convert another string?\n(Please type 'Y' for yes, or 'N' for no.)");
}
} while (ans != "N");
}//END MAIN
public static void Conversion() {
out.println("Please enter string to be converted to lowercase: ");
String str = new String(s.nextLine());
out.println("Your new string is: " + str.toLowerCase());
out.println("Would you like to convert another string? (Y/N)");
}
}
I notice a few issues; Conversion looks like a class-name (Java naming convention would be conversion) and ans != "N" is using == instead of .equals - and wouldn't ignore case (!ans.equalsIgnoreCase("N")). Globals (e.g. static) are bad (pass the Scanner to the methods that need it), and the static import just makes the code more difficult to reason about (in my opinion). Your current loop doesn't really follow a conventional form, I would extract the prompt and loop for "another" conversion to a new method and if you must print a thank you I'd do so after the "main loop". Something like,
public static void main(String args[]) {
Scanner sc = new Scanner(System.in);
do {
conversion(sc);
} while (another(sc));
System.out.println("Thank you for using this program!");
}
public static void conversion(Scanner s) {
System.out.println("Please enter string to be converted to lowercase: ");
System.out.printf("Your new string is: %s%n", s.nextLine().toLowerCase());
}
public static boolean another(Scanner s) {
while (true) {
System.out.println("Would you like to convert another string? (Y/N)");
String ans = s.nextLine();
if (ans.equalsIgnoreCase("Y")) {
return true;
} else if (ans.equalsIgnoreCase("N")) {
return false;
}
System.out.println("Invalid entry!");
System.out.println("(Please type 'Y' for yes, or 'N' for no.)");
}
}
Answering your first question:
There are many design patterns and practices so many people can argue what I would recommend you to do. It's basically the same for all programming languages. Let's take your function "Conversion". The name itself says that you use it to convert stuff. Not to display, not to prompt - to convert. In this case, the only actual thing it should do is to convert upperCase to lowercase. In fact, you might want to specify what type of conversion it has in the name of the function (convertToLowerCase?). In fact, in Java, we use lowerCamelCase for all function names and UpperCamelCase for classes.
If you accept my previous suggestion, you could break the Conversion function into multiple ones like promptUserForInput, WrongInputHandler and so forth.
If I understood your second question correctly, you wonder about the way the code executed and how the variable ans is transferred further into the loop. Let's take a look at your code and what variables do:
You initialize your variable in the class MyClass by making it accessible to all methods in the class;
You prompt the user for the input to assign to this variable inside the do..while loop with this line ans = new String(s.nextLine()); which saves the value of the variable and, again, which can be accessed inside the whole class so its value is changed.
It goes into the if..else if...else statement. The way it works, it goes line by line - if the first if-statement fails, it goes on until it finds a truthy statement and it doesn't go any further. In your case, if the ans is not equal to either y/Y/ it will go to else if statement and if it's not n/N, it will go to else (so basically whatever except y/Y/n/N) and will be executed. After that, it jumps into the while (ans!= "N"); line where it compares your class-member variable ans and if it's not equal to "N" it starts over the loop right after the do{ part until you type in the "N".
I hope that makes sense. Whenever the program is asking you for input, it does not execute code further but is stuck until you provide any input. The value from input itself isn't passed throughout the loop and the program. The reason why you can use it because you created a higher-scope variable ans where you saved the result of your input.
IMPORTANT: if you've declared the ans inside the do..while loop, you would've not been able to have accessed it in the while (ans...) because it will 'die' right before the curly brace between do { ...here} while(). If you want to learn more about the scope and variables in general, you can read this article.
Here is my code example:
public static void main(String args[]) {
//declare before entering the loop to have higher scope
String ans = "y";
do {
//we get the given string to convert from the user
String str = promptForString();
//we convert the string
str = converseStringToLowerCase(str);
//display the string (could use another function for that: easier to debug and locate problems and in bigger projects)
out.println("Your new string is: " + str);
//prompt user for respond to continue or not
ans = promptForContinue();
handleResponse(ans);
} while (!ans.equals("n"));
}//END MAIN
//prompts user for an input string
public static String promptForString() {
out.println("Please enter string to be converted to lowercase: ");
String str = new String(s.nextLine());
return str;
}
//converts any given string to lower case
public static String converseStringToLowerCase(String str) {
return str.toLowerCase();
}
//is used to prompt user for reply
public static String promptForContinue() {
out.println("Would you like to convert another string? (Y/N)");
String str = new String(s.nextLine());
//is good to make if...else statements easier - it will always be lower case (or upper if you prefer)
return str.toLowerCase();
}
//easier to locate other response scenarios
public static void handleResponse(String response) {
if (response.equals("n")) {
out.println("Thank you for using this program!");
//it's not a very good practice to innaturally break loops. Use input for that in while(..) statement
// break;
} else if (!response.equals("y")) {
out.println("Invalid entry!");
out.println("Would you like to convert another string?\n(Please type 'Y' for yes, or 'N' for no.)");
}
}
Can someone please advise why the inner loop of the code below will not exit?
I've added an inner loop to check if input from the user of of a particular value and if not prompts for the correct input. When debugging the code and passing in a value which should force the loop to end it doesn't although I can see the correct value in the variable:
while (finished.equalsIgnoreCase("n")) {
System.out.println("Enter a persons name");
names = in.nextLine();
writer.println(names);
System.out.println("Finished? (Y/N)");
finished = in.nextLine();
while( !finished.equalsIgnoreCase("y") || !finished.equalsIgnoreCase("n")) {
System.out.println("Invalid choice; (Y/N)");
finished = in.nextLine();
}
}
Every string is either not not equal to y or not equal to n. You probably meant to use the && operator:
while(!finished.equalsIgnoreCase("y") &&
!finished.equalsIgnoreCase("n")) {
// Code...
Trying to print a file based off the user's input as mentioned in the title. Basically, my program has been altered from one that I previously created which reads data from a file, so I know that the file has been imported correctly (not the problem).
The problem I have is that I'm trying to make the program print the entirety of the .txt file if the user chooses a specific number, in this case '1'. My current code so far is:
import java.io.FileReader;
import java.util.Scanner;
public class InputOutput {
public static void main(String[] args) throws Exception {
// these will never change (be re-assigned)
final Scanner S = new Scanner(System.in);
final Scanner INPUT = new Scanner(new FileReader("C:\\Users\\JakeWork\\workspace\\Coursework\\input.txt"));
System.out.print("-- MENU -- \n");
System.out.print("1: Blahblahblah \n");
System.out.print("2: Blahblahblah \n");
System.out.print("Q: Blahblahblah \n");
System.out.print("Pick an option: ");
if (S.nextInt() == 1) {
String num = INPUT.nextLine();
System.out.println(num);
}
I feel as if my if statement is totally off and I'm heading in the entire wrong direction, could anyone point me in the right and give me a helping hand?
You're close, but not quite there.
You a reading the user input correctly, but now you need the file contents in a loop.
if(S.nextInt() == 1) {
while (INPUT.hasNextLine()) {
System.out.println(INPUT.nextLine());
}
}
This will keep looking as long as the file contents hasNextLine
You can safely remove the String option = S.next();
Also, just a small bit of naming convention nitpicking, don't use all upper case letters for variable names unless they are meant to be static. Also, the first letter of a variable is generally lower case.
if (S.nextInt() == 1) {
// check if there is input ,if true print it
while((INPUT.hasNextLine())
System.out.println(INPUT.nextLine());
}
Also, for menu scenarios like this, consider using a switch statement, then place a call to the menu-printing (that you move to a separate method) in the default case, so that if you enter something wrong, you can reprint the menu choices. Also, the switch statement is more readable (arguably) than a bunch of if's, like this:
int option = S.nextInt();
switch(option) {
case 1 :
while(INPUT.hasNextLine()) {
System.out.println(INPUT.nextLine());
}
break;
case 2 :
//Do stuff
break;
default :
//Default case, reprint menu?
}
}
Everything of my guessing game is alright, but when it gets to the part of asking the user if he/she wants to play again, it repeats the question twice. However I found out that if I change the input method from nextLine() to next(), it doesn't repeat the question. Why is that?
Here is the input and output:
I'm guessing a number between 1-10
What is your guess? 5
You were wrong. It was 3
Do you want to play again? (Y/N) Do you want to play again? (Y/N) n
Here is the code:(It is in Java)
The last do while loop block is the part where it asks the user if he/she wants to play again.
import java.util.Scanner;
public class GuessingGame
{
public static void main(String[] args)
{
Scanner input = new Scanner(System.in);
boolean keepPlaying = true;
System.out.println("Welcome to the Guessing Game!");
while (keepPlaying) {
boolean validInput = true;
int guess, number;
String answer;
number = (int) (Math.random() * 10) + 1;
System.out.println("I'm guessing a number between 1-10");
System.out.print("What is your guess? ");
do {
validInput = true;
guess = input.nextInt();
if (guess < 1 || guess > 10) {
validInput = false;
System.out.print("That is not a valid input, " +
"guess again: ");
}
} while(!validInput);
if (guess == number)
System.out.println("You guessed correct!");
if (guess != number)
System.out.println("You were wrong. It was " + number);
do {
validInput = true;
System.out.print("Do you want to play again? (Y/N) ");
answer = input.nextLine();
if (answer.equalsIgnoreCase("y"))
keepPlaying = true;
else if (answer.equalsIgnoreCase("n"))
keepPlaying = false;
else
validInput = false;
} while (!validInput);
}
}
}
In your do while loop, you don't want the nextLine(), you just want next().
So change this:
answer = input.nextLine();
to this:
answer = input.next();
Note, as others have suggested, you could convert this to a while loop. The reason for this is that do while loops are used when you need to execute a loop at least once, but you don't know how often you need to execute it. Whilst it's certainly doable in this case, something like this would suffice:
System.out.println("Do you want to play again? (Y/N) ");
answer = input.next();
while (!answer.equalsIgnoreCase("y") && !answer.equalsIgnoreCase("n")) {
System.out.println("That is not valid input. Please enter again");
answer = input.next();
}
if (answer.equalsIgnoreCase("n"))
keepPlaying = false;
The while loop keeps looping as long as "y" or "n" (ignoring case) isn't entered. As soon as it is, the loop ends. The if conditional changes the keepPlaying value if necessary, otherwise nothing happens and your outer while loop executes again (thus restarting the program).
Edit: This explains WHY your original code didn't work
I should add, the reason your original statement didn't work was because of your first do while loop. In it, you use:
guess = input.nextInt();
This reads the number off the line, but not the return of the line, meaning when you use:
answer = input.nextLine();
It immediately detects the leftover carriage from the nextInt() statement. If you don't want to use my solution of reading just next() you could swallow that leftover by doing this:
guess = input.nextInt();
input.nextLine();
rest of code as normal...
The problem really lies in a completely different segment of code. When in the previous loop guess = input.nextInt(); is executed, it leaves a newline in the input. Then, when answer = input.nextLine(); is executed in the second loop, there already is a newline waiting to be read and it returns an empty String, which activates the final else and validInput = false; is executed, to repeat the loop (and the question).
One solution is to add an input.nextLine(); before the second loop. Another is to read guess with nextLine() and then parse it into an int. But this complicates things as the input could not be a correct int. On a second thought, the code already presents this issue. Try entering a non-numeric response. So, define a function
public static int safeParseInt(String str) {
int result;
try {
result= Integer.parseInt(str) ;
} catch(NumberFormatException ex) {
result= -1 ;
}
return result ;
}
And then replace your first loop with:
do {
validInput= true ;
int guess= safeParseInt( input.nextLine() ) ;
if( guess < 1 || guess > 10 ) {
validInput= false ;
System.out.print("That is not a valid input, guess again: ");
}
} while( !validInput );
PS: I don't see any problem with do-while loops. They are part of the language, and the syntax clearly indicates that the condition is evaluated after the body is executed at least one time. We don't need to remove useful parts of the language (at least from practice) just because others could not know them. On the contrary: if we do use them, they will get better known!
validInput = false;
do {
System.out.print("Do you want to play again? (Y/N) ");
answer = input.next();
if(answer.equalsIgnoreCase("y")){
keepPlaying = true;
validInput = true;
} else if(answer.equalsIgnoreCase("n")) {
keepPlaying = false;
validInput = true;
}
} while(!validInput);
I changed the coding style as I find this way more readable.
Your problem is that nextInt will stop as soon as the int ends, but leaves the newline in the input buffer. To make your code correctly read the answer, you'd have to enter it on the same line as your guess, like 5SpaceYReturn.
To make it behave more than one would expect, ignore the first nextLine result if it contains only whitespace, and just call nextLine again in that case without printing a message.
I believe the output of input.nextLine() will include the newline character at the end of the line, whereas input.next() will not (but the Scanner will stay on the same line). This means the output is never equal to "y" or "n". Try trimming the result:
answer = input.nextLine().trim();