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why does it not break out of the loop whatever is the input

whatever the input is, it never breaks the loop i tried.
i tried with the switch same result; i tried with only one condition (!answer.equalsIgnoreCase("y")) without the OR and it worked but nott with two conditions.
public static void main(String[] args) {
Scanner scn = new Scanner(System.in);
//checking if the dog is barking
System.out.println("is the dog is barking ?");
String answer;
do {
System.out.println("write 'n' for no and 'y' for yes");
answer = scn.next();
} while( !answer.equalsIgnoreCase("y") || !answer.equalsIgnoreCase("n"));
boolean dogBark=answer.equalsIgnoreCase("y");
i expect it to finish the while loos as soon as i enter 'y' or 'n' but its asks me for input over and over.

While answer is not yes or answer is not no?
If it's yes, then it isn't no. If it's no, then it isn't yes. So the condition is always true and it loops.
You probably want to use the ! operators.

while( !answer.equalsIgnoreCase("y") || !answer.equalsIgnoreCase("n"));
The issue is occurring because the do-while loop is checking both conditions with the OR statement.
Think about it like this, if Answer equals "Y" that condition is met to break the loop but the OR condition "N" is still active therefore the loop will continue.
The do while is only checking to ensure one of the conditions above is false before continuing the loop.
To fix this replace the || with && and the loop will break if either Y or N is entered as the loop will only continue if both conditions are false.

How to make do-loop conditions "when string does not equal"

I want to make my do loop run while the input the user made is not equal to the required letters (a-i) For some reason,even when i input the proper letters, it loops forever.
I've tried using switch cases as well as != inside the comparison.
Here is my code:
do {
System.out.println("Please enter the location of your battleship, starting with the first letter value. Make sure it is from the letters a-i.");
lL1=in.nextLine();
if (!lL1.equals("a")||!lL1.equals("b")||!lL1.equals("c")||!lL1.equals("d")||!lL1.equals("e")||!lL1.equals("f")||!lL1.equals("g")||!lL1.equals("h")||!lL1.equals("i")){
System.out.println("Invalid Input. Try again.");
}//End if statement
}while(!lL1.equals("a") || !lL1.equals("b") || !lL1.equals("c") || !lL1.equals("d") || !lL1.equals("e") || !lL1.equals("f") || !lL1.equals("g") || !lL1.equals("h") || !lL1.equals("i"));
My skills in Java are limited but this should work, unless i'm missing something obvious. Any help would be amazing!

Instead of using an operator for each case of the input, you might want to create a list of the accepted answers and then your condition will look like:
while answer is not in accepted answers, ask another input
An example would be:
Scanner scanner = new Scanner(System.in);
List<String> acceptedAnswers = Arrays.asList("a", "b", "c", "d", "e", "f", "g", "h", "i");
String input;
do {
System.out.println(
"Please enter the location of your battleship, starting with the first letter value. Make sure it is from the letters a-i.");
input = scanner.nextLine();
} while (!acceptedAnswers.contains(input));
scanner.close();
System.out.println("Got correct input: " + input);

If you have a negation you need AND to join the conditions not OR.

That's because if you or some not-ed values, they form an and.
Let me explain better.
If you input a, then the first is false (because you not it), and the others are true, so the or condition make the result be true.
You should instead group all the ors and then not it.
e.g.
!(lL1.equals("a") || lL1.equals("b") || lL1.equals("c") || lL1.equals("d") || lL1.equals("e") || lL1.equals("f") || lL1.equals("g") || lL1.equals("h") || lL1.equals("i"))

Please try this:
char lL1;
Scanner scanner = new Scanner(System.in);
do {
System.out.println("Please enter the location of your battleship, starting with the first letter value. Make sure it is from the letters a-i.");
lL1=scanner.next().charAt(0);
}while(lL1!='a' && lL1!='b' && lL1!='c' && lL1!='d' && lL1!='e' && lL1!='f' && lL1!='g' && lL1!='h' && lL1!='i');
Since you are only getting a single character, you can check that it does not match either [a to i] characters as shown above. This is the shortest way to do so by making the check as the condition of the loop. If it fails then the loop will be called.

Java NumberFormatException when using Integer.parseInt

I have truly searched for the answer all over the Internet before coming here and I think that the answer will have something to do with the try/catch statements, but even after watching a couple tutorials on the topic I am not sure on how to implement that.
Anyways, I am trying to do a simple thing in my newbie reminders app that I am making (I am learning Java as my first language for about 3 months now).
I want the program to check the user's input and if it is a certain letter ("R") I want the program to do a certain stuff. If it is an integer from 0 to 100 then I want to do other stuff. And if its neither of them, then I want the "else" statement to work.
The issue that I can't get the "else" statement to work as I get the NumberFormatException error. For example if I enter some other letter i.e. "d" - I get this error message:
Exception in thread "main" java.lang.NumberFormatException: For input
string: "d" at
java.lang.NumberFormatException.forInputString(NumberFormatException.java:65)
at java.lang.Integer.parseInt(Integer.java:580) at
java.lang.Integer.parseInt(Integer.java:615) at
mash.Dialogue.startDialogue(Dialogue.java:51) at
mash.Dialogue.newRem(Dialogue.java:27) at
mash.Dialogue.startDialogue(Dialogue.java:38) at
mash.Dialogue.start(Dialogue.java:13) at mash.Main.main(Main.java:9)
Here is the code (I am sorry for any readability issues, this is the first time ever I am showing my code to somebody). You don't have to read the else if statement, as the issue seems to not depend on the text inside of that statement.
I would really appreciate if anybody could point me what is wrong with the code and how I would get to do what I intended. Some newcomer-friendly solution will be much appreciated.
Thank you in advance!
String secondLetter = mash.nextLine();
if(secondLetter.equals("r") || secondLetter.equals("R")) { //if the user enters R - create a new Reminder
newRem();
}
else if((Integer.parseInt(secondLetter) >= 0) && (Integer.parseInt(secondLetter) < maximum)) { //if the user enters number - check task list
tasks.remText(Integer.parseInt(secondLetter));
System.out.println("Enter 'D' to set the reminder as Done. Or enter 'T' to return to the list");
String v = mash.nextLine();
System.out.println(v);
if(v.equals("d")|| v.equals("D")) { //if user enters D - set the reminder as done
tasks.setDone(Integer.parseInt(secondLetter));
System.out.println("The reminder is now added to 'Done' list");
}
else if(v.equals("t")|| v.equals("T")) { //if user enters T - return to the list of reminders
tasks.display();
}
else {
System.out.println("Please enter the correct symbol");
}
}
else {
System.out.println("Enter the correct symbol");
}

You can check your input if it's a valid number before attempting to convert it. For example:
if(!secondLetter.matches("[0-9]+")) {
//it's not a number, so dont attempt to parse it to an int
}
place it in your if/else like this:
if(secondLetter.equals("r") || secondLetter.equals("R")) {
newRem();
} else if(!secondLetter.matches("[0-9]+")){
System.out.println("please type r or R or a number");
} else if((Integer.parseInt(secondLetter) >= 0) && ...

Short answer: docs.oracle.
Complete answer:
You can use Integer.parsInt (String s) only on a string that can be parserized into an integer. The letter "R" can not be a number, so it generates an exception.
if(Character.isLetter(secondLetter) && "R".equalsIgnoreCase(secondLetter)){
do code with "R"
}else if(Integer.parseInt(secondLetter) > 0 && Integer.parseInt(secondLetter) < 100){
do code with 0 < number < 100
}else{
do something else
}

Loop until valid input is reached [duplicate]

This question already has answers here:
How do I compare strings in Java?
(23 answers)
Closed last year.
It executes correctly the first time, but:
It keeps printing "Please try again (Y/N)?" no matter what the
input is after asking to continue.
I am unsure if != is appropriate to use for String comparison. I want to say while
loopChoice "is not" Y or N, keep asking.
while(isLoop) {
// Ask for user input
System.out.print("Enter hours worked: ");
hoursWorked = scn.nextInt();
System.out.print("Enter rate per hour: ");
payRate = scn.nextInt();
scn.nextLine();
// Call functions to compute stuff
...
// Print results
...
System.out.print("\nDo you want to continue (Y/N)? ");
loopChoice = scn.nextLine().toUpperCase();
while(loopChoice != "Y" || loopChoice != "N") {
System.out.print("\nPlease try again (Y/N)? ");
loopChoice = scn.nextLine().toUpperCase();
}
switch(loopChoice) {
case "Y":
isLoop = true;
System.out.print("\n");
break;
case "N":
isLoop = false;
System.out.println("Terminating program...");
scn.close();
System.exit(0);
break;
default:
System.out.println("Your input is invalid!");
isLoop = false;
System.out.println("Terminating program...");
scn.close();
System.exit(0);
break;
}
}

You should compare with String equals
while (!loopChoice.equals("Y") && !loopChoice.equals("N"))
Also, replace the or operator with and operator

That's not how you compare strings in Java.
There is also a logical error in your code, as the string can't be both Y and N at the same time, you have to use && instead of ||. As long as the choice is neither Y or N, you want to continue the loop. If it is any of them, you want to stop. So && is the correct choice here.
To check if two strings are equal, you have to use .equals(obj)
while (!loopChoice.equals("Y") && !loopChoice.equals("N")) {
The reason for this is that == compares object references, and two Strings are most often not the same object reference. (Technically, they can be the same object reference, but that's not something you need to worry about now) To be safe, use .equals to compare Strings.
To avoid a possible NullPointerException in other situations, you can also use:
while (!"Y".equals(loopChoice) && !"N".equals(loopChoice)) {

You cannot use loopChoice != "Y", since "Y" is a String. Either use:
loopChoice != 'Y', or
"Y".equals(loopChoice)
Alternatively, use "Y".equalsIgnoreCase(loopChoice).
Case switching is also not possible for Strings if you use Java 1.6 or earlier. Be careful.

You need to know that OR Operation will return true if one of the two condition is true , so logically if you Enter Y , so you ask if the input is not equal Y so the answer is false then you will go to the next part in your condition if the input not equal N so the answer is True , so your finally result will be (True || False = True ) and then you will entered to while loop again
so the true condition is (the input not equal Y && not equal N)

You have fallen into the common early gap between checking equality of objects versus the values of objects. (You can see a quick list of string comparison information [here]
(http://docs.oracle.com/javase/tutorial/java/data/comparestrings.html)
What you wrote asks whether the object loopChoice is the same object as the string constant "Y" or the string constant "N" which will always return false. You want to ask whether the value of object loopChoice is the same as the value of string constant "Y".
You could rewrite your code as follows:
System.out.print("\nDo you want to continue (Y/N)? ");
// get value of next line, and trim whitespace in case use hit the spacebar
loopChoice = scn.nextLine().trim();
while (!("Y".equalsIgnoreCase(loopChoice) || "N".equalsIgnoreCase(loopChoice)) {
System.out.print("\nPlease try again (Y/N)? ");
loopChoice = scn.nextLine().toUpperCase();
}
Note, I like to put the constant value first for clarity. The general form for determining whether the value of two strings is the same is String1.equalsIgnoreCase(String2).

Putting 2 Boolean controllers in a do while expression?

newbie programmer here. I'm using Java to try and create a number guessing game. I want my do while loop to continue looping until the user inputs the correct number OR they run out of guesses. This is what I have and I can't figure out how to use 2 boolean controllers for the life of me.
do
{System.out.println("Enter guess #1");
userGuess = keyboard.nextInt();
} while ((userGuess != actualNumber) || (remainingGuesses = guesses; remainingGuesses >= 1; remainingGuesses--));
Any help is appreciated!

I think that you want something closer to this effect:
remainingGuess = guesses;
do {
System.out.println("Enter guess #1");
userGuess = keyboard.nextInt();
} while ( userGuess != actualNumber || remainingGuesses-- > 0 )
Line by line:
remainingGuesses = guesses;
assigns guesses to remainingGuesses once, if you were to do it every iteration of your loop, it would never end
userGuess != actualNumber || remainingGuesses-- > 0
Keep iterating while the user has guessed incorrectly OR remainingGuesses is more than 0.
remainingGuesses--
Evaluates to the current value of the variable remainingGuesses, then after the expression decrements it by 1.

Initialize remaining guesses before entering the loop. Only check the condition in the while paranthesis "()"
reaminingGuesses = guesses;
do
{System.out.println("Enter guess #1");
userGuess = keyboard.nextInt();
remainingGuess--;
} while ((userGuess != actualNumber) || (remainingGuesses >= 1));

If you want it as a do while I would make a few minor changes, and assuming you have constants defined for things like actualNumber and the total number of allowed guesses:
// break the calls into seperate methods, to make them easier to read in the long term and to
// seperate logical concerns into discrete elements.
do {
// get the user guess
guess = getGuess();
// if shouldStop returns true you would stop, if it returns false you should continue, thus the !..
} while (!shouldStop(guess)) {
// decrement the guesses.
--remainingGuesses;
}
/// define these methods somewhere..,
// get your guess
public int getGuess() {
System.out.println("Enter guess #1");
return keyboard.nextInt();
}
public boolean shouldStop(int guess) {
// condensed if statement syntax in java, very handy for this kind of thing
// do we have enough remaining guess to keep going? If so, compare the guess to the actual number
// otherwise return true (the stop signal
return remainingGuesses > 0 ? guess == actualNumber : true;
}
If you really wanted to follow the whole thing through you could break the guess==actual number into a method as well, but its probably not needed in this case since its a simple equality check.
The methos shouldStop could be defined a number of ways however...
Early on I think it's helpful to write these logical blocks out fully, and condense from there, for example:
public boolean shouldStop(int guess) {
if(remainingGuesses <=0) {
return true;
} else if(guess == actualNumber) {
return true;
} else {
return false;
}
}