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Taking websites as keyboard input?

Here is the problem I was given:
Write a program that takes website names as keyboard input until the user types the word 'stop'. The program m just also count how many of the website names are commercial website names (i.e., end with .com), and output that count.
The problem that keeps occurring is even if I type the word stop as input, it is still saying to "enter the next site." I'm not sure where I went wrong.
Can anyone help? Here is my code.
import java.util.Scanner;
public class NewClass
{
public static void main( String [] args)
{
int numberOfComSites = 0;
String commercialNames = "com";
final String SENTINEL = "stop";
String website;
Scanner scan = new Scanner(System.in);
System.out.print("Enter a website, or 'stop' to stop > ");
website = scan.next();
String substring = website.substring(website.length()-3);
while (website != SENTINEL)
{
if(substring == commercialNames)
{ numberOfComSites++;
}
System.out.print( "Enter the next site > ");
website = scan.next();
}
System.out.println( "You entered" + numberOfComSites + "commercial websites.");
}
}
Thanks!

You are using reference equality == to compare strings. You strings are from different sources. SENTINEL comes from constant pool, while website comes from user input. They are always different as references.
To compare strings by value, the equals method should be used. In your case you, should replace
while (website != SENTINEL)
by
while (!SENTINEL.equals(website))
Notice that we compares constant with user input. This address a problem when website is null. This is not the case in your code, but it is a sign of good style.
See What is the difference between == vs equals() in Java? for more information.

replace
while (website != SENTINEL)
with
while(!website.equals(SENTINEL))
website is of String type and is not a primitive type. So you need to use equals method to compare String. == is used for reference comparison.
Refer this for more details What is the difference between == vs equals() in Java?

Can somebody spot the error in this java program [duplicate]

This question already has answers here:
How do I compare strings in Java?
(23 answers)
Closed 8 years ago.
My program should check whether the input is a palindrome or not. The given program compiles and runs successfully. Program prints reverse string correctly but gives wrong output. Please help!
class Palindrome
{
public static void main(String[] args)
{
String str,revStr="";
System.out.println("Enter something to check if it is a palindrome");
Scanner sn = new Scanner(System.in);
str = sn.nextLine();
for(int i=str.length()-1;i>=0;i--)
{
revStr+=Character.toString(str.charAt(i));
}
if(revStr==str)
{
System.out.println("The string "+str+" is a Palindrome");
System.out.println(revStr);
}
else
{
System.out.println("The string "+str+" is not a Palindrome");
System.out.println(revStr);
}
}
}
output:
Enter something to check if it is a palindrome
nitin
The string nitin is not a Palindrome
nitin

Here change this line
if(revStr==str)
To
If ( revStr.equals(str))
The thing is == checks reference equality
Object.equals is the method given in java to define your object equality
String class overrides that and check if two Strings represent same char array

Your answer here:
import java.util.Scanner;
class Palindrome
{
public static void main(String[] args)
{
String str,revStr="";
System.out.println("Enter something to check if it is a palindrome");
Scanner sn = new Scanner(System.in);
str = sn.nextLine();
for(int i=str.length()-1;i>=0;i--)
{
revStr+=Character.toString(str.charAt(i));
System.out.println("revStr" + revStr);
}
if(revStr.equals(str))//Don't use ==
{
System.out.println("The string "+str+" is a Palindrome");
System.out.println(revStr);
}
else
{
System.out.println("The string "+str+" is not a Palindrome");
System.out.println(revStr);
}
}
}
The “==” operator
In Java, when the “==” operator is used to compare 2 objects, it checks to see if the objects refer to the same place in memory. In other words, it checks to see if the 2 object names are basically references to the same memory location.
Equals() method is defined in Object class in Java and used for checking equality of two object defined by business logic

your if condition should be like this
if(revStr.equals(str)){
System.out.println("The string "+str+" is a Palindrome");
System.out.println(revStr);
}
Because in java == check the address of object not content
for more details check below thread
What is the difference between == vs equals() in Java?

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).

Detecting Numbers in a String Variable (In Java) [duplicate]

This question already has answers here:
How do I compare strings in Java?
(23 answers)
Closed 8 years ago.
So, normally for detecting user input, I use int and double variable types.
Example:
Scanner in = new Scanner(System.in);
int selection;
System.out.println("Welcome to RPG! (prototype name)\nIn this game, you do stuff.\nChoose a class:\n1. Soldier\n2. Knight\n3. Paladin\n4. Heavy");
selection = in.nextInt();
if(selection == 1){
System.out.print("you are a soldier");
}
else{
System.out.print(selection);
}
}
This technique usually works fine for me, but I noticed that if the user inputs a letter into the int variable, the game will crash because integers can't store letters. (right?) So I tried using a String variable in its place, like this:
Scanner in = new Scanner(System.in);
String selection;
System.out.println("Welcome to RPG! (prototype name)\nIn this game, you do stuff.\nChoose a class:\n1. Soldier\n2. Knight\n3. Paladin\n4. Heavy");
selection = in.next();
if(selection == "1"){
System.out.print("you are a soldier");
}
else{
System.out.print(selection);
}
}
This seemed to work at first, but as you can see, I have it set so that if the variable "selection" is equal to 1, that it will print "you are a soldier", yet this did not work, instead it printed out the "selection" variables value (1). Did I do something wrong or should I use a different type of variable?

you can use something la this :
try{
int type = Integer.parseInt(selection);
switch(type){
case 1:{
//do soldier stuff
}
case 2:{
// do knight stuff
}
default:{
//do other stuff
}
}
}catch(NumberFormatException exc ){
System.out.println(selection + "is not a number, try again!!!");
}

selection == "1"
Compare strings with String#equals.
"1".equals(selection)

There are lots of ways to do this. A quick point I'd like to make is if you're comparing strings you should use:
var.equals("string");
Since you're only taking one character you could use a character in which case the correct syntax would be:
var == '1'
If you want to be fancy you can do a try catch around your read statement and just read in an string and parse it to an integer, but that is a bit more advanced.

change selection == "1" to "1".equals(selection)

use .equals() for comparing the string "1" and selection and read this A simple explanation would be
x == y returns true only when both x and y refer to same object which is not in your case. equals check if contents are equal meaning if contents of memory location that x and y are referring to are equal or not.

If statement String trouble [duplicate]

This question already has answers here:
How do I compare strings in Java?
(23 answers)
Closed 9 years ago.
I'm trying to create a program which takes user input of two words and determines whether or not these words are the same.
import java.util.Scanner;
public class L7E3 {
public static void main(String[] args) {
Scanner keyboard = new Scanner(System. in );
String word1, word2;
System.out.println("Please enter a word: ");
word1 = keyboard.nextLine();
System.out.println("Please enter a word: ");
word2 = keyboard.nextLine();
if (word1 == word2) {
System.out.println("The words are " + word1 + " and " + word2 + ". These words are the same.");
} else {
System.out.println("The words are " + word1 + " and " + word2 + ". These words are not the same.");
}
}
}
I figured that word1==word2 would have worked to determine whether the two strings were equal, I'm using JGrasp and it goes directly to my else option regardless of input. Am I doing something wrong with strings?

if(word1.equals(word2))
== doesn't do what you think it does. == essentially compares the memory locations of the two String variables and returns true only if they're located at the same memory location. The String.equals method compares the contents of the strings and returns true if they hold the same characters.

Short answer: use String#equals(Object) instead:
word1.equals(word2)
For a super detailed explanation to the reasoning why: check out this.

For Strings you need to use the .equals() function rather than the == equality operator.
if(word1.equals(word2))
If you wanted to test if two words are the same, while ignoring case ("This" is the same as "this) then you need to do something like this:
if (word1.toLowerCase().equals(word2.toLowerCase()))
Also in your specific example you might want to remove unnecesary whitespace from before and after the word (" word1 " should becomes "word1"). You can do this using the trim() function:
word1 = word.trim();