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Can't figure out of how to stop the application when a specific String value match a specific condition

It's my first time posting some topic on this website, and running on a problem that I can't get over it. The following problem that I'm stunning into is as followed:
When the total calculation of the above decimal numbers meets a specific target amount, print: "Congratulations" if not, print:"Calculation error", after they input the string word: "Ready". when a user inputs a string called: "I Quit!", the application will exit and prints: "Quitter".
Here is the Java code that I currently have:
public static void goal(double targetAmount) {
Scanner sc = new Scanner(System.in);
double total = 0;
while (scanner.hasNextDouble()) {
double input= sc.nextDouble();
total += input;
}
String inputString = sc.next();
}
I'm looking forward to see your response. Hope I've formulate my question properly?

I think the problem is the input you read using scanner.next(), because scanner.next() reads the input till the next blank character. Which means it will just read I when you enter I Quit!.
Printing the output of the variable input shows the problem:
public static void main(String[] args) {
salarisdoel(100);
}
public static void salarisdoel(double targetAmount) {
Scanner scanner= new Scanner(System.in);
double total = 0;
while (scanner.hasNextDouble()) {
double inputMoney= scanner.nextDouble();
total += inputMoney;
}
String input = scanner.next();
System.out.println(input);//print to check what was read from the console
switch (input) {
case "I Quit!":
System.out.println("Quitter");
break;
case "Ready":
if (total>= targetAmount) {
System.out.println("Congratulations");
} else {
System.out.println("Calculation Error");
}
break;
default:
System.out.println("Something went wrong. Try again!");
break;
}
}
If you use a string without a blank space for quitting (e.g. "I_Quit!") it will work.

You can use System.exit() to exit the program.
case "I Quit!":
System.out.println("Quitter");
System.exit();

How to validate for a non-integer being entered in an integer data type? [duplicate]

This question already has answers here:
Validating input using java.util.Scanner [duplicate]
(6 answers)
Closed 3 years ago.
I have to make a store that has items for purchase. After choosing an item, I prompt the user to enter the quantity of the item they would like to buy.
// 'input' is my Scanner object
int quantity;
quantity = input.nextInt();
If the user enters a non-integer (i.e. decimal, char...), it breaks the program.
Is there a way I can validate for this non-integer input?
Thank you

Sure, accept a String value instead of an int, check to see if you can parse that String value to an int, if you can, then do so. If not, sent a message stating that then entered value must be an number.
This could be done in a while loop.
import java.util.Scanner;
public class ScannerInputInt {
public static void main(String... args) {
Scanner in = new Scanner(System.in);
Integer input = null;
do {
System.out.println("Please enter number: ");
String s = in.nextLine();
try {
input = Integer.parseInt(s);
} catch (NumberFormatException e) {
System.out.println("ERROR: " + s + " is not a number.");
}
} while (input == null);
}
}

If you don't wanna use Exceptions method. You can try this.
This piece of code will continue to ask user input until user has entered correct input.
System.out.print("Enter quantity: ");
Scanner input = new Scanner(System.in);
boolean isInt = input.hasNextInt(); // Check if input is int
while (isInt == false) { // If it is not int
input.nextLine(); // Discarding the line with wrong input
System.out.print("Please Enter correct input: "); // Asking user again
isInt = input.hasNextInt(); // If this is true it exits the loop otherwise it loops again
}
int quantity = input.nextInt(); // If it is int. It reads the input
System.out.println("Quantity: " + quantity);
input.close();
Output:
Enter quantity: 12.2
Please Enter correct input: 12.6
Please Enter correct input: s
Please Enter correct input: s6
Please Enter correct input: as
Please Enter correct input: 2
Quantity: 2
I think it is slightly better approach because I think controlling flow of your program with Exceptions is bad practice and should be avoiding when there are other things that you can use.

When user presses "enter" key or space enter then error message pops up

I would like to print an error message when the user presses enter or space enter instead of a string. I have tried isEquals("") and isEmpty() but haven't found anything that works yet.
Here's my code:
import java.util.Scanner;
public class check{
public static void main(String args[]){
System.out.println("Enter a number: ");
Scanner keyboard = new Scanner(System.in);
String input = keyboard.next();
if(input.equals("")){
System.out.println("Empty");
} else {
System.out.println("number inputed");
}
}
}

One way to do this, change keyboard.next() to keyboard.nextLine(), use trim() to remove unnecessary spaces, check with isEmpty().
String input = keyboard.nextLine().trim();
if (input.isEmpty()) {
// error message
} else {
// good to go
}

import java.util.Scanner;
public class check{
public static void main(String args[]){
System.out.println("Enter a number: ");
Scanner keyboard = new Scanner(System.in);
String input = keyboard.nextLine();
if(input.trim().equals("")){
System.out.println("Empty");
} else {
System.out.println("number inputed");
}
}
}

Strangely, I don't get an error when running your code. However, I noticed that your code simply doesn't react to an empty input (just pressing enter). If you want to check for that, you can use keyboard.nextLine().
Judging by the rest of your code, it seems like you want the user to input only a number. An easy way to check if the user entered an integer if you're using Scanner is keyboard.hasNextInt().
Meaning you can do something like this:
if(keyboard.hasNextInt()) {
int yourNumber = keyboard.nextInt();
System.out.println("Your number is: " + your Number);
}
else {
System.out.println("Please enter a valid integer");
}

To check whether the string input is empty, you can use the String.isEmpty() method. Look below:
String input = keyboard.nextLine();
if(!input.isEmpty()) {
//the input is not empty!
}
else {
//the input is empty!
}
Note, however, that since you want to receive numbers as inputs you should not retrieve them as strings. Below is an example where the program retrieves a double from the user. Scanner provides many methods to validate the user's input. In this case, I'm using hasNextDouble() to check whether the input is a number.
Scanner scanner = new Scanner(System.in);
System.out.println("Enter a number:");
while(!scanner.hasNextDouble()) {
System.out.println("That's not a number!");
scanner.next();
}
double numberInput = scanner.nextDouble();
System.out.println("The entered number was " + numberInput);

I made a sample program similar to yours and used nextLine() instead of next(). When user enters space and clicks enter he will print "space" else "a number".

How to insist that a users input is an int?

Basic problem here.. I will start off by asking that you please not respond with any code, as that likely will only confuse me further (programming noob). I am looking for a clear explanation on how to solve this issue that I'm having.
I have a scanner that reads input from the user. The user is prompted to enter an int value between 1 to 150 (whole numbers only). I obtain the value as follows:
Scanner scan = new Scanner(System.in);
int input = scan.nextInt();
And continue on with my program, and everything works fine.
Unfortunately, the code isn't exactly bulletproof, since any input that is not an integer can break it (letters, symbols, etc).
How can I make the code more robust, where it would verify that only an int was entered?
These are the results I'm hoping for:
Lets say the input was:
23 -> valid
fx -> display an error message, ask the user for input again (a while loop would do..)
7w -> error, again
3.7 -> error
$$ -> error
etc

Scanner.hasNextInt() returns true if the next token is a number, returns false otherwise.
In this example, I call hasNextInt(). If it returns true, I go past the while and set the input; if it returns false, then I discard the input (scanner.next();) and repeat.
Scanner scan = new Scanner(System.in);
while(!scan.hasNextInt()) {
scan.next();
}
int input = scan.nextInt();

Here's a simple example with prompts and comments.
Scanner scan = new Scanner(System.in);
System.out.print("Enter an integer: "); // Initial prompt for input
// Repeat until next item is an integer
while (!scan.hasNextInt())
{
scan.next(); // Read and discard offending non-int input
System.out.print("Please enter an integer: "); // Re-prompt
}
// At this point in the code, the user has entered an integer
int input = scan.nextInt(); // Get the integer
// And now you can use the input variable.

Use scan.hasNextInt() to make sure the next input is an int.

I have written an example that ensures that the program will continue only if a number and not an invalid value is entered. Do not worry, I added the desired explanation.
The program asks the user to input a number. A loop ensures that the processing will not go on until a valid number is entered. Before that I have defined a variable "inputAccepted" that has false as default value. If he enters a number, the variable "inputAccepted" is set to true and the program leaves the loop. But if he enters something else than a number, an exception is thrown right in this moment, and the line that sets the variable "inputAccepted" to true will not be executed. Instead a message will be printed out that tells the user that his input is not valid. Since "inputAccepted" could not be set to true, the loop will do the same stuff again until the string can be converted to a number.
You can test the program here.
import java.util.Scanner;
public class Test {
public static void main(String[] args) {
Scanner input = new Scanner(System.in);
boolean inputAccepted = false;
while (!inputAccepted) {
try {
System.out.print("Please enter a number: ");
Integer.valueOf(input.nextLine());
inputAccepted = true;
} catch (NumberFormatException e) {
System.out.println("Not a valid number.");
}
}
System.out.println("Thank you!");
}
}

Just get "anything" and parse it:
Scanner scan = new Scanner(System.in);
Integer number = null;
while (number == null) {
try {
number = Integer.parseInt(scan.next());
} catch (NumberParseException e) {
System.out.println("bad input: " + input);
}
}

Without any code and just in English, I'd say there's two things you have to test or look out for. First that the input is an int, second that the int is within the correct range.
In terms of pseudocode, the first thing to do is make sure it's an int. Declaring an int named "input", I would put a try / catch block, where you try to scan in the user input as an int, with parseInt(). If the try part fails, you know it's not an int and can return an error message.
Then, now that you know that "input" is an int, you can test whether it is less than 1 or more than 150, and return an error message if so!

public class Sample {
/**
* author CLRZ
*/
public static void main(String[] args) {
int a; // variable
Scanner in = new Scanner(System.in); // scans your input
System.out.println("Enter your number's choice:");
int sem1 = in.nextInt(); // reads next integer
if (sem1 == 1) // conditioned if your choice number is equal to 1
System.out.println("Hello World1"); // output wil be Hello World
int b;
System.out.println("Enter your number's choice:");
int sem2 = in.nextInt();
if (sem2 == 2)
System.out.println("Hello World2");
int c;
System.out.println("Enter your number's choice:");
int sem3 = in.nextInt();
if (sem3 == 3)
System.out.println("Hello World3");
}
}

In java Scanner.next()

public class Calculator
{
public static void main(String[] args)
{
boolean isValid = false;
Scanner myScanner = new Scanner(System.in);
String customerType = null;
System.out.print("Customer Type? (C/R) ");
customerType = myScanner.next();
while (isValid == false)
{
System.out.print("Enter Subtotal: ");
if (myScanner.hasNextDouble())
{
double sobTotal = myScanner.nextDouble();
isValid = true;
}
else
{
System.out
.println("Hay! Entry error please enter a valid number");
}
myScanner.nextLine();
}
}
}
Hi, im new to java, as usual im trying out a few things in the Scanner Class.
is there a way to see the input of the scanner?
Cuz i have a problem here with the code above as you'll see. this is the output of my console window after i entered wrong data. instead of numbers i entered KKK so can somebody explain me why i got this error message 2 times?
"this is the console"
Customer Type? (C/R) R
Enter Subtotal: KKK
Hay! Entry error please enter a valid number
Enter Subtotal: Hay! Entry error please enter a valid number
Enter Subtotal:

Change your code as follows.
boolean isValid = false;
Scanner myScanner = new Scanner(System.in);
String customerType = null;
System.out.print("Customer Type? (C/R) ");
customerType = myScanner.next();
while (!isValid)
{
System.out.print("Enter Subtotal: ");
if (myScanner.hasNext())
{
try
{
double sobTotal =Double.parseDouble(myScanner.next());
isValid = true;
}
catch(Exception e)
{
System.out.println("Hay! Entry error please enter a valid number");
}
}
}
When you use Scanner.nextDouble(), it does not consume the new line (or other delimiter) itself so the next token returned will typically be an empty string.

The problem is you're calling scanner.nextdouble() which will work fine if and when there is a double in the input. When there is not a double in the input your input that was "KKK" is ignored by the call to nextDouble() and your error message is displayed then when you call nextLine() that same input "KKK" is still there waiting wo when you loop back throught the while the "KKK" is then passed back to your program and since it is still not a double you get the repeated error message.
Try this:
boolean isValid = false;
Scanner myScanner = new Scanner(System.in).useDelimiter("\\n");
String customerType = null;
System.out.print("Customer Type? (C/R) ");
customerType = myScanner.next();
while (!isValid)
{
System.out.print("Enter Subtotal: ");
if (myScanner.hasNextDouble())
{
double sobTotal = myScanner.nextDouble();
isValid = true;
}
else
{
System.out.println("Hay! Entry error please enter a valid number");
if(myScanner.hasNext()){
myScanner.next();
}
}
}
That will consume the invalid input of "KKK" and let the program continue properly.

I think the real problem is that you have some obfuscation in your loop structure that makes it hard to understand the in and out conditions of the scanner with each iteration of the loop.
Heres what is really happening. We enter the loop with one token in the Scanner which is not a double. We ask is there a double next in the scanner with hasNextDouble(), which since the one token can not be converted to a double, this returns false. We report an error and isValid remains false. After the branch we call scanner.next() and consumer our non double token. We renter the loop since isValid remains false. We again ask if there is a double next in the Scanner. Now the scanner is empty, since we consumed out token, and this call returns false, so we report another error. Now we call scanner.next() again. Since the scanner is now empty, and scanner.next() implements synchronous message passing (I/O), we now block until the scanner recieves a new token, so your program will stall until you input a new token. Try puting in more into your console. You would probably repeat the loop and see another error.
What you want to accomplish is with each loop, consume a token. If that token is a double, accept the input and continue, if not report an error and repeat. Try something like this instead, it should present a more clear picture of what the scanner is doing every loop.
public class Calculator
{
public static void main(String[] args)
{
Scanner scan = new Scanner(System.in);
String customerType = null;
System.out.print("Customer Type? (C/R) ");
customerType = scan.next();
Double test = null;
do{
try{
test = scan.nextDouble();
}catch (InputMismatchException ime){ //will be thrown if the token we input is not a double
System.err.println("Error Message");
}
}while(test == null);
}
}
Notice i use the wrapper class Double not the primative double. This is so that I can set the double to null, since Double is an Object and not a primitive.