I'm creating an application for a homework, the problem is that I am trying to create a do-while loop to exit the application (Using the question "Do you want to exit (Y/N)"). To work with the do-while loop, I created a method to store the app and then called the method in the do-while loop, so that when I try to stop the loop, the method loops once more. I want when I type "Y" to the console the whole program stops and doesn't loop one more time.
I created a simple example to explain my problem.
Here's the method:
public static void App(){
Scanner sc = new Scanner(System.in);
System.out.print("Write a number: ");
int num1 = sc.nextInt();
System.out.print("Write another number: ");
int num2 = sc.nextInt();
System.out.println("\nResult: "+(num1+num2));
}
And here I'm trying to create the loop in the main method:
public static void main(String[] args) {
Scanner sc2 = new Scanner(System.in);
App();
String answer;
do {
System.out.println("Do you want to exit (Y/N)?");
answer = sc2.next();
App();
} while (answer.equalsIgnoreCase("N")) ;
}
the problem is that I am trying to create a do-while loop to exit the application
You already have that in your program.
so that when I try to stop the loop, the method loops once more...
That doesn't fit the goal of your program.
I want when I type "Y" to the console the whole program stops and doesn't loop one more time
A lot of context that doesn't fit right in.
But anyway, you just have to reorganize your program.
In other words, just move your App() method.
public static void main(String[] args) {
Scanner sc2 = new Scanner(System.in);
String answer;
do {
App();
System.out.println("Do you want to exit (Y/N)?");
answer = sc2.next();
} while (answer.equalsIgnoreCase("N")) ;
}
Also, I spotted a lot of bad practices, so I kind of fixed them:
public static void main(String[] args) throws Exception {
try(Scanner sc2 = new Scanner(System.in)){
String answer;
do {
App();
System.out.print("Do you want to exit (Y/N)?");
answer = sc2.nextLine();
} while (answer.equalsIgnoreCase("N")) ;
}
}
Lastly, maybe (just maybe) try to solve your problem first before seeking help for your homework.
The reason why your program is running again after you type n is because the App() method is ran after the question is asked within the do part of the loop.
This code below is the simplest fix I could think of.
public static void main(String[] args) {
Scanner sc2 = new Scanner(System.in);
// I removed the line 'App();' as the App method will always run at least one time. Therefore putting that method within the 'do' part of the loop allows us to ask the user if they wish to exit or not after they have received their answer.
String answer;
do {
App();
System.out.print("Do you want to exit (Y/N)?"); //I changed the 'println' to 'print' here
answer = sc2.next();
} while (answer.equalsIgnoreCase("N")) ;
}
As a side note, methods in java should be lower-case when following typical Java naming conventions. While this will not affect how your code runs, I would suggest renaming the method from App() to app().
Everything looks good in your code, Just change the execution logic as shown in code blocks.
public static void main(String[] args) {
Scanner sc2 = new Scanner(System.in);
App(); //remove this line from here
String answer;
do {
App(); //call App function here so that it got executed at least one time
System.out.println("Do you want to exit (Y/N)?");
answer = sc2.next();
App(); //remove this as well
} while (answer.equalsIgnoreCase("N")) ;
}
Here is yet another approach except it uses a while loops instead of do/while loops. Two different approaches are provided and both provide User entry validation:
Approach #1:
public static void appMethod() {
Scanner sc = new Scanner(System.in);
int num1 = Integer.MIN_VALUE; // Initialize with some obscure value.
int num2 = Integer.MIN_VALUE; // Initialize with some obscure value.
while (num1 == Integer.MIN_VALUE) {
System.out.print("Write a number: ");
try {
num1 = sc.nextInt();
} catch ( java.util.InputMismatchException ex) {
System.out.println("Invalid Entry! Try again..."
+ System.lineSeparator());
sc.nextLine(); // consume the ENTER key hit otherwise this error will keep cycling.
num1 = Integer.MIN_VALUE;
}
}
while (num2 == Integer.MIN_VALUE) {
System.out.print("Now, write yet another number: ");
try {
num2 = sc.nextInt();
} catch ( java.util.InputMismatchException ex) {
System.out.println("Invalid Entry! Try again..."
+ System.lineSeparator());
sc.nextLine(); // consume the ENTER key hit otherwise this error will keep cycling.
num2 = Integer.MIN_VALUE;
}
}
System.out.println("\nResult: " + num1 +" + " + num2 + " = " + (num1 + num2));
}
Approach #2:
This next approach makes use of the Scanner#nextLine() method. The thing to remember about nextLine() is that, if you use it in your console application then basically recommend you use it for everything (all prompts). A 'quit' mechanism is also available in this version. Read the comments in code:
public static void appMethod() {
Scanner sc = new Scanner(System.in);
// Retrieve first number...
String num1 = "";
while (num1.isEmpty()) {
System.out.print("Write a number (q to quit): ");
// Making use of the Scanner#nextLine() method
num1 = sc.nextLine();
// Has 'q' been supplied to Quit?
if (num1.equalsIgnoreCase("q")) {
return;
}
/* Validate the fact that a signed or unsigned Integer or
Floating Point value has been entered. If not show Msg. */
if (!num1.matches("-?\\d+(\\.\\d+)?")) {
System.out.println("Invalid Entry! (" + num1 + ") Try again..."
+ System.lineSeparator());
num1 = ""; // empty num1 so as to re-loop.
}
}
// Retrieve second number...
String num2 = "";
while (num2.isEmpty()) {
System.out.print("Now, write yet another number (q to quit): ");
num2 = sc.nextLine();
if (num2.equalsIgnoreCase("q")) {
return;
}
if (!num2.matches("-?\\d+(\\.\\d+)?")) {
System.out.println("Invalid Entry! (" + num2 + ") Try again..."
+ System.lineSeparator());
num2 = "";
}
}
// Convert the numerical strings to double data type values.
double number1 = Double.parseDouble(num1);
double number2 = Double.parseDouble(num2);
// Display the result.
System.out.println("\nResult: " + num1 +" + " + num2 + " = " + (number1 + number2));
}
I attempted to create a calculator, but I can not get it to work because I don't know how to get user input.
How can I get the user input in Java?
One of the simplest ways is to use a Scanner object as follows:
import java.util.Scanner;
Scanner reader = new Scanner(System.in); // Reading from System.in
System.out.println("Enter a number: ");
int n = reader.nextInt(); // Scans the next token of the input as an int.
//once finished
reader.close();
You can use any of the following options based on the requirements.
Scanner class
import java.util.Scanner;
//...
Scanner scan = new Scanner(System.in);
String s = scan.next();
int i = scan.nextInt();
BufferedReader and InputStreamReader classes
import java.io.BufferedReader;
import java.io.InputStreamReader;
//...
BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
String s = br.readLine();
int i = Integer.parseInt(s);
DataInputStream class
import java.io.DataInputStream;
//...
DataInputStream dis = new DataInputStream(System.in);
int i = dis.readInt();
The readLine method from the DataInputStream class has been deprecated. To get String value, you should use the previous solution with BufferedReader
Console class
import java.io.Console;
//...
Console console = System.console();
String s = console.readLine();
int i = Integer.parseInt(console.readLine());
Apparently, this method does not work well in some IDEs.
You can use the Scanner class or the Console class
Console console = System.console();
String input = console.readLine("Enter input:");
You can get user input using BufferedReader.
BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
String accStr;
System.out.println("Enter your Account number: ");
accStr = br.readLine();
It will store a String value in accStr so you have to parse it to an int using Integer.parseInt.
int accInt = Integer.parseInt(accStr);
Here is how you can get the keyboard inputs:
Scanner scanner = new Scanner (System.in);
System.out.print("Enter your name");
String name = scanner.next(); // Get what the user types.
The best two options are BufferedReader and Scanner.
The most widely used method is Scanner and I personally prefer it because of its simplicity and easy implementation, as well as its powerful utility to parse text into primitive data.
Advantages of Using Scanner
Easy to use the Scanner class
Easy input of numbers (int, short, byte, float, long and double)
Exceptions are unchecked which is more convenient. It is up to the programmer to be civilized, and specify or catch the exceptions.
Is able to read lines, white spaces, and regex-delimited tokens
Advantages of BufferedInputStream
BufferedInputStream is about reading in blocks of data rather than a single byte at a time
Can read chars, char arrays, and lines
Throws checked exceptions
Fast performance
Synchronized (you cannot share Scanner between threads)
Overall each input method has different purposes.
If you are inputting large amount of data BufferedReader might be
better for you
If you are inputting lots of numbers Scanner does automatic parsing
which is very convenient
For more basic uses I would recommend the Scanner because it is easier to use and easier to write programs with. Here is a quick example of how to create a Scanner. I will provide a comprehensive example below of how to use the Scanner
Scanner scanner = new Scanner (System.in); // create scanner
System.out.print("Enter your name"); // prompt user
name = scanner.next(); // get user input
(For more info about BufferedReader see How to use a BufferedReader and see Reading lines of Chars)
java.util.Scanner
import java.util.InputMismatchException; // import the exception catching class
import java.util.Scanner; // import the scanner class
public class RunScanner {
// main method which will run your program
public static void main(String args[]) {
// create your new scanner
// Note: since scanner is opened to "System.in" closing it will close "System.in".
// Do not close scanner until you no longer want to use it at all.
Scanner scanner = new Scanner(System.in);
// PROMPT THE USER
// Note: when using scanner it is recommended to prompt the user with "System.out.print" or "System.out.println"
System.out.println("Please enter a number");
// use "try" to catch invalid inputs
try {
// get integer with "nextInt()"
int n = scanner.nextInt();
System.out.println("Please enter a decimal"); // PROMPT
// get decimal with "nextFloat()"
float f = scanner.nextFloat();
System.out.println("Please enter a word"); // PROMPT
// get single word with "next()"
String s = scanner.next();
// ---- Note: Scanner.nextInt() does not consume a nextLine character /n
// ---- In order to read a new line we first need to clear the current nextLine by reading it:
scanner.nextLine();
// ----
System.out.println("Please enter a line"); // PROMPT
// get line with "nextLine()"
String l = scanner.nextLine();
// do something with the input
System.out.println("The number entered was: " + n);
System.out.println("The decimal entered was: " + f);
System.out.println("The word entered was: " + s);
System.out.println("The line entered was: " + l);
}
catch (InputMismatchException e) {
System.out.println("\tInvalid input entered. Please enter the specified input");
}
scanner.close(); // close the scanner so it doesn't leak
}
}
Note: Other classes such as Console and DataInputStream are also viable alternatives.
Console has some powerful features such as ability to read passwords, however, is not available in all IDE's (such as Eclipse). The reason this occurs is because Eclipse runs your application as a background process and not as a top-level process with a system console. Here is a link to a useful example on how to implement the Console class.
DataInputStream is primarily used for reading input as a primitive datatype, from an underlying input stream, in a machine-independent way. DataInputStream is usually used for reading binary data. It also provides convenience methods for reading certain data types. For example, it has a method to read a UTF String which can contain any number of lines within them.
However, it is a more complicated class and harder to implement so not recommended for beginners. Here is a link to a useful example how to implement a DataInputStream.
You can make a simple program to ask for user's name and print what ever the reply use inputs.
Or ask user to enter two numbers and you can add, multiply, subtract, or divide those numbers and print the answers for user inputs just like a behavior of a calculator.
So there you need Scanner class. You have to import java.util.Scanner; and in the code you need to use
Scanner input = new Scanner(System.in);
Input is a variable name.
Scanner input = new Scanner(System.in);
System.out.println("Please enter your name : ");
s = input.next(); // getting a String value
System.out.println("Please enter your age : ");
i = input.nextInt(); // getting an integer
System.out.println("Please enter your salary : ");
d = input.nextDouble(); // getting a double
See how this differs: input.next();, i = input.nextInt();, d = input.nextDouble();
According to a String, int and a double varies same way for the rest. Don't forget the import statement at the top of your code.
Also see the blog post "Scanner class and getting User Inputs".
To read a line or a string, you can use a BufferedReader object combined with an InputStreamReader one as follows:
BufferedReader bufferReader = new BufferedReader(new InputStreamReader(System.in));
String inputLine = bufferReader.readLine();
Here, the program asks the user to enter a number. After that, the program prints the digits of the number and the sum of the digits.
import java.util.Scanner;
public class PrintNumber {
public static void main(String[] args) {
Scanner scan = new Scanner(System.in);
int num = 0;
int sum = 0;
System.out.println(
"Please enter a number to show its digits");
num = scan.nextInt();
System.out.println(
"Here are the digits and the sum of the digits");
while (num > 0) {
System.out.println("==>" + num % 10);
sum += num % 10;
num = num / 10;
}
System.out.println("Sum is " + sum);
}
}
Here is your program from the question using java.util.Scanner:
import java.util.Scanner;
public class Example {
public static void main(String[] args) {
int input = 0;
System.out.println("The super insano calculator");
System.out.println("enter the corrosponding number:");
Scanner reader3 = new Scanner(System.in);
System.out.println(
"1. Add | 2. Subtract | 3. Divide | 4. Multiply");
input = reader3.nextInt();
int a = 0, b = 0;
Scanner reader = new Scanner(System.in);
System.out.println("Enter the first number");
// get user input for a
a = reader.nextInt();
Scanner reader1 = new Scanner(System.in);
System.out.println("Enter the scend number");
// get user input for b
b = reader1.nextInt();
switch (input){
case 1: System.out.println(a + " + " + b + " = " + add(a, b));
break;
case 2: System.out.println(a + " - " + b + " = " + subtract(a, b));
break;
case 3: System.out.println(a + " / " + b + " = " + divide(a, b));
break;
case 4: System.out.println(a + " * " + b + " = " + multiply(a, b));
break;
default: System.out.println("your input is invalid!");
break;
}
}
static int add(int lhs, int rhs) { return lhs + rhs; }
static int subtract(int lhs, int rhs) { return lhs - rhs; }
static int divide(int lhs, int rhs) { return lhs / rhs; }
static int multiply(int lhs, int rhs) { return lhs * rhs; }
}
Scanner input = new Scanner(System.in);
String inputval = input.next();
Scanner input=new Scanner(System.in);
int integer=input.nextInt();
String string=input.next();
long longInteger=input.nextLong();
Just one extra detail. If you don't want to risk a memory/resource leak, you should close the scanner stream when you are finished:
myScanner.close();
Note that java 1.7 and later catch this as a compile warning (don't ask how I know that :-)
Here is a more developed version of the accepted answer that addresses two common needs:
Collecting user input repeatedly until an exit value has been entered
Dealing with invalid input values (non-integers in this example)
Code
package inputTest;
import java.util.Scanner;
import java.util.InputMismatchException;
public class InputTest {
public static void main(String args[]) {
Scanner reader = new Scanner(System.in);
System.out.println("Please enter integers. Type 0 to exit.");
boolean done = false;
while (!done) {
System.out.print("Enter an integer: ");
try {
int n = reader.nextInt();
if (n == 0) {
done = true;
}
else {
// do something with the input
System.out.println("\tThe number entered was: " + n);
}
}
catch (InputMismatchException e) {
System.out.println("\tInvalid input type (must be an integer)");
reader.nextLine(); // Clear invalid input from scanner buffer.
}
}
System.out.println("Exiting...");
reader.close();
}
}
Example
Please enter integers. Type 0 to exit.
Enter an integer: 12
The number entered was: 12
Enter an integer: -56
The number entered was: -56
Enter an integer: 4.2
Invalid input type (must be an integer)
Enter an integer: but i hate integers
Invalid input type (must be an integer)
Enter an integer: 3
The number entered was: 3
Enter an integer: 0
Exiting...
Note that without nextLine(), the bad input will trigger the same exception repeatedly in an infinite loop. You might want to use next() instead depending on the circumstance, but know that input like this has spaces will generate multiple exceptions.
import java.util.Scanner;
class Daytwo{
public static void main(String[] args){
System.out.println("HelloWorld");
Scanner reader = new Scanner(System.in);
System.out.println("Enter the number ");
int n = reader.nextInt();
System.out.println("You entered " + n);
}
}
Add throws IOException beside main(), then
DataInputStream input = new DataInputStream(System.in);
System.out.print("Enter your name");
String name = input.readLine();
It is very simple to get input in java, all you have to do is:
import java.util.Scanner;
class GetInputFromUser
{
public static void main(String args[])
{
int a;
float b;
String s;
Scanner in = new Scanner(System.in);
System.out.println("Enter a string");
s = in.nextLine();
System.out.println("You entered string " + s);
System.out.println("Enter an integer");
a = in.nextInt();
System.out.println("You entered integer " + a);
System.out.println("Enter a float");
b = in.nextFloat();
System.out.println("You entered float " + b);
}
}
import java.util.Scanner;
public class Myapplication{
public static void main(String[] args){
Scanner in = new Scanner(System.in);
int a;
System.out.println("enter:");
a = in.nextInt();
System.out.println("Number is= " + a);
}
}
You can get user input like this using a BufferedReader:
InputStreamReader inp = new InputStreamReader(System.in);
BufferedReader br = new BufferedReader(inp);
// you will need to import these things.
This is how you apply them
String name = br.readline();
So when the user types in his name into the console, "String name" will store that information.
If it is a number you want to store, the code will look like this:
int x = Integer.parseInt(br.readLine());
Hop this helps!
Can be something like this...
public static void main(String[] args) {
Scanner reader = new Scanner(System.in);
System.out.println("Enter a number: ");
int i = reader.nextInt();
for (int j = 0; j < i; j++)
System.out.println("I love java");
}
You can get the user input using Scanner. You can use the proper input validation using proper methods for different data types like next() for String or nextInt() for Integer.
import java.util.Scanner;
Scanner scanner = new Scanner(System.in);
//reads the input until it reaches the space
System.out.println("Enter a string: ");
String str = scanner.next();
System.out.println("str = " + str);
//reads until the end of line
String aLine = scanner.nextLine();
//reads the integer
System.out.println("Enter an integer num: ");
int num = scanner.nextInt();
System.out.println("num = " + num);
//reads the double value
System.out.println("Enter a double: ");
double aDouble = scanner.nextDouble();
System.out.println("double = " + aDouble);
//reads the float value, long value, boolean value, byte and short
double aFloat = scanner.nextFloat();
long aLong = scanner.nextLong();
boolean aBoolean = scanner.nextBoolean();
byte aByte = scanner.nextByte();
short aShort = scanner.nextShort();
scanner.close();
import java.util.Scanner;
public class Main {
public static void main(String[] args) {
System.out.println("Welcome to the best program in the world! ");
while (true) {
System.out.print("Enter a query: ");
Scanner scan = new Scanner(System.in);
String s = scan.nextLine();
if (s.equals("q")) {
System.out.println("The program is ending now ....");
break;
} else {
System.out.println("The program is running...");
}
}
}
}
This is a simple code that uses the System.in.read() function. This code just writes out whatever was typed. You can get rid of the while loop if you just want to take input once, and you could store answers in a character array if you so choose.
package main;
import java.io.IOException;
public class Root
{
public static void main(String[] args)
{
new Root();
}
public Root()
{
while(true)
{
try
{
for(int y = 0; y < System.in.available(); ++y)
{
System.out.print((char)System.in.read());
}
}
catch(IOException ex)
{
ex.printStackTrace(System.out);
break;
}
}
}
}
I like the following:
public String readLine(String tPromptString) {
byte[] tBuffer = new byte[256];
int tPos = 0;
System.out.print(tPromptString);
while(true) {
byte tNextByte = readByte();
if(tNextByte == 10) {
return new String(tBuffer, 0, tPos);
}
if(tNextByte != 13) {
tBuffer[tPos] = tNextByte;
++tPos;
}
}
}
and for example, I would do:
String name = this.readLine("What is your name?")
Keyboard entry using Scanner is possible, as others have posted. But in these highly graphic times it is pointless making a calculator without a graphical user interface (GUI).
In modern Java this means using a JavaFX drag-and-drop tool like Scene Builder to lay out a GUI that resembles a calculator's console.
Note that using Scene Builder is intuitively easy and demands no additional Java skill for its event handlers that what you already may have.
For user input, you should have a wide TextField at the top of the GUI console.
This is where the user enters the numbers that they want to perform functions on.
Below the TextField, you would have an array of function buttons doing basic (i.e. add/subtract/multiply/divide and memory/recall/clear) functions.
Once the GUI is lain out, you can then add the 'controller' references that link each button function to its Java implementation, e.g a call to method in your project's controller class.
This video is a bit old but still shows how easy Scene Builder is to use.
The most simple way to get user input would be to use Scanner. Here's an example of how it's supposed to be used:
import java.util.Scanner;
public class main {
public static void main(String[]args) {
Scanner sc=new Scanner(System.in);
int a;
String b;
System.out.println("Type an integer here: ");
a=sc.nextInt();
System.out.println("Type anything here:");
b=sc.nextLine();
The line of code import java.util.Scanner; tells the program that the programmer will be using user inputs in their code. Like it says, it imports the scanner utility. Scanner sc=new Scanner(System.in); tells the program to start the user inputs. After you do that, you must make a string or integer without a value, then put those in the line a=sc.nextInt(); or a=sc.nextLine();. This gives the variables the value of the user inputs. Then you can use it in your code. Hope this helps.
Using JOptionPane you can achieve it.
Int a =JOptionPane.showInputDialog(null,"Enter number:");
import java.util.Scanner;
public class userinput {
public static void main(String[] args) {
Scanner input = new Scanner(System.in);
System.out.print("Name : ");
String name = input.next();
System.out.print("Last Name : ");
String lname = input.next();
System.out.print("Age : ");
byte age = input.nextByte();
System.out.println(" " );
System.out.println(" " );
System.out.println("Firt Name: " + name);
System.out.println("Last Name: " + lname);
System.out.println(" Age: " + age);
}
}
class ex1 {
public static void main(String args[]){
int a, b, c;
a = Integer.parseInt(args[0]);
b = Integer.parseInt(args[1]);
c = a + b;
System.out.println("c = " + c);
}
}
// Output
javac ex1.java
java ex1 10 20
c = 30
Using Java to compile a code that takes in any input, from strings to doubles, floats, integers, etc., but can only process based on integers. If it receives a double, float, or a string, it will just produce a message prompting the user to try again. I'm not sure exactly how to get this rolling, so here's my current code.
import java.util.Scanner;
public class Sand
{
public static void main(String[] args)
{
int firstInt, secondInt, thirdInt;
Scanner keyboard = new Scanner(System.in);
System.out.println("Hi. I'm going to be asking you for three integers in just a moment.");
System.out.println("An integer is defined as a WHOLE number that's not a fraction or decimal.");
System.out.println("I swear to you, I will go off if you put in a non-whole number...");
System.out.println("Okay go ahead and type in the first of the three integers:");
if (keyboard.hasNextInt())
{
firstInt = keyboard.nextInt();
System.out.println("Red.");
System.out.println("So the first integer is " + firstInt);
System.out.println("Okay go ahead and type in the second of the three integers: ");
if (keyboard.hasNextInt())
{
secondInt = keyboard.nextInt();
System.out.println("Green.");
System.out.println("So the first and second integers are " + firstInt + " and " + secondInt);
}
else
{
System.out.println("Orange.");
System.out.println("Please try again.");
}
}
else
{
System.out.println("Blue.");
System.out.println("Please try again.");
}
}
}
I have orange and blue representing times in which there are some input errors, but it's not complete. I'm not sure how to approach this, be it a while loop or a for loop or a try/catch. I'm new when it comes to learning Java so some #notes would be helpful along the way. The code is to designed to read three numbers that are integers from the user in a string. That's straightforward, but analyzing the input is where I'm struggling.
This is not meant to answer your question totally but to point you in the right direction. There should be enough here to help you figure out what else is needed for your program.
import java.util.Scanner;
public class Sandbox {
// arguments are passed using the text field below this editor
public static void main(String[] args) {
Scanner keyboard = new Scanner(System.in);
int num = 0;
System.out.println("enter an integer");
while (true) { //keep prompting the user until they comply
try {
num = Integer.parseInt(keyboard.nextLine());
keyboard.close();
break;
} catch (NumberFormatException e) {
System.out.println("You must enter an integer");
}
}
System.out.println("num: " + num);
}
}
I, personally, would use an array or arrayList (based on your needs) to store the int values and then use a while loop to check and accept the numbers. while (array.length < 3) do {int checking}. This way the script will not run if there are already 3 values in the array and will run until you get 3 values.
this should do the job..
import java.util.Scanner;
public class Main {
public static void main(String[] args) {
int[] intArray = new int[3];
Scanner keyboard = new Scanner(System.in);
for (int i = 0; i < intArray.length; i++) {
System.out.print("input integer #"+(i+1)+": ");
if (keyboard.hasNextInt()) {
intArray[i] = keyboard.nextInt();
System.out.println("value for #"+(i+1)+": " + intArray[i]);
} else
{
System.out.println("not an integer");
break;
}
}
keyboard.close();
}
}
or maybe another solution that asks again when the number was no integer
import java.util.Scanner;
public class Main2 {
public static void main(String[] args) {
int[] intArray = new int[3];
Scanner keyboard = new Scanner(System.in);
for (int i = 0; i < intArray.length; i++) {
boolean isInputCorrect = false;
do {
System.out.print("input integer #" + (i + 1) + ": ");
if (keyboard.hasNextInt()) {
intArray[i] = keyboard.nextInt();
System.out.println("value for #" + (i + 1) + ": " + intArray[i]);
isInputCorrect = true;
} else {
System.out.println("not an integer, try again");
keyboard.nextLine();
}
} while (!isInputCorrect);
}
keyboard.close();
}
}
I'm fairly new to java, so don't think this is some idiot. Anyways, I've been trying to make a program that can read a certain letter from the console and then decide which operation to use, let's say to add. However, I can't get an If loop to read the variable that decides which operator to use, here is the code, and please help.
import java.util.Scanner;
class Main {
public static void main(String[] args) {
Scanner user_input = new Scanner( System.in );
int number;
String function;
System.out.println("What Do You Want to Do? (a to add; s to" +
" subrtact; d to divited; m to multiply, and sq to square your nummber.)" );
function = user_input.next();
if (function == "sq"){
System.out.print("Enter your number: ");
number = user_input.nextInt();
System.out.print(number * number);
} else {
System.out.println("Unidentified Function!");
}
}
}
(I made the description shorter so that it would fit).
This is just an example to get you started in the right direction.
import java.util.Scanner;
public class Example {
public static void main(String[] args) {
Scanner user_input = new Scanner(System.in);
int num1, num2, result;
System.out.println("What Do You Want to Do? (a to add; s to"
+ " subrtact; d to divited; m to multiply, and s to square your nummber.)");
String choice = user_input.next();
// Add
if (Character.isLetter('a')) {
System.out.println("Enter first number: ");
num1 = user_input.nextInt();
System.out.println("Enter second number: ");
num2 = user_input.nextInt();
result = num1 + num2;
System.out.println("Answer: " + result);
}
}
}
If you use hasNext() on a scanner it will wait for an input until you stop the program. Also using equals() is a better way of comparing strings.
while(user_input.hasNext()){
function = user_input.next();
if (function.equals("s")){
System.out.print("Enter your number: ");
number = user_input.nextInt();
System.out.print(number * number);
} else {
System.out.println("Unidentified Function!");
}
}
Scanner s = new Scanner(System.in);
String str = s.nextLine();
int a=s.nextInt();
int b=s.nextInt();
if(str.equals("+"))
c=a+b;
else if(str.equals("-"))
c=a-b;
else if(str.equals("/"))
c=a/b;
// you can add operators as your use
else
System.out.println("Unidentified operator" );
I hope it helps!