i've just started java programming and was wondering on how to approach or solve this problem i'm faced with.
I have to write a program that asks a user for a number and continually sums the numbers inputted and print the result.
This program stops when the user enters "END"
I just can't seem to think of a solution to this problem, any help or guidance throughout this problem would be much appreciated and would really help me understand problems like this. This is the best i could do
public static void main(String[] args) {
Scanner scan = new Scanner(System.in);
while (true) {
System.out.print("Enter a number: ");
int x = scan.nextInt();
System.out.print("Enter a number: ");
int y = scan.nextInt();
int sum = x + y;
System.out.println("Sum is now: " + sum);
}
}
}
The output is supposed to look like this:
Enter a number: 5
Sum is now: 5
Enter a number: 10
Sum is now: 15
Enter a number: END
One solution would be to not use the Scanner#nextInt() method at all but instead utilize the Scanner#nextLine() method and confirm the entry of the numerical entry with the String#matches() method along with a small Regular Expression (RegEx) of "\d+". This expression checks to see if the entire string contains nothing but numerical digits. If it does then the matches() method returns true otherwise it returns false.
Scanner scan = new Scanner(System.in);
int sum = 0;
String val = "";
while (val.equals("")) {
System.out.print("Enter a number (END to quit): ");
val = scan.nextLine();
// Was the word 'end' in any letter case supplied?
if (val.equalsIgnoreCase("end")) {
// Yes, so break out of loop.
break;
}
// Was a string representation of a
// integer numerical value supplied?
else if (val.matches("\\-?\\+?\\d+")) {
// Yes, convert the string to integer and sum it.
sum += Integer.parseInt(val);
System.out.println("Sum is now: " + sum); // Display Sum
}
// No, inform User of Invalid entry
else {
System.err.println("Invalid number supplied! Try again...");
}
val = ""; // Clear val to continue looping
}
// Broken out of loop with the entry of 'End"
System.out.println("Application ENDED");
EDIT: Based on Comment:
Since since an integer can be signed (ie: -20) or unsigned (ie: 20) and the fact that an Integer can be prefixed with a + (ie: +20) which is the same as unsigned 20, the code snippet above takes this into consideration.
Do it like this:
public static void main(String[] args) throws Exception {
int sum = 0;
Scanner scan = new Scanner(System.in);
while (scan.hasNext()) {
System.out.print("Enter a number: ");
if (scan.hasNextInt())
sum += scan.nextInt();
else
break;
System.out.println("Sum is now: " + sum);
}
System.out.print("END");
}
This will end if the input is not a number (int).
As pointed out in the comments, if you want the program to stop when the user specifically enters "END", change the else-statement to:
else if (scanner.next().equals("END"))
break;
New to java. I need to ask the user the number of strings (consisting only of upper and lowercase letters, spaces, and numbers) they want to input. These strings need to be stored in an array. Then I created a boolean method to be able to tell if those strings are palindromic (ignoring spaces and cases). If it is palindromic then I add to the result list to print later on. I am confused on how to ask the user to input that exact amount of strings and how to check each individual string. I must use StringBuilder. This is what I have so far (it's kind of a mess, sorry). I feel like I'm using the StringBuilder/array wrong, how can I fix this?
public class Palindromes {
public static void main(String[] args) {
int numOfStrings;
Scanner scan = new Scanner(System.in); // Creating Scanner object
System.out.print("Enter the number of strings: ");
numOfStrings = scan.nextInt();
System.out.print("Enter the strings: ");
StringBuilder paliString = new StringBuilder(numOfStrings);
for(int n=0; n < paliString; n++){
paliString[n] = scan.nextLine();
scan.nextLine();
String[] stringPali = new String[numOfStrings];
StringBuilder str = paliString;
if(isPali(userString)){
paliString = append.userString;
}
System.out.println("The palindromes are: " + userString ";");
}
static boolean isPali(String userString) {
int l = 0;
int h = userString.length() - 1;
// Lowercase string
userString = userString.toLowerCase();
// Compares character until they are equal
while (l <= h) {
char getAtl = userString.charAt(l);
char getAth = userString.charAt(h);
// If there is another symbol in left
// of sentence
if (!(getAtl >= 'a' && getAtl <= 'z'))
l++;
// If there is another symbol in right
// of sentence
else if (!(getAth >= 'a' && getAth <= 'z'))
h--;
// If characters are equal
else if (getAtl == getAth) {
l++;
h--;
}
// If characters are not equal then
// sentence is not palindrome
else
return false;
}
// Returns true if sentence is palindrome
return true;
}
}
SAMPLE RESULT:
Enter the number of strings: 8
Enter the strings:
Race Car
Mountain Dew
BATMAN
Taco Cat
Stressed Desserts
Is Mayonnaise an instrument
swap paws
A Toyotas a Toyota
The palindromes are: Race Car; Taco Cat; Stressed Desserts; swap paws; A Toyotas a Toyota
As I think the best way to answer this is to help you learn in small steps, I tried to stick with your initial idea on how to solve this and edited your main method with minimal changes.
This one does the trick.
public static void main(String[] args) {
int numOfStrings;
Scanner scan = new Scanner(System.in); // Creating Scanner object
System.out.print("Enter the number of strings: ");
numOfStrings = scan.nextInt();
scan.nextLine(); // you need this to catch the enter after the integer you entered
System.out.print("Enter the strings: ");
StringBuilder paliString = new StringBuilder();
for (int n = 0; n < numOfStrings; n++) {
String userString = scan.nextLine();
if (isPali(userString)) {
if (paliString.length() > 0) {
paliString.append("; ");
}
paliString.append(userString);
}
}
System.out.println("The palindromes are: " + paliString);
}
Key changes:
I added scan.nextLine(); right after reading the number of strings. This handles the newline you get when the user hits enter.
You don't need to initialize the StringBuilder with numOfStrings. This just preallocates the size of the StringBuilder in characters. Not the number of strings. Either way, it's not necessary. StringBuilder grows as needed.
I suggest you inspect what I did inside the for-loop. This was the biggest mess and changed significantly.
Last but not least: Writing the result needs to be outside of the for-loop, after all palindromes have been added to the StringBuilder.
Edit
Based on your comment, in this next iteration, I changed the usage of StringBuilder to the usage of an ArrayList. (Which is something completely different)
I am using it here because Lists in Java grow on demand. And since the number of palindromes is probably not equal to the number of input strings, this is the way to go. To really assign it to an array, one could always call String[] paliStringsArray = paliStrings.toArray(new String[]{}); but as ArrayLists already use an underlying array and are not necessary to to generate the output you want, I didn't put it into the new version.
Please compare the differences of this step to the previous version. I also added this String.join("; ", paliStrings) part, which creates the output you want.
public static void main(String[] args) {
int numOfStrings;
Scanner scan = new Scanner(System.in); // Creating Scanner object
System.out.print("Enter the number of strings: ");
numOfStrings = scan.nextInt();
scan.nextLine(); // you need this to catch the enter after the integer you entered
System.out.print("Enter the strings: ");
List<String> paliStrings = new ArrayList<>();
for (int n = 0; n < numOfStrings; n++) {
String userString = scan.nextLine();
if (isPali(userString)) {
paliStrings.add(userString);
}
}
System.out.println("The palindromes are: " + String.join("; ", paliStrings));
}
And now to the last step. Arvind Kumar Avinash actually solved a part that I also missed in the initial question. (I'll read more carefully in the future). He was validating the user input. So for the last iteration, I added his validation code in a modified way. I put it into a method as I think that makes things clearer and gets rid of the necessity of a the boolean valid variable.
public static void main(String[] args) {
int numOfStrings;
Scanner scan = new Scanner(System.in); // Creating Scanner object
System.out.print("Enter the number of strings: ");
numOfStrings = scan.nextInt();
scan.nextLine(); // you need this to catch the enter after the integer you entered
System.out.print("Enter the strings: ");
List<String> paliStrings = new ArrayList<>();
for (int n = 0; n < numOfStrings; n++) {
String userString = readNextLine(scan);
if (isPali(userString)) {
paliStrings.add(userString);
}
}
System.out.println("The palindromes are: " + String.join("; ", paliStrings));
}
static String readNextLine(Scanner scanner) {
while (true) {
String userString = scanner.nextLine();
if (userString.matches("[A-Za-z0-9 ]+")) {
return userString;
} else {
System.out.println("Error: invalid input.");
}
}
}
I need to ask the user the number of strings (consisting only of upper
and lowercase letters, spaces, and numbers) they want to input. These
strings need to be stored in an array.
I have done the above part of your question. I hope, this will give you direction to move forward.
import java.util.Scanner;
public class Main {
public static void main(String args[]) {
Scanner scan = new Scanner(System.in);
boolean valid = true;
int numOfStrings = 0;
do {
valid = true;
System.out.print("Enter the number of strings: ");
try {
numOfStrings = Integer.parseInt(scan.nextLine());
} catch (NumberFormatException e) {
System.out.println("Error: invalid input.");
valid = false;
}
} while (!valid);
String[] stringPali = new String[numOfStrings];
String input;
for (int i = 0; i < numOfStrings; i++) {
do {
valid = true;
System.out.print("Enter a string consisting of only letters and digits: ");
input = scan.nextLine();
if (!input.matches("[A-Za-z0-9 ]+")) {
System.out.println("Error: invalid input.");
valid = false;
}
} while (!valid);
stringPali[i] = input;
}
}
}
A sample run:
Enter the number of strings: a
Error: invalid input.
Enter the number of strings: 3
Enter a string consisting of only letters and digits: Arvind
Enter a string consisting of only letters and digits: Kumar Avinash
Enter a string consisting of only letters and digits: !#£$%^&*()_+
Error: invalid input.
Enter a string consisting of only letters and digits: Hello #
Error: invalid input.
Enter a string consisting of only letters and digits: Hello 123
Feel free to comment in case of any doubt/issue.
Wish you all the best!
[Update]
Based on your request, I have posted the following update which asks for the strings only once and then allows the user to enter all the strings one-by-one:
import java.util.Scanner;
public class Main {
public static void main(String args[]) {
Scanner scan = new Scanner(System.in);
boolean valid = true;
int numOfStrings = 0;
do {
valid = true;
System.out.print("Enter the number of strings: ");
try {
numOfStrings = Integer.parseInt(scan.nextLine());
} catch (NumberFormatException e) {
System.out.println("Error: invalid input.");
valid = false;
}
} while (!valid);
String[] stringPali = new String[numOfStrings];
String input;
System.out.println("Enter " + numOfStrings + " strings consisting of only letters and digits: ");
for (int i = 0; i < numOfStrings; i++) {
do {
valid = true;
input = scan.nextLine();
if (!input.matches("[A-Za-z0-9 ]+")) {
System.out.println("Error: invalid input.");
valid = false;
}
} while (!valid);
stringPali[i] = input;
}
}
}
A sample run:
Enter the number of strings: 3
Enter 3 strings consisting of only letters and digits:
Arvind
Kumar
He$ll0
Error: invalid input.
Avinash
Feel free to comment in case of any doubt.
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
Very Frustrated at my professor, because she did not teach try and catch concepts, neither did she teach us about throw exceptions either, so it is very difficult for me to do this program. The objective is to make a program where the user is asked to input an integer that prints "Hello World" that many times of the integer. The problem is I cannot check to make sure the user input is an integer. For instance, if the user chose to type a character or a double, how do I implement that into my code? And I cannot use throw exceptions or try and catch because we did not learn them yet.Thanks guys!!!
import java.util.Scanner;
public class PrintHelloWorld
{
public static void main( String[] args )
{
Scanner scan = new Scanner(System.in);
int number;
System.out.println("Please enter an integer that shows the " +
"number of times to print \"Hello World\" : ");
//store count
number = scan.nextInt();
System.out.print("Your integer is " + number);
int remainder = number%1;
int counts = 0;
if( number>0 && remainder == 0)
{
while(counts <= number)
{
System.out.println("Hello World!");
counts++;
}
}
else
System.out.print("Wrong, choose an integer!");
}
}
scan.hasNextInt()
will check to see if the next value in the input stream is an integer.
as such:
int number = -1;
System.out.println("Please enter an integer that shows the " +
"number of times to print \"Hello World\" : ");
//store count
if (scan.hasNextInt()) number = scan.nextInt();
if (number != -1) System.out.print("Your integer is " + number);
You can use a loop and validate the input with a regex, like this:
Scanner scan = new Scanner(System.in);
String input = null;
while (true) {
input = scan.nextLine();
if (input.matches("\\d+")) {
break;
}
System.out.println("Invalid input, please enter an integer!");
}
int number = Integer.parseInt(input);
This will keep asking for input until a valid integer is entered.
And I cannot use throw exceptions or try and catch because we did not
learn them yet.
For a first attempt, you could create a method that accepts a String as parameter. You will loop through all the chars of this String and check if each char is a digit. While this method returns false, re-ask the user for a new input.
Then use Integer.valueOf to get the int value..
public static boolean isNumber(String input)
You will have to use sc.nextLine() to get the input
The method Character.isDigit and toCharArray() will be useful