How to put Input on next line in Java's Scanner utility - java

I have a basic java question about the scanner utility. Let's say i have a simple program that takes the user input and stores it in a variable. My question is when i run the program that asks for multiple inputs, the cursor starts at the beginning of the question and not after it.
My code is:
public class question3 {
public static void main(String[] args){
Scanner s = new Scanner(System.in);
System.out.println("Enter the first number:");
Float a = s.nextFloat();
System.out.println("Enter the second number:");
Float b = s.nextFloat();
System.out.println("Sum = " + (a+b));
System.out.println("Difference = " + (a-b));
System.out.println("Product = " + (a*b));
}
}
When I run this program it will look like Enter First Number then i type the number, and then |Enter Second Number. "|" meaning where the blinking cursor is. When I type it'll show up underneath, but it could confuse the user so I was wondering what the solution could be.
It is an IDE problem, since nothing else is wrong with the code.

Instead of println(String) before each input, change it to print(String). So it would look something like this:
public class question3{
public static void main(String[] args){
Scanner s = new Scanner(System.in);
System.out.print("Enter the first number:");
Float a = s.nextFloat();
System.out.print("Enter the second number:");
Float b = s.nextFloat();
System.out.println("Sum = " + (a+b));
System.out.println("Difference = " + (a-b));
System.out.println("Product = " + (a*b));
}
}
Also, just a note, you should use proper/appropriate naming conventions for your variables. Like for your Scanner, you should call it reader or input; something which represents its function. The same idea goes for the rest of your variables. Also, class names start with a capital.
Here is what the finished result looks like:

System.out.println prints out string then a new line, so your input is being placed on a new line. Try making it read
System.out.print("Enter the first number:");
Float a = s.nextFloat();
System.out.println();
System.out.print("Enter the second number:");
Float b = s.nextFloat();
System.out.println();

This can save you some seconds, by typing few lines:
System.out.print("Enter the first number:");
Float a = s.nextFloat();
System.out.print("\nEnter the second number:");
Float b = s.nextFloat();
System.out.println("\nSum = " + (a+b)
+"\nDifference = " + (a-b)
+"\nProduct = " + (a*b));

Related

I can't print out letters in my code? If I remove the 0 from the circle brackets the code won't run

So this question is to let the user enter the product id, the name, how much in stock and how much is bought from the store. Everything in the code works just fine but when I enter the letters for the name, only the first letter shows up.
import java.util.Scanner;
public class ProductStock {
public static void main(String[] args) {
{
Scanner sc = new Scanner(System.in);
int id = sc.nextInt();
char name = sc.next().charAt(0);
int stock = sc.nextInt();
int sold = sc.nextInt();
sc.close();
System.out.print(id + " " + name + " " + stock + " " + sold);
}
}
}
A char in Java represents a single character. To have a sequence of characters, you should use a String. With using charAt(0) in your snippet, you only take the first character of the name you input, but you want to keep the whole String. By calling next() on the scanner, you actually get a String as the return value, so there is nothing more to do.
The fixed example should look like this:
public static void main(String[] args) {
Scanner sc = new Scanner(System.in);
int id = sc.nextInt();
String name = sc.next();
int stock = sc.nextInt();
int sold = sc.nextInt();
System.out.print(id + " " + name + " " + stock + " " + sold);
sc.close();
}
Just a heads up: once you should be aware that calling close() on a Scanner using System.in, you won't be able to use System.in for the rest of the runtime of your program.
in code char name was storing only the first character and because of char which can only store one character and print that one, so you have to use String for storing and displaying a complete text.
Scanner sc = new Scanner(System.in);
int id = sc.nextInt();
String name = sc.next();
int stock = sc.nextInt();
int sold = sc.nextInt();
sc.close();
System.out.print(id + " " + name + " " + stock + " " + sold);

using the command line prompt in Java Eclipse [duplicate]

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

Java: How to have two or more inputs in one line in the console

I am new to java and I wrote a Basic Input/Output program and in this program I want the user to provide 3 different inputs and be saved in 3 different variables, they are two int and one char.
public static void ForTesting() {
Scanner newScanTest = new Scanner(System.in);
ٍٍٍٍٍٍٍٍٍٍٍSystem.out.print("Please type two numbers: ");
int numberOne = newScanTest.nextInt();
int numberTwo = newScanTest.nextInt();
System.out.println("First Nr.: " + numberOne + " Second Nr.: " + numberTwo);
}
In the consol
What I get:
Please type two number: 4
5
First Nr.: 4 Second Nr.: 5
What I want:
Please type two number: 4 5
First Nr.: 4 Second Nr.: 5
(The bold numbers are the user input)
nextLine(), as the name suggests, reads a single line. If both numbers are contained on this single line, you should read both integers from that line. For example:
Scanner scanner = new Scanner(System.in);
String line = scanner.nextLine(); // reads the single input line from the console
String[] strings = line.split(" "); // splits the string wherever a space character is encountered, returns the result as a String[]
int first = Integer.parseInt(strings[0]);
int second = Integer.parseInt(strings[1]);
System.out.println("First number = " + first + ", second number = " + second + ".");
Note, this will fail if you don't provide 2 integers in the input.
Please try the bellow code, i try to convert integers to strings.
public static void ForTesting() {
Scanner newScanTest = new Scanner(System.in);
ٍٍٍٍٍٍٍٍٍٍٍSystem.out.print("Please type two numbers: ");
int number = newScanTest.nextInt();
int firstDigit = Integer.parseInt(Integer.toString(number).substring(0, 1));
int secondDigit = Integer.parseInt(Integer.toString(number).substring(1, 2));
System.out.println("First Nr.: " + firstDigit + " Second Nr.: " + secondDigit);
}
If you are very sure about input like (2+2) or (2-2) or (2*2) or (2/2)
below should work -
Scanner scanner = new Scanner(System.in);
Integer first = scanner.nextInt();
String operator = scanner.next();
Integer second = scanner.nextInt();
System.out.println("First number = " + first + ", second number = " + second + ".");

How to make a string equal a scanner and "if" statement?

I'm having a lot of trouble having the user input a string into the console and the specific string they typed equal an if statement. I want to input "SquareRoot" into the console and it go to the if statement, but when I type it in, nothing happens. What can I do to fix this? How do I make the user input equal to the string and the if statement? Is there something wrong with my "if" statement?
Scanner userInput = new Scanner(System.in);
String SquareRoot;
System.out.println("Type 'SquareRoot' - find the square root of (x)")
SquareRoot = userInput.next();
if(SquareRoot.equals("SquareRoot")) {
Scanner numInput = new Scanner(System.in);
System.out.println("Enter a number - ");
double sR;
sR = numInput.nextDouble();
System.out.println("The square root of " + sr + "is " + Math.sqrt(sR));
Your code was mostly correct:
You had some typos that would prevent your code compiling successfully. You
should consider using an IDE such as Eclipse, as it will highlight these
kinds of issues for you as you type.
You shouldn't create a 2nd Scanner object, reuse the existing one
Be sure to close your Scanner when done
Here's your corrected code:
public static void main(String[] args)
{
Scanner userInput = new Scanner(System.in);
String SquareRoot;
System.out.println("Type 'SquareRoot' - find the square root of (x)");
SquareRoot = userInput.next();
if (SquareRoot.equals("SquareRoot"))
{
// You shouldn't create a new Scanner
// Scanner numInput = new Scanner(System.in);
System.out.println("Enter a number - ");
double sR;
// Reuse the userInput Scanner
sR = userInput.nextDouble();
System.out.println("The square root of " + sR + " is " + Math.sqrt(sR));
}
// Be sure to close your Scanner when done
userInput.close();
}

tantamount of while(cin>>something){some task ;} in java?

I'm a beginner in java. But I know that we have some trick for take input since user add input in c++.
while(cin>>x)
{
g++;
}
I want use from this in java ,thanks.
Use either command line arguments or the Scanner class.
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);
}
}
You can put any loop in here. Search loops in java. And use this to perform what you're asking.

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