Why string inputs after integer input gets skipped in Java? [duplicate] - java

This question already exists:
Scanner issue when using nextLine after nextXXX [duplicate]
Closed 8 years ago.
I am trying to input values of certain string and integer variables in Java.
But if I am taking the input of string after the integer, in the console the string input is just skipped and moves to the next input.
Here is the code
String name1;
int id1,age1;
Scanner in = new Scanner(System.in);
//I can input name if input is before all integers
System.out.println("Enter id");
id1 = in.nextInt();
System.out.println("Enter name"); //Problem here, name input gets skipped
name1 = in.nextLine();
System.out.println("Enter age");
age1 = in.nextInt();

This is a common problem, and it happens because the nextInt method doesn't read the newline character of your input, so when you issue the command nextLine, the Scanner finds the newline character and gives you that as a line.
A workaround could be this one:
System.out.println("Enter id");
id1 = in.nextInt();
in.nextLine(); // skip the newline character
System.out.println("Enter name");
name1 = in.nextLine();
Another way would be to always use nextLine wrapped into a Integer.parseInt:
int id1;
try {
System.out.println("Enter id");
id1 = Integer.parseInt(input.nextLine());
} catch (NumberFormatException e) {
e.printStackTrace();
}
System.out.println("Enter name");
name1 = in.nextLine();
Why not just Scanner.next() ?
I would not use Scanner.next() because this will read only the next token and not the full line. For example the following code:
System.out("Enter name: ");
String name = in.next();
System.out(name);
will produce:
Enter name: Mad Scientist
Mad
It will not process Scientist because Mad is already a completed token per se.
So maybe this is the expected behavior for your application, but it has a different semantic from the code you posted in the question.

This is your updated working code.
package myPackage;
import java.util.Scanner;
public class test {
/**
* #param args
*/
public static void main(String[] args) {
String name1;
int id1,age1;
Scanner in = new Scanner(System.in);
//I can input name if input is before all integers
System.out.println("Enter id");
id1 = in.nextInt();
System.out.println("Enter name"); //Problem here, name input gets skipped
name1 = in.next();
System.out.println("Enter age");
age1 = in.nextInt();
}
}

May be you try this way..
Instead of this code
System.out.println("Enter name"); //Problem here, name input gets skipped
name1 = in.nextLine();
try this
System.out.println("Enter name");
name1 = in.next();

Related

How do I get the name input to read whitespaces

So im having trouble getting the first read input to read all inputs on the line. But for some reason it doesnt take into consideration of whitespaces. in fact, it considers the whitespace from the print information as part of the whitespace. only name has this problem. ID does not have this problem. I would just like to know how to fix this one problem since it is giving me the most trouble.
public class Project2 {
public static Scanner sc = new Scanner (System.in);
public static void main(String[] args) {
PersonList PersonList = new PersonList();
System.out.print("Welcome to my personal Management Program\n\n");
System.out.print("\nChoose one of the following options: \n\n");
//print out options for user
for (;;) {
int input = 0;
System.out.print("1- Enter the information of a faculty\n");
System.out.print("2- Enter the information of a student\n");
System.out.print("3- Print tuition person for a student\n");
System.out.print("4- Print faculty information\n");
System.out.print("5- Enter the information of a staff member\n");
System.out.print("6- Print the information of a staff member\n");
System.out.print("7-Exit the program\n\n");
System.out.print("\tEnter a selection: ");
input = sc.nextInt();
if (input == 1) {
faculty f = new faculty();
System.out.print("Enter the faculty info:\n");
System.out.print("\tName of Faculty: ");
f.name = sc.nextLine();
System.out.print("\tID: ");
f.ID = sc.nextLine();
String rank, department;
for(;;) {
System.out.print("\n\tRank: ");
rank = sc.nextLine();
if (rank.equalsIgnoreCase("professor") || rank.equalsIgnoreCase("adjunct")) {
f.rank = rank;
break;
}
else {
System.out.print("\"" + rank +"\" is invalid");
}
}
for(;;) {
System.out.print("\tDepartment: ");
department = sc.nextLine();
if (department.equalsIgnoreCase("mathematics") || department.equalsIgnoreCase("engineering") || department.equalsIgnoreCase("sciences")) {
f.Department = department;
break;
}
else {
System.out.print("\"" + department +"\" is invalid");
}
}
System.out.println("Faculty added!");
PersonList.addPerson(f);
}
When you read the selection input = sc.nextInt() it returns the next integer it finds, stopping right after that. It returns the choice typed but it does not read the newline. If you typed 1 (space one space) it would skip over the leading space then read the 1 and stop; it would not read the following space or the newline.
So when you get to the faculty name f.name = sc.nextLine() it reads whatever was remaining from the input = line, which is probably just a newline, and returns that for the name.
Now when you get to f.ID = sc.nextLine() the input buffer is clear so it reads ID as you expect it to.
A simple solution is to just finish reading the line by adding a nextLine() after reading the selection:
System.out.print("\tEnter a selection: ");
input = sc.nextInt();
sc.nextLine(); // Read the remainder of the line and throw it away
Read the javadoc for java.util.Scanner and for nextInt() and nextInt(radix)

Java giving error when trying to use multiple inputs in while loop [duplicate]

This question already has answers here:
How do I compare strings in Java?
(23 answers)
java.util.NoSuchElementException - Scanner reading user input
(5 answers)
Closed 1 year ago.
The following java code gives the following error: Exception in thread "main" java.util.NoSuchElementException: No line found.
boolean running = true;
while (running) {
Scanner sc2 = new Scanner(System.in);
System.out.println("Enter an item to order:");
String name = sc2.nextLine();
System.out.println("Enter the price:");
String price = sc2.nextLine();
System.out.println("Enter the quantity:");
String quantity = sc2.nextLine();
orderItems.add(name);
orderItems.add(price);
orderItems.add(quantity);
orderItems.add(";");
System.out.println("Would you like to add another item?: (y/n)");
if (sc2.nextLine() != "y") {
running = false;
}
}
Any suggestions? I've gotten a similar error before, which was caused by using a new scanner for each input instead of using the same one for each input, but this error seems to be caused by a different issue. Thanks in advance.
A similar issue is here, but you create a new instance every iteration of the while loop.
Create a single Scanner in the first line of the method.
Declare your scanner object only once. Your while loop was creating multiple scanners (see Multiple scanner objects)
You were not performing the String comparison in the right way (see How do I compare strings in Java?)
boolean running = true;
Scanner sc2 = new Scanner(System.in);
while (running) {
System.out.println("Enter an item to order:");
String name = sc2.nextLine();
System.out.println("Enter the price:");
String price = sc2.nextLine();
System.out.println("Enter the quantity:");
String quantity = sc2.nextLine();
orderItems.add(name);
orderItems.add(price);
orderItems.add(quantity);
orderItems.add(";");
System.out.println("Would you like to add another item?: (y/n)");
if (sc2.nextLine().equals("y")) {
running = false;
}
}
Create only a single instance for the Scanner by providing it before the while loop.
And also a suggestion, use !sc2.nextLine().equals("y") instead of sc2.nextLine() != "y"
boolean running = true;
Scanner sc2 = new Scanner(System.in);
while (running) {
System.out.println("Enter an item to order:");
String name = sc2.nextLine();
System.out.println("Enter the price:");
String price = sc2.nextLine();
System.out.println("Enter the quantity:");
String quantity = sc2.nextLine();
orderItems.add(name);
orderItems.add(price);
orderItems.add(quantity);
orderItems.add(";");
System.out.println("Would you like to add another item?: (y/n)");
if (!sc2.nextLine().equals("y")) {
running = false;
}
}

How can I get multiple user inputs and save their value? [duplicate]

How could I read input from the console using the Scanner class? Something like this:
System.out.println("Enter your username: ");
Scanner = input(); // Or something like this, I don't know the code
Basically, all I want is have the scanner read an input for the username, and assign the input to a String variable.
A simple example to illustrate how java.util.Scanner works would be reading a single integer from System.in. It's really quite simple.
Scanner sc = new Scanner(System.in);
int i = sc.nextInt();
To retrieve a username I would probably use sc.nextLine().
System.out.println("Enter your username: ");
Scanner scanner = new Scanner(System.in);
String username = scanner.nextLine();
System.out.println("Your username is " + username);
You could also use next(String pattern) if you want more control over the input, or just validate the username variable.
You'll find more information on their implementation in the API Documentation for java.util.Scanner
Scanner scan = new Scanner(System.in);
String myLine = scan.nextLine();
Reading Data From The Console
BufferedReader is synchronized, so read operations on a BufferedReader can be safely done from multiple threads. The buffer size may be specified, or the default size(8192) may be used. The default is large enough for most purposes.
readLine() « just reads data line by line from the stream or source. A line is considered to be terminated by any one these: \n, \r (or) \r\n
Scanner breaks its input into tokens using a delimiter pattern, which by default matches whitespace(\s) and it is recognised by Character.isWhitespace.
« Until the user enters data, the scanning operation may block, waiting for input.
« Use Scanner(BUFFER_SIZE = 1024) if you want to parse a specific type of token from a stream.
« A scanner however is not thread safe. It has to be externally synchronized.
next() « Finds and returns the next complete token from this scanner.
nextInt() « Scans the next token of the input as an int.
Code
String name = null;
int number;
java.io.BufferedReader in = new BufferedReader(new InputStreamReader(System.in));
name = in.readLine(); // If the user has not entered anything, assume the default value.
number = Integer.parseInt(in.readLine()); // It reads only String,and we need to parse it.
System.out.println("Name " + name + "\t number " + number);
java.util.Scanner sc = new Scanner(System.in).useDelimiter("\\s");
name = sc.next(); // It will not leave until the user enters data.
number = sc.nextInt(); // We can read specific data.
System.out.println("Name " + name + "\t number " + number);
// The Console class is not working in the IDE as expected.
java.io.Console cnsl = System.console();
if (cnsl != null) {
// Read a line from the user input. The cursor blinks after the specified input.
name = cnsl.readLine("Name: ");
System.out.println("Name entered: " + name);
}
Inputs and outputs of Stream
Reader Input: Output:
Yash 777 Line1 = Yash 777
7 Line1 = 7
Scanner Input: Output:
Yash 777 token1 = Yash
token2 = 777
There is problem with the input.nextInt() method - it only reads the int value.
So when reading the next line using input.nextLine() you receive "\n", i.e. the Enter key. So to skip this you have to add the input.nextLine().
Try it like that:
System.out.print("Insert a number: ");
int number = input.nextInt();
input.nextLine(); // This line you have to add (it consumes the \n character)
System.out.print("Text1: ");
String text1 = input.nextLine();
System.out.print("Text2: ");
String text2 = input.nextLine();
There are several ways to get input from the user. Here in this program we will take the Scanner class to achieve the task. This Scanner class comes under java.util, hence the first line of the program is import java.util.Scanner; which allows the user to read values of various types in Java. The import statement line should have to be in the first line the java program, and we proceed further for code.
in.nextInt(); // It just reads the numbers
in.nextLine(); // It get the String which user enters
To access methods in the Scanner class create a new scanner object as "in". Now we use one of its method, that is "next". The "next" method gets the string of text that a user enters on the keyboard.
Here I'm using in.nextLine(); to get the String which the user enters.
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 ScannerDemo {
public static void main(String[] arguments){
Scanner input = new Scanner(System.in);
String username;
double age;
String gender;
String marital_status;
int telephone_number;
// Allows a person to enter his/her name
Scanner one = new Scanner(System.in);
System.out.println("Enter Name:" );
username = one.next();
System.out.println("Name accepted " + username);
// Allows a person to enter his/her age
Scanner two = new Scanner(System.in);
System.out.println("Enter Age:" );
age = two.nextDouble();
System.out.println("Age accepted " + age);
// Allows a person to enter his/her gender
Scanner three = new Scanner(System.in);
System.out.println("Enter Gender:" );
gender = three.next();
System.out.println("Gender accepted " + gender);
// Allows a person to enter his/her marital status
Scanner four = new Scanner(System.in);
System.out.println("Enter Marital status:" );
marital_status = four.next();
System.out.println("Marital status accepted " + marital_status);
// Allows a person to enter his/her telephone number
Scanner five = new Scanner(System.in);
System.out.println("Enter Telephone number:" );
telephone_number = five.nextInt();
System.out.println("Telephone number accepted " + telephone_number);
}
}
You can make a simple program to ask for the user's name and print whatever the reply use inputs.
Or ask the user to enter two numbers and you can add, multiply, subtract, or divide those numbers and print the answers for user inputs just like the behavior of a calculator.
So there you need the 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 the same way for the rest. Don't forget the import statement at the top of your code.
A simple example:
import java.util.Scanner;
public class Example
{
public static void main(String[] args)
{
int number1, number2, sum;
Scanner input = new Scanner(System.in);
System.out.println("Enter First multiple");
number1 = input.nextInt();
System.out.println("Enter second multiple");
number2 = input.nextInt();
sum = number1 * number2;
System.out.printf("The product of both number is %d", sum);
}
}
When the user enters his/her username, check for valid entry also.
java.util.Scanner input = new java.util.Scanner(System.in);
String userName;
final int validLength = 6; // This is the valid length of an user name
System.out.print("Please enter the username: ");
userName = input.nextLine();
while(userName.length() < validLength) {
// If the user enters less than validLength characters
// ask for entering again
System.out.println(
"\nUsername needs to be " + validLength + " character long");
System.out.print("\nPlease enter the username again: ");
userName = input.nextLine();
}
System.out.println("Username is: " + userName);
To read input:
Scanner scanner = new Scanner(System.in);
String input = scanner.nextLine();
To read input when you call a method with some arguments/parameters:
if (args.length != 2) {
System.err.println("Utilizare: java Grep <fisier> <cuvant>");
System.exit(1);
}
try {
grep(args[0], args[1]);
} catch (IOException e) {
System.out.println(e.getMessage());
}
import java.util.*;
class Ss
{
int id, salary;
String name;
void Ss(int id, int salary, String name)
{
this.id = id;
this.salary = salary;
this.name = name;
}
void display()
{
System.out.println("The id of employee:" + id);
System.out.println("The name of employye:" + name);
System.out.println("The salary of employee:" + salary);
}
}
class employee
{
public static void main(String args[])
{
Scanner sc = new Scanner(System.in);
Ss s = new Ss(sc.nextInt(), sc.nextInt(), sc.nextLine());
s.display();
}
}
Here is the complete class which performs the required operation:
import java.util.Scanner;
public class App {
public static void main(String[] args) {
Scanner input = new Scanner(System.in);
final int valid = 6;
Scanner one = new Scanner(System.in);
System.out.println("Enter your username: ");
String s = one.nextLine();
if (s.length() < valid) {
System.out.println("Enter a valid username");
System.out.println(
"User name must contain " + valid + " characters");
System.out.println("Enter again: ");
s = one.nextLine();
}
System.out.println("Username accepted: " + s);
Scanner two = new Scanner(System.in);
System.out.println("Enter your age: ");
int a = two.nextInt();
System.out.println("Age accepted: " + a);
Scanner three = new Scanner(System.in);
System.out.println("Enter your sex: ");
String sex = three.nextLine();
System.out.println("Sex accepted: " + sex);
}
}
There is a simple way to read from the console.
Please find the below code:
import java.util.Scanner;
public class ScannerDemo {
public static void main(String[] args) {
Scanner sc = new Scanner(System.in);
// Reading of Integer
int number = sc.nextInt();
// Reading of String
String str = sc.next();
}
}
For a detailed understanding, please refer to the below documents.
Doc
Now let's talk about the detailed understanding of the Scanner class working:
public Scanner(InputStream source) {
this(new InputStreamReader(source), WHITESPACE_PATTERN);
}
This is the constructor for creating the Scanner instance.
Here we are passing the InputStream reference which is nothing but a System.In. Here it opens the InputStream Pipe for console input.
public InputStreamReader(InputStream in) {
super(in);
try {
sd = StreamDecoder.forInputStreamReader(in, this, (String)null); // ## Check lock object
}
catch (UnsupportedEncodingException e) {
// The default encoding should always be available
throw new Error(e);
}
}
By passing the System.in this code will opens the socket for reading from console.
You can flow this code:
Scanner obj= new Scanner(System.in);
String s = obj.nextLine();
You can use the Scanner class in Java
Scanner scan = new Scanner(System.in);
String s = scan.nextLine();
System.out.println("String: " + s);
import java.util.Scanner; // Import the Scanner class
class Main { // Main is the class name
public static void main(String[] args) {
Scanner myObj = new Scanner(System.in); // Create a Scanner object
System.out.println("Enter username");
String userName = myObj.nextLine(); // Read user input
System.out.println("Username is: " + userName); // Output user input
}
}
you have wrote
Scanner = input()
this is wrong method, you have to make an integer or a string, i would like to prefer string, and then give a string any name that can be i that can be n or anything else, remember that you are giving name to username you can also give name username also, and the code is
String username = sc.nextline();
System.our.println("the username is" + username);
I hope you understand now

Ensuring user input is a single integer while using Scanner [duplicate]

This question already has answers here:
How do I ensure that Scanner hasNextInt() asks for new input?
(2 answers)
Closed 6 years ago.
Desired outcome:
Accepts user input
Makes sure user inputs only 1 integer value at a time
Stores that integer in a variable
I tried to achieve this by doing the following:
Store user input in variable
Count number of tokens in variable
If there's not one token, reject the input
If the input is not of data type int, reject the input
Code:
Scanner scan = new Scanner(System.in);
System.out.println("Enter an integer:");
String myString = scan.nextLine();
int tokens = new StringTokenizer(myString, " ").countTokens();
while (tokens != 1 && !scan.hasNextInt()) {
System.out.println("Enter a single integer");
myString = scanner.nextLine();
tokens = new StringTokenizer(myString, " ").countTokens();
}
int number = scanner.nextInt();
System.out.println(number);
This code is full of holes. The output is inconsistent and undesired. It typically ends by throwing a java.util.InputMismatchException error, indicating the value it's trying to store isn't an int. I've experienced this error occur after one loop and after multiple loops, even with the same type and quantity of input (e.g. 2 strings).
Should I keep going with this code, or should I try to approach this problem from a different angle?
I've modified your program a little bit. My approach was to accept a single line of input. If the input contains more than one token, ask the user to re-enter input. If there is only one input, check if the input is an integer, if not, as the user to again provide input.
Seems to work for me:
Scanner scanner = new Scanner(System.in);
String myString;
int tokens;
int number;
do {
System.out.println("Enter a single integer");
myString = scanner.nextLine();
tokens = new StringTokenizer(myString, " ").countTokens();
try {
number = Integer.parseInt(myString);
} catch(NumberFormatException e) {
tokens = 0;
number = -1;
}
}while (tokens != 1);
scanner.close();
System.out.println(number);
Update: Alternate approach without using StringTokenizer
Scanner scanner = new Scanner(System.in);
String myString;
boolean validInput;
int number;
do {
System.out.println("Enter a single integer");
myString = scanner.nextLine();
try {
number = Integer.parseInt(myString);
validInput = true;
} catch(NumberFormatException e) {
validInput = false;
number = -1;
}
}while (validInput == false);
scanner.close();
System.out.println(number);
Update 2: Another approach using regular expressions to validate input before accepting it.
The Scanner allows us to use a regular expression to match the input. If the input matches the pattern, you can use it to accept the input. Otherwise, discard it and ask user to provide input again.
Here's the code:
Scanner scanner = new Scanner(System.in);
System.out.println("Enter a single integer");
String integerPattern = "[+-]?\\d+$"; // positive or negative and digits only
while(scanner.hasNext(integerPattern) == false) {
String x = scanner.nextLine();// capture and discard input.
System.out.println("Enter a single integer. '" + x + "' is an invalid input.");
}
int number = scanner.nextInt(); // capture input only if it matches pattern.
scanner.close();
System.out.println("number: " + number);
Hope this helps!

Scanner input validation in while loop

I've got to show Scanner inputs in a while loop: the user has to insert inputs until he writes "quit". So, I've got to validate each input to check if he writes "quit". How can I do that?
while (!scanner.nextLine().equals("quit")) {
System.out.println("Insert question code:");
String question = scanner.nextLine();
System.out.println("Insert answer code:");
String answer = scanner.nextLine();
service.storeResults(question, answer); // This stores given inputs on db
}
This doesn't work. How can I validate each user input?
The problem is that nextLine() "Advances this scanner past the current line". So when you call nextLine() in the while condition, and don't save the return value, you've lost that line of the user's input. The call to nextLine() on line 3 returns a different line.
You can try something like this
Scanner scanner=new Scanner(System.in);
while (true) {
System.out.println("Insert question code:");
String question = scanner.nextLine();
if(question.equals("quit")){
break;
}
System.out.println("Insert answer code:");
String answer = scanner.nextLine();
if(answer.equals("quit")){
break;
}
service.storeResults(question, answer);
}
Try:
while (scanner.hasNextLine()) {
System.out.println("Insert question code:");
String question = scanner.nextLine();
if(question.equals("quit")){
break;
}
System.out.println("Insert answer code:");
String answer = scanner.nextLine();
service.storeResults(question, answer); // This stores given inputs on db
}
always check if scanner.nextLine is not "quit"
while (!scanner.nextLine().equals("quit")) {
System.out.println("Insert question code:");
String question = scanner.nextLine();
if(question.equals("quit"))
break;
System.out.println("Insert answer code:");
String answer = scanner.nextLine();
if(answer.equals("quit"))
break;
service.storeResults(question, answer); // This stores given inputs on db
}

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