I am currently using a Scanner to record the user input which is a String and print it out. If the user input is a single name such as Alan, it works fine. If I enter a name with spacing such as Alan Smith, it returns an error saying InputMisMatchException.
I read around similar cases here and they advised to use nextLine() instead of next(). It made sense but that doesn't work for me either. When I use a nextLine(), it immediately skips the step where I enter the name and goes back to the starting of the loop asking me to input choice again. Please advice how I can correct this. Thank you.
import java.io.IOException;
import java.util.Scanner;
public class ScannerTest {
static String name;
static Scanner in = new Scanner(System.in);
static int choice;
public static void main(String[] args) {
while(choice != 5){
System.out.print("\nEnter Choice :> ");
choice = in.nextInt();
if(choice == 1){
try{
printName();
}
catch(IOException e){
System.out.println("IO Exception");
}
}
}
}
private static void printName()throws IOException{
System.out.print("\nEnter name :> ");
name = in.next();
//name = in.nextLine();
if (name != null){
System.out.println(name);
}
}
}
Try this instead: add name = in.nextLine(); after choice = in.nextInt();.
Then try replacing name = in.next(); with name = in.nextLine();
Explanation: After the scanner calls nextInt() it gets the first value and leaves the rest of the string to the \n. We then consume the rest of the string with nextLine().
The second nextLine() is then used to get your string parameters.
The problem is easy: when you prompt the user to enter his/her choice, the choice will be an int followed by a new line (the user will press enter). When you use in.nextInt() to retrieve the choice, only the number will be consumed, the new line will still be in the buffer, and, so, when you call in.nextLine(), you will get whatever is between the number and the new line (usually nothing).
What you have to do, is call in.nextLine() just after reading the number to empty the buffer:
choice = in.nextInt();
if (in.hasNextLine())
in.nextLine();
before to call name = in.next(); do this in = new Scanner(System.in);
the object need rebuild itself because already has value.
good luck
Related
This question already exists:
Scanner issue when using nextLine after nextXXX [duplicate]
Closed 8 years ago.
Just one question: why I must type answer = in.nextLine(); twice? If this line is single the program doesn't work as expected. Without second line the program doesn't ask you to enter a string.
public class Main {
public static void main(String[] args) {
Scanner in = new Scanner(System.in);
String answer = "Yes";
while (answer.equals("Yes")) {
System.out.println("Enter name and rating:");
String name = in.nextLine();
int rating = 0;
if (in.hasNextInt()) {
rating = in.nextInt();
} else {
System.out.println("Error. Exit.");
return;
}
System.out.println("Name: " + name);
System.out.println("Rating: " + rating);
ECTS ects = new ECTS();
rating = ects.checkRating(rating);
System.out.println("Enter \"Yes\" to continue: ");
answer = in.nextLine();
answer = in.nextLine();
}
System.out.println("Bye!");
in.close();
}
}
The Scanner-Object has an internal cache.
You start the scann for nextInt().
You press the key 1
You press the key 2
You press return
Now, the internal Cache has 3 characters and the scanner sees the third character(return) is not a number, so the nextInt() will only return the integer from the 1st and 2nd character (1,2=12).
nextInt() returns 12.
Unfortunately the return is still part of the Scanner's cache.
You call nextLine(), but the Method scans its cache for a newline-marker that has been kept in the cache from before the nextInt() call returned.
The nextLine() returns a 0-length-string.
The next nextLine() has an empty cache! It will wait until the cache has been filled with the next newline-marker.
reset()
There is a more elegant way to clear the cache instead of using nextLine():
in.reset();
Because you are using nextInt() this method only grabs the next int it doesn't consume the \n character so the next time you do nextLine() it finishes consuming that line then moves to the next line.
import java.util.Scanner;
class Tutorial {
public static void main (String args[]){
System.out.println("Who goes there?");
Scanner name = new Scanner(System.in); ## I am asking for input form user but it does not take imput
if (name.equals("me") || name.equals("Me") ){
System.out.println("Well, good for you smartass.");
}else System.out.println("Well good meet");
}
}
Why does the program run the else and not ask for my input?
You should read your input by using scanner.nextLine():
Scanner scanner = new Scanner(System.in);
String name = scanner.nextLine();
if (name.equals("me") || name.equals("Me"))
{
System.out.println("Well, good for you smartass.");
} else {
System.out.println("Well good meet");
}
scanner.close();
You merely created a Scanner but did not tell it to read something from the standard input. You can do that by calling scanner.next() to read a token scanner.nextLine() to read a line, etc. As well you are comparing a Scanner to a String in the if-statement.
import java.util.Scanner;
class Tutorial {
public static void main (String args[]){
System.out.println("Who goes there?");
Scanner s = new Scanner(System.in);
String name = s.next(); // get the token
if (name.equals("me") || name.equals("Me") ){
System.out.println("Well, good for you smartass.");
} else System.out.println("Well good meet");
}
}
You've only created an instance of the Scanner object. You need to invoke a method such as Scanner#nextLine() to read input and then compare the read value to "me" or "Me".
Example:
Scanner name = new Scanner(System.in);
String input = name.nextLine();
if (...) // Compare input to something here.
You might want to use String#equalsIgnoreCase for case-insensitive matching too.
I'm writting a JAVA Class to validate input data, especifically integer numbers.
The class I develop is running fine but when I press more than one time enter and then a char type, it display several times " Error!! Invalid number. Try again. " and I would like to avoid it.
I have use nextLine() method but it doesn't seems to correct it.
Here is the Class:
package chapter07.libro;
import java.util.Scanner;
public class Validator_integer
{
public static int getInt (Scanner scanner, String promt)
{
int numberInteger = 0;
boolean isValid = false;
System.out.println(promt);
while(isValid == false)
{
if(scanner.hasNextInt())
{
numberInteger= scanner.nextInt();
isValid = true;
}//if
else
{
System.out.println("Error!! Invalid number. Try again.");
}//else
scanner.nextLine();
}//while
return numberInteger;
}//getInt
}//Validator_integer
and next is the app to use the class:
package chapter.prueba;
import java.util.Scanner;
import chapter07.libro.Validator_integer;
public class Test_Validator_Integer
{
public static void main(String[] args) {
Scanner sc = new Scanner (System.in);
String choice = "y";
while(choice.equalsIgnoreCase("y"))
{
int number = Validator_integer.getInt(sc, "Enter integer number: ");
System.out.println(number);
System.out.println("Continue (y/n): ");
choice = sc.next();
}//while
}//main
}//Test_Validator_Integer
The results I get are next:
Enter integer number:
2
2
Continue (y/n):
y
Enter integer number:
(Here I press several time enter)
xx
Error!! Invalid number. Try again.
Error!! Invalid number. Try again.
Error!! Invalid number. Try again.
Error!! Invalid number. Try again.
2
2
Continue (y/n):
n
So the part of (Error!! Invalid number. Try again.) displayed several times, is the one I would like to avoid.
Does any one know how to fix it???
Thanks in advance!!!
Before you read from System.in, make sure to "clear" it's contents to get rid of buffered/queued-up input. If there are n characters queued-up, then skip that many chars and you'll need to enter something new.
int n = System.in.available();
System.in.skip(n);
http://docs.oracle.com/javase/7/docs/api/java/io/InputStream.html
import java.util.Scanner;
public class ClassList {
static final int SIZE = 10;
public static void main(String Args[]) {
Student[] students = new Student[SIZE];
Scanner s = new Scanner(System.in);
String name, id;
float gpa;
do {
System.out.println("Enter a Student");
name = s.nextLine();
System.out.println("Enter a ID");
id = s.nextLine();
System.out.println("Enter a GPA");
gpa = s.nextFloat();
} while(!name.equals("quit"));
}
}
basically, what happened when you run this program is that it will run one fine, then on the 2nd iteration, it will ask for a student name. but it will just skip over where the user should input it and go onto "get ID". How do i stop that.
In the first iteration, nextFloat() doesn't consume any new line characters. In the second iteration, the nextLine() method will consume the newline characters left from the first iteration. As a result, it appears that the nextLine() method in the second iteration skips over user input, while it's just consuming the newline characters left from the previous iteration.
One way to fix this problem is to call nextLine() after nextFloat() to consume the leftover new line character.
You can place
s.nextLine();
at the end of the loop to consume the newline character. Currently Scanner.getFloat() is passing the newline character through to your statement:
name = s.nextLine();
which is not blocking as it has been supplied with this character.
Here is my code:
public static void main(String[] args) {
Scanner in = new Scanner(System.in);
String question;
question = in.next();
if (question.equalsIgnoreCase("howdoyoulikeschool?") )
/* it seems strings do not allow for spaces */
System.out.println("CLOSED!!");
else
System.out.println("Que?");
When I try to write "how do you like school?" the answer is always "Que?" but it works fine as "howdoyoulikeschool?"
Should I define the input as something other than String?
in.next() will return space-delimited strings. Use in.nextLine() if you want to read the whole line. After reading the string, use question = question.replaceAll("\\s","") to remove spaces.
Since it's a long time and people keep suggesting to use Scanner#nextLine(), there's another chance that Scanner can take spaces included in input.
Class Scanner
A Scanner breaks its input into tokens using a delimiter pattern, which by default matches whitespace.
You can use Scanner#useDelimiter() to change the delimiter of Scanner to another pattern such as a line feed or something else.
Scanner in = new Scanner(System.in);
in.useDelimiter("\n"); // use LF as the delimiter
String question;
System.out.println("Please input question:");
question = in.next();
// TODO do something with your input such as removing spaces...
if (question.equalsIgnoreCase("howdoyoulikeschool?") )
/* it seems strings do not allow for spaces */
System.out.println("CLOSED!!");
else
System.out.println("Que?");
I found a very weird thing in Java today, so it goes like -
If you are inputting more than 1 thing from the user, say
Scanner sc = new Scanner(System.in);
int i = sc.nextInt();
double d = sc.nextDouble();
String s = sc.nextLine();
System.out.println(i);
System.out.println(d);
System.out.println(s);
So, it might look like if we run this program, it will ask for these 3 inputs and say our input values are 10, 2.5, "Welcome to java"
The program should print these 3 values as it is, as we have used nextLine() so it shouldn't ignore the text after spaces that we have entered in our variable s
But, the output that you will get is -
10
2.5
And that's it, it doesn't even prompt for the String input.
Now I was reading about it and to be very honest there are still some gaps in my understanding, all I could figure out was after taking the int input and then the double input when we press enter, it considers that as the prompt and ignores the nextLine().
So changing my code to something like this -
Scanner sc = new Scanner(System.in);
int i = sc.nextInt();
double d = sc.nextDouble();
sc.nextLine();
String s = sc.nextLine();
System.out.println(i);
System.out.println(d);
System.out.println(s);
does the job perfectly, so it is related to something like "\n" being stored in the keyboard buffer in the previous example which we can bypass using this.
Please if anybody knows help me with an explanation for this.
Instead of
Scanner in = new Scanner(System.in);
String question;
question = in.next();
Type in
Scanner in = new Scanner(System.in);
String question;
question = in.nextLine();
This should be able to take spaces as input.
This is a sample implementation of taking input in java, I added some fault tolerance on just the salary field to show how it's done. If you notice, you also have to close the input stream .. Enjoy :-)
/* AUTHOR: MIKEQ
* DATE: 04/29/2016
* DESCRIPTION: Take input with Java using Scanner Class, Wow, stunningly fun. :-)
* Added example of error check on salary input.
* TESTED: Eclipse Java EE IDE for Web Developers. Version: Mars.2 Release (4.5.2)
*/
import java.util.Scanner;
public class userInputVersion1 {
public static void main(String[] args) {
System.out.println("** Taking in User input **");
Scanner input = new Scanner(System.in);
System.out.println("Please enter your name : ");
String s = input.nextLine(); // getting a String value (full line)
//String s = input.next(); // getting a String value (issues with spaces in line)
System.out.println("Please enter your age : ");
int i = input.nextInt(); // getting an integer
// version with Fault Tolerance:
System.out.println("Please enter your salary : ");
while (!input.hasNextDouble())
{
System.out.println("Invalid input\n Type the double-type number:");
input.next();
}
double d = input.nextDouble(); // need to check the data type?
System.out.printf("\nName %s" +
"\nAge: %d" +
"\nSalary: %f\n", s, i, d);
// close the scanner
System.out.println("Closing Scanner...");
input.close();
System.out.println("Scanner Closed.");
}
}