Develop Reference

For loop skipping index 0 when user's input and print in Java

For example, I entered a size of 3 Students. It skips index 0 in the console also in printing.
Please refer to this image, I have a sample size of 3 students and its output.
I don't have the slightest idea of why it skips index 0? Thanks for the help!
import java.util.Arrays;
import java.util.Scanner;
class string{
public static void main(String [] args){
Scanner console = new Scanner(System.in);
System.out.print("Enter Student Size: ");
int studentSize = console.nextInt();
String [] arrName = new String[studentSize];
for (int i=0; i<arrName.length; i++){
System.out.print("Enter student name: ");
String nameString = console.nextLine();
arrName[i] = nameString;
}
System.out.print(Arrays.toString(arrName));
//Closing Braces for Class and Main
}
}

The problem is with the console.nextInt(), this function only reads the int value.So In your code inside the loop console.nextLine() first time skip the getting input.just puting console.nextLine()afterconsole.nextInt()
you can solve the problem.
public static void main(String [] args){
Scanner console = new Scanner(System.in);
System.out.print("Enter Student Size: ");
int studentSize = console.nextInt();
console.nextLine();
String [] arrName = new String[studentSize];
for (int i=0; i<arrName.length; i++){
System.out.print("Enter student name: ");
String nameString = console.nextLine();
arrName[i] = nameString;
}
System.out.print(Arrays.toString(arrName));
//Closing Braces for Class and Main
}

The reason for this skip is due to the different behavior of the console.nextInt() and console.nextLine() as:
console.nextInt() reads the integer value entered, regardless of whether you hit the enter for new-line or not.
console.nextLine() reads the whole line, but as you previously hit thee enter when you give the size of Array.
3 was accepted as the size of the Array and when you hit enter it was accepted as the first value for you array which is a blank space or I can say it is referred as "".
Following are the two resolution for this:
Either put a console.nextLine() call after each console.nextInt() to consume rest of that line including newline
Or, even better, read the input through Scanner.nextLine and convert your input to the proper format you need. you may convert to an integer using int studentSize = Integer.parseInt(console.nextLine()) method. (Surround it with try-catch)

Java : Get only 1 line of integers from the console?

I recently picked up Java and I am having an issue with some console input.
Basically, I want to read in an array of ints from the console in a format like this :
1 2 3 4 5 6
I looked through some examples on the forums and decided to do this by using the scanner nextInt() method.
My code currently looks like this :
Scanner get = new Scanner(System.in);
List<Integer> elements = new ArrayList<>();
while (get.hasNextInt()) {
elements.add(get.nextInt());
}
The problem with this code is that the while loop doesn't stop when I hit "Enter" on the console.
Meaning that after I enter some numbers (1 3 5 7) and then hit enter, the program doesn't continue with execution, but instead waits for more integers. The only way it stops is if I enter a letter to the console.
I tried adding !get.hasNextLine() as a condition in my while loop, but this didn't help.
I would be very greatful, if anyone has an idea how can I fix this.

If you want to read only one line the simpliest answer may be the best :)
Scanner in = new Scanner(System.in);
String hString = in.nextLine();
String[] hArray = hString.split(" ");
Now, in array hArray you have all elements from input and you can call them like hArray[0]

You can read one line, and then use that to construct another Scanner. Something like,
if (get.hasNextLine()) {
String line = get.nextLine();
Scanner lineScanner = new Scanner(line);
while (lineScanner.hasNextInt()) {
elements.add(lineScanner.nextInt());
}
}
The Scanner(String) constructor (per the Javadoc) constructs a new Scanner that produces values scanned from the specified string.

Scanner get = new Scanner(System.in);
String arrSt = get.next();
StringTokinizer stTokens = new StringTokinizer(arrSt," ");
int [] myArr = new Int[stTokens.countTokens()];
int i =0;
while(stTokens.hasMoreTokens()){
myArr[i++]=Integer.parseInt(stTokens.nextToken());
}

java-8
You may use the following. User just has to enter each integer without pressing enter and press enter at the end.
Scanner get = new Scanner(System.in);
List<Integer> elements = Stream.of(get.nextLine().split(" "))
.map(Integer::parseInt)
.collect(Collectors.toList());

Accepting Multiple Inputs into an Array

So, I want to ask the user to input three values (all double) with only one prompt and have them stored in an array (I just started getting familiar with creating objects and arrays). This is what I've got so far:
import java.util.Scanner;
final int ARRAY_LIMIT=3;
Scanner input=new Scanner(System.in);
System.out.println("Enter the points and speed: ");
double entryVal=input.nextDouble();
double[] array=new double[ARRAY_LIMIT];
for(int entriesCounter=0; entriesCounter<array.length; entriesCounter++)
{
array[entriesCounter]=entryVal;
if(array[2]==0)
break;
}
So when I try to run it and I input something like 1.0, 1.0, 10.0, but I get all sorts of code spewed out. The point is to input all 3 values at the same time, eventually use these values in separate equations and terminate if the third number is 0. But I want to first be able to get these values to be stored correctly. I've tried to look at other previously answered questions but couldn't find something that helped me. I'm very new to Java so I'm still bumbling about. Any detailed help would be greatly appreciated.

The best way I think is just read the whole line to String :
public static void main(String[] args) {
Scanner input = new Scanner(System.in);
System.out.println("Enter the points and speed: ");
String entryVal = input.nextLine();
String[] stringArray = entryVal.split(" ");
double[] array = new double[stringArray.length];
for (int i = 0; i < array.length; i++) {
array[i] = Double.valueOf(stringArray[i]);
}
}
Note that each value has to be seperated only by space.

Scanner NoSuchElementException

I made a program that asks for 3 integers to output type of triangle. Everything runs and compiled successfully, however, it seems the part where it asks the user to see if they want to loop it again, the online compiler outputs the error:
Exception in thread "main" java.util.NoSuchElementException
at java.util.Scanner.throwFor(Scanner.java:838)
at java.util.Scanner.next(Scanner.java:1347)
at Assignment5.main(Assignment5.java:56)
import java.util.Scanner;
public class Assignment5 {
public static void main (String[]args)
{
for (int a = 0; a < Integer.MAX_VALUE; a++)
{
Scanner userInput = new Scanner(System.in);
Scanner answer = new Scanner(System.in);
int x,y,z;
System.out.println("Enter the sides of the triangle: ");
x = userInput.nextInt();
y = userInput.nextInt();
z = userInput.nextInt();
Tri isos = new Tri(x,y,z);
Tri equal = new Tri(x,y,z);
Tri scalene = new Tri(x,y,z);
// check the equilateral triangle
System.out.println(equal.toString() + " triangle:");
if (equal.is_isosceles())
System.out.println("\tIt is isosceles");
else
System.out.println("\tIt is not isosceles");
if (equal.is_equilateral())
System.out.println("\tIt is equilateral");
else
System.out.println("\tIt is not a equilateral");
if (equal.is_scalene())
System.out.println("\tIt is scalene");
else
System.out.println("\tIt is not scalene");
System.out.println("Would you like to enter values again? (y/n)" );
String input = answer.next(); //Exception is thrown from here
if (input.equals("y"))
{
System.out.println("ok");
}
else if(!input.equals("y"))
{
System.out.println("Ok, bye.");
break;
}
}
}
}

NoSuchElementException:
Thrown by the nextElement method of an Enumeration to indicate that
there are no more elements in the enumeration.
You're getting this exception because Scanner#next doesn't read the new line character, which is the character when you press enter (\n), so in the next for iteration, you're trying to read it, which causes the exception.
One possible solution is to add answer.nextLine() right after answer.next() in order to swallow this extra \n.
Example of your code:
Iteration (a) | input for scanner | Data for scanner
--------------+-----------------------+-------------------
0 | "Hello" (And enter) | Hello
1 | \n | PROBLEM!

to me it seems that answer.next() does not actually have any value assigned to it
usually int name = answer.next() name is assigned what ever answer is. What i mean is that name cant be assigned a value because answer.next() doesn't have one.
At least this is my understanding. The alternative is the get rid of answer.next and use the other scanner.
actually an edit to this.
a scanner reads from files or the console. You have one scanner already (userInput) the second scanner isn't actually doing anything as well as it being an actual scanner, it doesn't have anything to read.
get rid of answer as a scanner, replace is with an int, String, double and have
int answer = userInput.nextInt();
or
double answer = userInput.nextDouble();
or
String answer = userInput.nextLine();

As you said the code runs for you but doesn't when compiled and executed on an online compiler. The answer scanner is exhausted because it doesn't have any elements.
It's embarrassing but i once got the same error when compiling my code on an online compiler, it turned out i wasn't supplying input beforehand to the input section and was expecting the online compiler to ask for the input.
Since you are using two scanners to get input from console, try using the scanner userInput to take the input from a file instead. (It may vary for different online compilers, but there will be an option to provide input from file)

How do I make Java register a string input with spaces?

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.");
}
}