New programmer here. This is probably a really basic question, but it's stumping me nevertheless.
What I'm trying to do is write a method that supplies only one integer input so I can use that input in my main program without having to mess around with non-integer inputs. However, even writing the method to do that in its own method seems to be problematic.
public static int goodInput () {
Scanner input = new Scanner (System.in); //construct scanner
boolean test = input.hasNextInt(); //set a sentinel value
while (test == false) { //enter a loop until I actually get an integer
System.out.println("Integers only please"); //tell user to give me an integer
test = input.hasNextInt(); //get new input, see if it's an integer
}
int finalInput = input.nextInt(); //once i have an integer, set it to a variable
input.close(); //closing scanner
return finalInput; //return my integer so I don't have to mess around with hasNextInt over there
}
This seems to be broken in multiple levels, but I'm not really sure why.
If I enter an integer value like 0 or 1 when I'm first asked for input, it should skip the loop entirely. But, instead, it enters the loop, and prints "Integers only please". Even worse, it doesn't actually ask for input while I'm in there, and just prints that line repeatedly.
I understand the latter problem is probably due to token issues, but I'm not necessarily sure how to solve them; closing and then reopening the scanner gets Eclipse to bug me over "duplicate objects", simply assigning the old input to a garbage String variable that is never used tells me that "No line was found" at runtime, and I'm not experienced enough to think of other ways to get new input.
Even once that's solved, I need to find some way to avoid entering the loop in the case of having an integer. I don't really understand why integer inputs inter the loop to begin with, so I'm not sure how this would be possible.
Please help? Sorry if this is an old question; tried looking at past questions but none of them seem to have the same problem that I have.
You were close: this works fine for me:
Scanner input = new Scanner(System.in); //construct scanner
while(!input.hasNextInt()) {
input.next(); // next input is not an int, so consume it and move on
}
int finalInput = input.nextInt();
input.close(); //closing scanner
System.out.println("finalInput: " + finalInput);
By calling input.next() in your while loop, you consume the non-integer content and try again, and again, until the next input is an int.
//while (test == false) { // Line #1
while (!test) { /* Better notation */ // Line #2
System.out.println("Integers only please"); // Line #3
test = input.hasNextInt(); // Line #4
} // Line #5
The problem is that in line #4 above, input.hasNextInt() only tests if an integer is inputted, and does not ask for a new integer. If the user inputs something other than an integer, hasNextInt() returns false and you cannot ask for nextInt(), because then an InputMismatchException is thrown, since the Scanner is still expecting an integer.
You must use next() instead of nextInt():
while (!input.hasNextInt()) {
input.next();
// That will 'consume' the result, but doesn't use it.
}
int result = input.nextInt();
input.close();
return result;
Related
I'm trying to put user input into an array but the hasNextInt() method will not return false and stop the input.
public static void main (String[] args) {
Scanner in = new Scanner(System.in);
int target = in.nextInt();
while(in.hasNextInt()) {
weights.insert(in.nextInt());
}
recKnapSack(target, 0);
}
Scanner.hasNextInt() will return false when it encounters a non-integer character in its buffer.
However, it may strip out whitespace when it's reading prompts. So Space+Enter or Enter will most likely not stop the loop. But any other character will.
Since you'd like to input any number of ints, you must instruct the user on what to type when they're done. In fact, if you're writing a console application, it's a good idea to always explain WHY a program is waiting for input.
Any non-integer will stop the loop condition. In this case the syntax will work as-is, the user just needs some instruction:
System.out.println("Please enter the target");
int target = in.nextInt();
System.out.println("Enter weights. Type 'X' to stop");
while(in.hasNextInt()) {
I want to read user input like: 11 12 13 14 15 16
Scanner sc = new Scanner(System.in);
while(sc.hasNext()){
System.out.println(sc.next());
}
System.out.println("Test");
but it newer goes out of while loop and prints "Test".
How could i read that input?
The method hasNext() works like this:
If it sees the end of the file, it returns false;
If it sees another valid, non-whitespace input, it returns true;
If neither of the above is true, it waits for the next input the user is going to enter, and doesn't return until he does.
Usually, if you use Scanner for files, such a loop will work correctly, because a file has a definite end, and it usually doesn't get stuck waiting for more input.
But when you are working with console input (System.in, not redirected), then usually the user does not send the end-of-file signal. He just presses Return, and so, hasNext() sits and waits to see if the user will enter more input on the next line and so on.
There are two general ways to deal with this:
The user has to actually terminate the input. After you finish entering all your numbers and press Return, you also need to send the end-of-file sequence, which is usually ctrlD or ctrlZ.
If you do that, you will not be able to enter any more input to that program.
The program tells the user to enter some particular value that will tell it that the input is over. For example, the string "DONE". When you do that, you have to change the loop to something like:
String nextInput;
while( sc.hasNext() && ! (nextInput = sc.next()).equals( "DONE" ) ){
System.out.println(nextInput);
}
You can break the loop depending whether you want to quit or not E.g.
public static void main(String[] args) {
Scanner sc = new Scanner(System.in);
while(sc.hasNext()){
String next = sc.next();
if (next.equals("q")) { //if user press q then break the loop
break;
}
System.out.println(next);
}
System.out.println("Test");
}
Use api like:
while(sc.hasNextInt()){
int aba= sc.nextInt();
if (aba == 0) {//or even non numeric value here would let this loop exit
break;
}
}
So you need to enter 0 or even in other way enter non numeric value inorder to come out of loop. nextLine method will read whole line just once and then you will need to parse it and then convert to integer so it's good to use sc.nextInt which will do the work for you.
If I comment out the line garbage = scan.nextLine();, the while-loop runs infinitely. Otherwise, it does not. I understand why it will run infinitely if there were only the print command, but I don't completely understand how the inclusion of the garbage variable stops it from running infintely. Can someone explain please?
import java.util.Scanner;
public class TypeSafeReadInteger
{
public static void main(String [] args)
{
Scanner scan = new Scanner(System.in);
String garbage;
System.out.print("Enter age as an integer > ");
while (! scan.hasNextInt())
{
garbage = scan.nextLine();
System.out.print("\nPlease enter an integer > ");
}
int age = scan.nextInt();
System.out.println("Your age is " + age);
}
}
garbage is just a variable, what 'stops' the while loop is the nextLine() It is a method that waits for user input. The while doesn't continue until your user inputs something using a keyboard and saves the input into the garbage variable.
You need to know two things:
hasNextLine() does not advance the Scanner instance.
nextLine() does advance the Scanner instance.
By "advance the Scanner instance", I mean "consume" input. Think of input as a stream, and think of a scanner object as something that is consuming that stream.
Things in a normal stream can only be consumed once. You captured your's in a variable called garbage, but you could just as easily have called scan.nextLine() without storing the result. I strongly advise you to read the Javadoc on Scanner to see which methods advance the Scanner instance and which do not.
To fix your code:
while (!scan.hasNextInt())
{
scan.nextLine(); // the order of the lines inside the loop makes the difference!
System.out.print("\nPlease enter an integer > ");
// when you perform nextLine() here - you reach the beginning of the loop
// without a token in the scanner - so you end up looping forever
}
int age = scan.nextInt();
By the way - as you can see from the example above, garbage is redundant.
If the user inputs an integer, then everything works. If they don't, then you get the infinite loop without the garbage = scan.nextLine(); line due to the way the Scanner class works.
When you do something like scan.hasNextInt();, no characters are actually read from the input. So if a user input something like "cat" in response to your prompt, then the input would be paused just before the first letter of that word. Since you are looping until there is an integer in the input, nothing further is read and you will loop infinitely because "cat" is just sitting in the input buffer.
By adding in the scan.nextLine() you will cause the Scanner to discard everything up to when the user hit <enter> and additional input could be processed.
I want to accept user input as either an integer or a string in Java but I always get an error no matter what I do.
My code is very simple (I'm a beginner):
System.out.println(
"Enter the row number (or enter e to quit the application):"
);
Scanner rowInput = new Scanner(System.in);
int row1 = rowInput.nextInt();
I want the user also to be able to press "e" to exit.
I have tried several things:
1) To convert row1 to a String and and say:
if((String)(row1).equals("e"){
System.out.println("You have quit") }
2) To convert "e" to an integer and say:
if(row1 == Integer.ParseInt("e"){
System.out.println("You have quit") }
3) To do the switch statement but it said I had incompatible types (String and an int).
My errors usually say: Exception in thread "main" java.util.InputMismatchException
Is there somebody who could help me?
Many thanks in advance!
You can try something like this
Scanner rowInput = new Scanner(System.in);
String inputStr = rowInput.nextLine();
try{
int row1 = Integer.parseInt(inputStr);
} catch (NumberFormatException e) //If exception occurred it means user has entered 'e'
{
if ("e".equals(inputStr)){
System.out.println("quiting application");
}
}
you should read the input from your scanner as String type and the try to convert that String input into integer using Integer.parseInt().
If its successful in parsing, it means user has input an Integer.
And If it fails, it will throw an exception NumberFormatException which will tell you that its not an Integer. So you can go ahead and check it for e if it is, do whatever you want as per your requirement.
You can't use nextInt() if you aren't sure that you will have a number coming in. Also, you can't parse e as an integer, because it is not an integer, it's either char or string (I'll suggest string in this case).
That means, your code should look like:
System.out.println("Enter the row number (or enter e to quit the application):");
Scanner rowInput = new Scanner(System.in);
string row1 = rowInput.next();
Then you have the data in a string. You can simply check if the string is e:
if (row1.Equals("e")) ...
It is also a good idea to check if the input is actually an integer then. Good way to check it is described here:
How to check if a String is numeric in Java
1) you should say Integer.parseInt() instead of Integer.ParseInt()
2) if you pass "e" to Integer.parseInt() you'll get java.lang.NumberFormatException
3) get your input as a string this way (cause it's safer)*:
Scanner rowInput = new Scanner(System.in);
String row = rowInput.next();
if (row.equals("e")) {
//do whatever
}
else if(row.matches("\\d+$")) {
//do whatever
}
*In your approach if user enters a non-integer input you'll encounter java.util.InputMismatchException
I have tried several things ....
Attempting to solve this problem by "trying things" is the wrong approach.
The correct approach is to understand what is going on, and then modify your solution to take this into account.
The problem you have is that you have a line of input that could be either a number or the special value e which means quit. (And in fact, it could be a couple of other things too ... but I will come to that.)
When you attempt to either:
call Scanner.nextInt() when the next input is not an integer, OR
call Integer.parseInt(...) on a String that is not an integer
you will get an exception. So what do you do?
One way is to change what you are doing so that you don't make those calls in those situations. For example:
call Scanner.hasInt() to see if the next token is an integer before calling nextInt(), or
test for the "e" case before you attempt to convert the String to an integer.
Another way to deal with this is to catch the exception. For example, you could do something like this:
String s = ...
try {
number = Integer.parseInt(s);
} catch (NumberFormatException ex) {
if (s.equals("e")) {
...
}
}
Either way will work, but I'm going to leave you to work out the details.
But the real point I'm trying to make is that the right way to solve this is to understand what is happening in your original version / versions, and based on your understanding, modify what you were doing. If you just randomly "try" alternatives that you found with Google, you won't progress to the point of being a productive programmer.
I said there were other cases. They are:
The user types something that is neither a number of your special e character. It could be anything. An empty line, hi mum ...
The user types the "end of file" character; e.g. CTRL-D on Linux or CTRL-Z on windows.
If you want your program to be robust you need to deal with these cases too.
Try this:
public class Demo{
public static void main(String[] args) {
Scanner s=new Scanner(System.in);
String choice = null;
System.out.println("Enter e to exit or some other key to stay:");
choice=s.next();
if(choice.equals("e"))
{
System.out.println("quits");
System.exit(0);
}
else
{
System.out.println("stays");
}
}
}
You can't convert a string into integer, use char ASCII codes instead.
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)