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What is memory buffer? And how different Scanner class methods operate on the probable issues of line breaks, white spaces etc.?

I am having trouble understanding how memory buffer works when I am working with Scanner class methods such as hasNextInt() hasNextDouble() etc. Considering the following code,
Scanner in = new Scanner(System.in);
int number;
do {
System.out.print("Enter a positive integer: ");
while (!in.hasNextInt()) {
System.out.println("It's not an integer!");
in.next();
}
number = in.nextInt();
} while (number <= 0);
System.out.println("Your number is " + number);
The output for some random values:
Enter a positive integer: five
It's not an integer!
-1
Enter a positive integer: 45
Your number is 45
What actually happens here? At line 1 when I enter five the nested while loop runs. What is the job of in.next()? After I enter five it says It's not an integer! But why doesn't it ask again: Enter a positive integer: ? Basically, I want the corresponding output to be like this:
Enter a positive integer: five
It's not an integer!
Enter a positive integer: -1
It's not a positive integer!
Enter a positive integer: 45
Your number is 45.
I would appreciate a brief and intuitive explanation how white spaces, line breaks are handled in input validation? And what is memory buffer? And how different methods of Scanner class like next(), nextLine(), nextInt(), nextDouble() etc. operate?
Also, how do I avoid repetition of It's not an integer!
Enter a positive number: five
It's not an integer!
one two three
It's not an integer!
It's not an integer!
It's not an integer!
10
Your number is 10
And finally, why many recommend try catch?

To start with, 0, -1, -66, 2352, +66, are all Integer values so you can't very well decide to designate them as otherwise. Your validation response should really be:
System.out.println("It's not a positive integer value!");
I personally never use those nextInt(), nextDouble(), etc methods unless I want blind validation. I just stick with a single loop, and utilize the nextLine() method along with the String#matches() method (with a small Regular Expression). I also don't really care for using a try/catch to solve a situation where I don't have to.
Scanner in = new Scanner(System.in);
int number = 0;
while (number < 1) {
System.out.print("Enter a positive integer (q to quit): ");
String str = in.nextLine();
if (!str.equals("") && String.valueOf(str.charAt(0)).equalsIgnoreCase("q")) {
System.exit(0);
}
// If a string representation of a positive Integer value
// is supplied (even if it's prefixed with the '+' character)
// then convert it to Integer.
if (str.matches("\\+?\\d+") && !str.equals("0")) {
number = Integer.parseInt(str);
}
// Otherwise...
else {
System.err.println(str + " is not considered a 'positive' integer value!");
}
}
System.out.println("Your number is " + number);
In this particular use-case, I actually find this more versatile but then, perhaps that's just me. It doesn't matter what is entered, you will always get a response of one form or another and, you have a quit option as well. To quit either the word quit or the letter q (in any letter case) can be supplied.
People like to utilize the try/catch in case a NumberFormatException is thrown by nextInt() because a white-space or any character other than a digit is supplied. This then allows the opportunity of displaying a message to console that an invalid input was supplied.
Because the Scanner class is passed System.in within its' constructor (in is an object of InputStream) it is a Stream mechanism and therefore contains a input (holding) buffer. When anything is typed to the Console Window it is place within the input buffer until the buffer is read by any one of the next...() methods.
Not all Scanner class methods like next(), nextInt(), nextDouble(), etc, completely utilize everything contained within the stream input buffer, for example, these methods do not consume whitespaces, tabs, and any newline characters when the ENTER key is hit. The nextLine() method however does consume everything within the input buffer.
This is exactly why when you have a prompt for a User to supply an Integer value (age) and you use the nextInt() method to get that data and then directly afterwords you prompt for a string like the User's name using the nextLine() method, you will notice that the nextLine() prompt is skipped over. This is because there is still a newline character within the input buffer that wasn't consumed by the nextInt() method and now forces the nextLine() method to consume it. That ENTER that was done in the previous nextInt() method is now passed into the nextLine() method thus giving the impression that the prompt was bypassed when in reality, it did receive a newline character (which in most cases is pretty much useless).
To overcome this particular situation the easiest thing to do is to consume the ENTER key newline character by adding scanner.nextLine(); directly after a int myVar = scanner.nextInt(); call. This then empties the input buffer before the String name = scanner.nextLine(); comes into play.

Is there a way to show an error message while the user has not entered a valid value, within the condition of a while loop

I pretty new to java programming so i was wondering if there is a way to use the condition of a while loop to stop an invalid value being used.
I am writing a program that prompts the user to enter an identification number as an integer then uses a scanner to store that value.
Just wanted to know if this is possible to put something in the condition of the for loop that prints an error message if the enter something like a string, double or char so i dont get the Input Mismatch Exception.
like this:
identification = userId(in); //scanner
while (identification (is not an integer)){
System.out.println("Invalid Value, Please enter an integer");
identification = userId(in);

Even better, you can write:
while ((identification = userId(in)) < 0) {
System.out.println("blah ...");
}
The assumption is that the userIn method returns some negative value if the input is not an integer. You can make the invalid return value whatever you want as long as it is not something that is a valid input.
Some people don't like this style because it has gone out of fashion and they are not used to it; however, it used to be common in C programming and there is nothing about it that is implicitly unclear or bad.

This should do what you are asking. It is basically a while loop that waits until the next input is an integer before continuing the code. The important part is making sure to use in.next() inside the while loop instead of in.nextInt() because the values could be anything.
Scanner in = new Scanner(System.in);
System.out.print("Enter an integer: ");
while (!(in.hasNextInt()))
{
System.out.print("Integer not entered, please enter an integer: ");
in.next();
}
int value = in.nextInt();
System.out.println("The int was " + value);
in.close();

Get user to input integers

I want to make a program which keeps prompting the user to input integers(from CUI) until it receives a 'X' or 'x' from the user.
The program then prints out the maximum number, minimum number and average value of the input numbers.
I did manage to get the user to input numbers until someone types 'X', but I can't seem to get it to stop if someone types 'x' and the second bit.
This is the code that I have managed to work out:
Scanner in = new Scanner(System.in);
System.out.println("Enter a number")
while(!in.hasNext("X") && !in.hasNext("x"))
s = in.next().charAt(0);
System.out.println("This is the end of the numbers");
Any hints on how I proceed further?

You will need to do something like this:
Scanner in = new Scanner(System.in);
System.out.println("Enter a number")
while(!(in.hasNext("X") || in.hasNext("x")))
s = in.next().charAt(0);
System.out.println("This is the end of the numbers");
Whenever you use while loop you have to use the {} in case the arguments in the while block are more than 1 line, but if they are just of a line then you can just go on without using the {}.
But the problem, you had I suppose is the use of && instead of ||. What the && (AND) operator does is execute if both the statements are true but a || (OR) Operator works if any of the conditions are true.
If you say while(!in.hasNext("X") && !in.hasNext("x")) it makes no sense as the user input is not both at the same time, but instead if you usewhile(!in.hasNext("X") || !in.hasNext("x"))` it makes sense. Understood?
And about sorry, im really new at this. but ive added the code No problem, you need not say sorry but there are a few things to keep in mind before asking a question. You must read this https://stackoverflow.com/help/how-to-ask and yeah one more thing, you should use proper English Grammar while framing your question.
Last of all, about how to calculate the average..., for that what you need to do is store all the input variables into an array and then take out the mean of that or alternatively you could think about it and code something up yourself. Like to take out mean, you could make a variable sum and then keep adding the integers the user enters and also keep a variable count which will keep the count of the number of integers entered and then at last you could divide both of them to have your answer
Update: For checking the minimum and the maximum, what you can do is make 2 new variables like int min=0, max=0; and when the user enters a new variable you can check
//Note you have to change the "userinput" to the actual user input
if(min>userinput){
min=userinput;
}
and
if(max<userinput){
max=userinput;
}
Note: At stackoverflow we are there to help you out with the problems you are facing BUT you cannot exploit this. You cannot just post your homework here. But if you are trying to code something up and are stuck at it and cannot find a answer at google/stackoverflow then you can ask a new question and in that you need to tell what all you have already tried. Welcome to SO! :D Hope you have a nice time here

This would fit your needs:
public void readNumbers() {
// The list of numbers that we read
List<Integer> numbers = new ArrayList<>();
// The scanner for the systems standard input stream
Scanner scanner = new Scanner(System.in);
// As long as there a tokens...
while (scanner.hasNext()) {
if (scanner.hasNextInt()) { // ...check if the next token is an integer
// Get the token converted to an integer and store it in the list
numbers.add(scanner.nextInt());
} else if (scanner.hasNext("X") || scanner.hasNext("x")) { // ...check if 'X' or 'x' has been entered
break; // Leave the loop
}
}
// Close the scanner to avoid resource leaks
scanner.close();
// If the list has no elements we can return
if (numbers.isEmpty()) {
System.out.println("No numbers were entered.");
return;
}
// The following is only executed if the list is not empty/
// Sort the list ascending
Collections.sort(numbers);
// Calculate the average
double average = 0;
for (int num : numbers) {
average += num;
}
average /= numbers.size();
// Print the first number
System.out.println("Minimum number: " + numbers.get(0));
// Print the last number
System.out.println("Maximum number: " + numbers.get(numbers.size() - 1));
// Print the average
System.out.println("Average: " + average);
}

Trouble with Scanner taking data from a command line using Java

I will admit, this is a school assignment... But I simply cannot figure out what I am doing wrong.
I have a hash table with an insert function. The following code is supposed to take a line of data from System.in in the format "Long String" (i.e. "32452 John"). The first token must be a Long for the ID number, and it must be followed by a String token for the name. When I run the program and I get to the portion where this must be executed (It is in a switch statement), I entered 'a' and hit enter. The command line immediately reads "Invalid value." (note: not VALUES, as that would mean it hit the nested if statement. It won't let me type in any data. Thank you in advance!
System.out.println("Enter ID and Name.");
//temp to take in the next line entered by the user
//inScan is the Scanner for System.in
temp = inScan.nextLine();
//Create Scanner for the line
Scanner tempScan = new Scanner(temp);
if(tempScan.hasNextLong()){
thisID = tempScan.nextLong();
if((tempScan.hasNext()) && (thisID>0)){
thisName = tempScan.next();
//The data will only be inserted if both segments of data are entered
myTable.insert(new Student(thisID, thisName));
}else{
System.out.println("Invalid values.");
}
}else{
System.out.println("Invalid value.");
}

Why do you need the second Scanner?
Example
String input = scanner.nextLine();
String[] tokens = input.split(" ");
Long id = Long.parseLong(tokens[0]);
String name = tokens[1];
And if you wanted to add your validation:
String input = scanner.nextLine();
if(input.contains(" ")) {
// You know there's a space in it.
String[] tokens = input.split(" ");
if(tokens.length == 2) {
// You know it's a value, followed by a space, followed by a value.
if(tokens[0].matches("[0-9]+")) {
// You know it only contains numbers.
Long id = Long.parseLong(tokens[0]);
}
}
}

I've not run it, but i guess your problem is that when you enter the text 'a' and hit enter, this line is false:
if(tempScan.hasNextLong()){
as you haven't entered a number. hence why it drops to the next block. If you enter something numerical first, i suspect your code with work. you probably need to add a 'while' loop around it, to run until it gets a number.

You already have a Scanner which reads from System.in, there's no need for another one. The second one you've made is a scanner for a String, which will never have a nextLong as it has nothing to scan after your String.
I won't write any code for you as this is homework, but stick to your original scanner when checking for user input instead.

Validating input using java.util.Scanner [duplicate]

This question already has an answer here:
How to use java.util.Scanner to correctly read user input from System.in and act on it?
(1 answer)
Closed 5 years ago.
I'm taking user input from System.in using a java.util.Scanner. I need to validate the input for things like:
It must be a non-negative number
It must be an alphabetical letter
... etc
What's the best way to do this?

Overview of Scanner.hasNextXXX methods
java.util.Scanner has many hasNextXXX methods that can be used to validate input. Here's a brief overview of all of them:
hasNext() - does it have any token at all?
hasNextLine() - does it have another line of input?
For Java primitives
hasNextInt() - does it have a token that can be parsed into an int?
Also available are hasNextDouble(), hasNextFloat(), hasNextByte(), hasNextShort(), hasNextLong(), and hasNextBoolean()
As bonus, there's also hasNextBigInteger() and hasNextBigDecimal()
The integral types also has overloads to specify radix (for e.g. hexadecimal)
Regular expression-based
hasNext(String pattern)
hasNext(Pattern pattern) is the Pattern.compile overload
Scanner is capable of more, enabled by the fact that it's regex-based. One important feature is useDelimiter(String pattern), which lets you define what pattern separates your tokens. There are also find and skip methods that ignores delimiters.
The following discussion will keep the regex as simple as possible, so the focus remains on Scanner.
Example 1: Validating positive ints
Here's a simple example of using hasNextInt() to validate positive int from the input.
Scanner sc = new Scanner(System.in);
int number;
do {
System.out.println("Please enter a positive number!");
while (!sc.hasNextInt()) {
System.out.println("That's not a number!");
sc.next(); // this is important!
}
number = sc.nextInt();
} while (number <= 0);
System.out.println("Thank you! Got " + number);
Here's an example session:
Please enter a positive number!
five
That's not a number!
-3
Please enter a positive number!
5
Thank you! Got 5
Note how much easier Scanner.hasNextInt() is to use compared to the more verbose try/catch Integer.parseInt/NumberFormatException combo. By contract, a Scanner guarantees that if it hasNextInt(), then nextInt() will peacefully give you that int, and will not throw any NumberFormatException/InputMismatchException/NoSuchElementException.
Related questions
How to use Scanner to accept only valid int as input
How do I keep a scanner from throwing exceptions when the wrong type is entered? (java)
Example 2: Multiple hasNextXXX on the same token
Note that the snippet above contains a sc.next() statement to advance the Scanner until it hasNextInt(). It's important to realize that none of the hasNextXXX methods advance the Scanner past any input! You will find that if you omit this line from the snippet, then it'd go into an infinite loop on an invalid input!
This has two consequences:
If you need to skip the "garbage" input that fails your hasNextXXX test, then you need to advance the Scanner one way or another (e.g. next(), nextLine(), skip, etc).
If one hasNextXXX test fails, you can still test if it perhaps hasNextYYY!
Here's an example of performing multiple hasNextXXX tests.
Scanner sc = new Scanner(System.in);
while (!sc.hasNext("exit")) {
System.out.println(
sc.hasNextInt() ? "(int) " + sc.nextInt() :
sc.hasNextLong() ? "(long) " + sc.nextLong() :
sc.hasNextDouble() ? "(double) " + sc.nextDouble() :
sc.hasNextBoolean() ? "(boolean) " + sc.nextBoolean() :
"(String) " + sc.next()
);
}
Here's an example session:
5
(int) 5
false
(boolean) false
blah
(String) blah
1.1
(double) 1.1
100000000000
(long) 100000000000
exit
Note that the order of the tests matters. If a Scanner hasNextInt(), then it also hasNextLong(), but it's not necessarily true the other way around. More often than not you'd want to do the more specific test before the more general test.
Example 3 : Validating vowels
Scanner has many advanced features supported by regular expressions. Here's an example of using it to validate vowels.
Scanner sc = new Scanner(System.in);
System.out.println("Please enter a vowel, lowercase!");
while (!sc.hasNext("[aeiou]")) {
System.out.println("That's not a vowel!");
sc.next();
}
String vowel = sc.next();
System.out.println("Thank you! Got " + vowel);
Here's an example session:
Please enter a vowel, lowercase!
5
That's not a vowel!
z
That's not a vowel!
e
Thank you! Got e
In regex, as a Java string literal, the pattern "[aeiou]" is what is called a "character class"; it matches any of the letters a, e, i, o, u. Note that it's trivial to make the above test case-insensitive: just provide such regex pattern to the Scanner.
API links
hasNext(String pattern) - Returns true if the next token matches the pattern constructed from the specified string.
java.util.regex.Pattern
Related questions
Reading a single char in Java
References
Java Tutorials/Essential Classes/Regular Expressions
regular-expressions.info/Character Classes
Example 4: Using two Scanner at once
Sometimes you need to scan line-by-line, with multiple tokens on a line. The easiest way to accomplish this is to use two Scanner, where the second Scanner takes the nextLine() from the first Scanner as input. Here's an example:
Scanner sc = new Scanner(System.in);
System.out.println("Give me a bunch of numbers in a line (or 'exit')");
while (!sc.hasNext("exit")) {
Scanner lineSc = new Scanner(sc.nextLine());
int sum = 0;
while (lineSc.hasNextInt()) {
sum += lineSc.nextInt();
}
System.out.println("Sum is " + sum);
}
Here's an example session:
Give me a bunch of numbers in a line (or 'exit')
3 4 5
Sum is 12
10 100 a million dollar
Sum is 110
wait what?
Sum is 0
exit
In addition to Scanner(String) constructor, there's also Scanner(java.io.File) among others.
Summary
Scanner provides a rich set of features, such as hasNextXXX methods for validation.
Proper usage of hasNextXXX/nextXXX in combination means that a Scanner will NEVER throw an InputMismatchException/NoSuchElementException.
Always remember that hasNextXXX does not advance the Scanner past any input.
Don't be shy to create multiple Scanner if necessary. Two simple Scanner is often better than one overly complex Scanner.
Finally, even if you don't have any plans to use the advanced regex features, do keep in mind which methods are regex-based and which aren't. Any Scanner method that takes a String pattern argument is regex-based.
Tip: an easy way to turn any String into a literal pattern is to Pattern.quote it.

Here's a minimalist way to do it.
System.out.print("Please enter an integer: ");
while(!scan.hasNextInt()) scan.next();
int demoInt = scan.nextInt();

For checking Strings for letters you can use regular expressions for example:
someString.matches("[A-F]");
For checking numbers and stopping the program crashing, I have a quite simple class you can find below where you can define the range of values you want.
Here
public int readInt(String prompt, int min, int max)
{
Scanner scan = new Scanner(System.in);
int number = 0;
//Run once and loop until the input is within the specified range.
do
{
//Print users message.
System.out.printf("\n%s > ", prompt);
//Prevent string input crashing the program.
while (!scan.hasNextInt())
{
System.out.printf("Input doesn't match specifications. Try again.");
System.out.printf("\n%s > ", prompt);
scan.next();
}
//Set the number.
number = scan.nextInt();
//If the number is outside range print an error message.
if (number < min || number > max)
System.out.printf("Input doesn't match specifications. Try again.");
} while (number < min || number > max);
return number;
}

One idea:
try {
int i = Integer.parseInt(myString);
if (i < 0) {
// Error, negative input
}
} catch (NumberFormatException e) {
// Error, not a number.
}
There is also, in commons-lang library the CharUtils class that provides the methods isAsciiNumeric() to check that a character is a number, and isAsciiAlpha() to check that the character is a letter...

If you are parsing string data from the console or similar, the best way is to use regular expressions. Read more on that here:
http://java.sun.com/developer/technicalArticles/releases/1.4regex/
Otherwise, to parse an int from a string, try
Integer.parseInt(string). If the string is not a number, you will get an exception. Otherise you can then perform your checks on that value to make sure it is not negative.
String input;
int number;
try
{
number = Integer.parseInt(input);
if(number > 0)
{
System.out.println("You positive number is " + number);
}
} catch (NumberFormatException ex)
{
System.out.println("That is not a positive number!");
}
To get a character-only string, you would probably be better of looping over each character checking for digits, using for instance Character.isLetter(char).
String input
for(int i = 0; i<input.length(); i++)
{
if(!Character.isLetter(input.charAt(i)))
{
System.out.println("This string does not contain only letters!");
break;
}
}
Good luck!

what i have tried is that first i took the integer input and checked that whether its is negative or not if its negative then again take the input
Scanner s=new Scanner(System.in);
int a=s.nextInt();
while(a<0)
{
System.out.println("please provide non negative integer input ");
a=s.nextInt();
}
System.out.println("the non negative integer input is "+a);
Here, you need to take the character input first and check whether user gave character or not if not than again take the character input
char ch = s.findInLine(".").charAt(0);
while(!Charcter.isLetter(ch))
{
System.out.println("please provide a character input ");
ch=s.findInLine(".").charAt(0);
}
System.out.println("the character input is "+ch);