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I want to read those inputs in a single line separated by space in java without using String?

System.out.println("Number of pages + Number of lost pages + Number of Readers");
int n = s.nextInt();
int m = s.nextInt();
int q = s.nextInt();
I want to read input values all the values are going to be integer but I want to read it in a same line with changing it form Integer.

Assuming s is an instance of Scanner: Your code, as written, does exactly what you want.
scanners are created by default with a delimiter configured to be 'any whitespace'. nextInt() reads the next token (which are the things in between the delimiter, i.e. the whitespace), and returns it to you by parsing it into an integer.
Thus, your code as pasted works fine.
If it doesn't, stop setting up a delimiter, or reset it back to 'any whitespace' with e.g. scanner.reset(); or scanner.useDelimiter("\\s+");.
class Example {
public static void main(String[] args) {
var in = new Scanner(System.in);
System.out.println("Enter something:");
System.out.println(in.nextInt());
System.out.println(in.nextInt());
System.out.println(in.nextInt());
}
}
works fine here.

Why doesn't hasNextInt() method work for large integers?

I'm writing a simple program which receives integer inputs from a Scanner object, determines whether it's a palindrome or not, and returns the boolean value.
For most numbers, it works well. However, at this code snippet:
private static void programRunner() {
System.out.print("Insert your number:");
Scanner in = new Scanner(System.in);
if (in.hasNextInt()) {
int testNumber = in.nextInt();
boolean result = palindromeTester(testNumber);
System.out.println(result);
programRunner();
} else {
System.exit(0);
}
}
I added the "System.exit(0)" expression to make users easily terminate the program by intentionally typing any non-integer value. The problem is that when "considerably large" integers, such as "1234567654321," are provided, the code launches System.exit(0) which means it's not recognized as an integer?
I believe it's the problem lies in the "default radii" of the hasNextInt method, which probably limits the size of integer values it recognizes. (The program runs fine up to 9-digit integers) But I'm not sure. Or is there something wrong with the recursion?

Because an int in Java is 32 bit and can only hold 2^31 - 1 (2147483647) as maximum value (see also Integer.MAX_VALUE).
Anything bigger than that is not int, but long (except if it's even bigger than Long.MAX_VALUE, in which case you need to get the value as BigInteger.)
See Integer.MAX_VALUE, Long.MAX_VALUE,
Scanner.nextInteger(), Scanner.nextLong(),
Scanner.nextBigInteger() and BigInteger.

You can use nextLong() if you have larg "long" integers to read like so :
private static void programRunner() {
System.out.print("Insert your number:");
Scanner in = new Scanner(System.in);
if (in.hasNextLong())
{
long testNumber = in.nextLong();
boolean result = palindromeTester(testNumber);
System.out.println(result);
programRunner();
}
else
{
System.exit(0);
}
}

It seems that you've exceeded the range limit for the int type. the long type seems like what you're looking for.
So, you can either use the hasNextLong() and nextLong() methods of the Scanner class or as #Hovercraft Full Of Eels has suggested in the comments since you're not using the data in a numerical way, it may be better to receive the data as a String instead.
Lastly, but not least I find it inefficient to use recursion here as each recursive call creates a new stack frame along with that for each recursive invocation you're newing up a Scanner instance. it's better to use a while loop instead.

Confusion with Scanners (Big Java Ex 6.3)

Currently reading Chapter 6 in my book. Where we introduce for loops and while loops.
Alright So basically The program example they have wants me to let the user to type in any amount of numbers until the user types in Q. Once the user types in Q, I need to get the max number and average.
I won't put the methods that actually do calculations since I named them pretty nicely, but the main is where my confusion lies.
By the way Heres a simple input output
Input
10
0
-1
Q
Output
Average = 3.0
Max = 10.0
My code
public class DataSet{
public static void main(String [] args)
{
DataAnalyze data = new DataAnalyze();
Scanner input = new Scanner(System.in);
Scanner inputTwo = new Scanner(System.in);
boolean done = false;
while(!done)
{
String result = input.next();
if (result.equalsIgnoreCase("Q"))
{
done = true;
}
else {
double x = inputTwo.nextDouble();
data.add(x);
}
}
System.out.println("Average = " + data.getAverage());
System.out.println("Max num = " + data.getMaximum());
}
}
I'm getting an error at double x = inputTwo.nextDouble();.
Heres my thought process.
Lets make a flag and keep looping asking the user for a number until we hit Q. Now my issue is that of course the number needs to be a double and the Q will be a string. So my attempt was to make two scanners
Heres how my understanding of scanner based on chapter two in my book.
Alright so import Scanner from java.util library so we can use this package. After that we have to create the scanner object. Say Scanner input = new Scanner(System.in);. Now the only thing left to do is actually ASK the user for input so we doing this by setting this to another variable (namely input here). The reason this is nice is that it allows us to set our Scanner to doubles and ints etc, when it comes as a default string ( via .nextDouble(), .nextInt());
So since I set result to a string, I was under the impression that I couldn't use the same Scanner object to get a double, so I made another Scanner Object named inputTwo, so that if the user doesn't put Q (i.e puts numbers) it will get those values.
How should I approach this? I feel like i'm not thinking of something very trivial and easy.

You are on the right path here, however you do not need two scanners to process the input. If the result is a number, cast it to a double using double x = Double.parseDouble(result) and remove the second scanner all together. Good Luck!

How do i declare and read values into integer values?

I'm really new to java and i'm taking an introductory class to computer science. I need to know how to Prompt the user to user for two values, declare and define 2 variables to store the integers, and then be able to read the values in, and finally print the values out. But im pretty lost and i dont even know how to start i spent a whole day trying.. I really need some help/guidance. I need to do that for integers, decimal numbers and strings. Can someone help me?

You can do this by using Scanner class :
A simple text scanner which can parse primitive types and strings using regular expressions.
A Scanner breaks its input into tokens using a delimiter pattern, which by default matches whitespace. The resulting tokens may then be converted into values of different types using the various next methods.
For example, this code allows a user to read a number from System.in:
Scanner scan = new Scanner(System.in);
int i = scan.nextInt();
int j = scan.nextInt();
System.out.println("i = "+i +" j = "+j);
nextInt() : -Scans the next token of the input as an int and returns the int scanned from the input.
For more.
or to get user input you can also use the Console class : provides methods to access the character-based console device, if any, associated with the current Java virtual machine.
Console console = System.console();
String s = console.readLine();
int i = Integer.parseInt(console.readLine());
or you can also use BufferedReader and InputStreamReader classes and
DataInputStream class to get user input .

Use the Scanner class to get the values from the user. For integers you should use int, for decimal numbers (also called real numbers) use double and for strings use Strings.
A little example:
Scanner scan = new Scanner(System.in);
int intValue;
double decimalValue;
String textValue;
System.out.println("Please enter an integer value");
intValue = scan.nextInt(); // see how I use nextInt() for integers
System.out.println("Please enter a real number");
decimalValue = scan.nextDouble(); // nextDouble() for real numbers
System.out.println("Please enter a string value");
textValue = scan.next(); // next() for string variables
System.out.println("Your integer is: " + intValue + ", your real number is: "
+ decimalValue + " and your string is: " + textValue);
If you still don't understand something, please look further into the Scanner class via google.

As you will likely continue to run into problems like this in your class and in your programming career:
Lessons on fishing.
Learn to explore the provided tutorials through oracle.
Learn to read the Java API documentation
Now to the fish.
You can use the Scanner class. Example provided in the documentation.
Scanner sc = new Scanner(System.in);
int i = sc.nextInt();

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);