I am unable to read A 1 2 3 from the same line, 1 2 3 can be read from one line but because of A in the start, scanner only reads A.
String command = "";
int numbers;
List<Integer> ints = new ArrayList<Integer>();
Scanner scan = new Scanner(System.in);
command = scan.next();
while (scan.hasNextInt())
{
numbers = scan.nextInt();
ints.add(numbers);
}
scan.close();
I would use nextLine() (and hasNextLine()), split on one or more consecutive white-space characters (to create an array of tokens), set the command as the first token and then Stream the tokens - skip the first one (it's command) and collect to a List. Like,
Scanner scan = new Scanner(System.in);
while (scan.hasNextLine()) {
String line = scan.nextLine();
String[] tokens = line.split("\\s+");
String command = tokens[0];
List<Integer> ints = Stream.of(tokens).skip(1)
.mapToInt(Integer::parseInt).boxed().collect(Collectors.toList());
System.out.printf("command = %s, ints = %s%n", command, ints.toString());
}
I have problem with my Java program. I am running the program on the console (CMD).
I would like, after I entered input, that the console stays on the same line (currently it goes to the next line automatically).
This is my current program:
int data[];
Scanner in = new Scanner(System.in);
data = new int[10];
System.out.println("Please Insert Numbers : ");
for(int i=0;i<5;i++)
{
data[i] = in.nextInt();
System.out.print("/t");
}
How can I return to the start of the line instead of going to the next?
Scanner in = new Scanner(System.in);
System.out.println("Please Insert Numbers Separated By Comma: ");
String input = in.nextLine();
input = input.replaceAll("\\s",""); //remove whitespaces
String[] numbers= input.split(","); //build string array using ',' as delimiter between numbers
int data[]= Arrays.asList(numbers).stream().mapToInt(Integer::parseInt).toArray(); //Available since Java8
Like Oscar Martinez said, you could read a string of numbers delimited by white space and build your integer array like this example (before parsing the string to int, I verify if the string can be converted to int or not using a regex ) :
import java.util.Arrays;
import java.util.Scanner;
public class Main {
public static void main(String[] args){
int data[];
Scanner in = new Scanner(System.in);
data = new int[10];
System.out.println("Please Insert Numbers : ");
String line = in.nextLine();
String[] stringsNumber = line.split("\\s");
for(int i=0;i<stringsNumber.length;i++){
if(stringsNumber[i].matches("^\\d+$") && i<10){// \\d is a regex to verify if s is a number
data[i]= Integer.parseInt(stringsNumber[i]);
}
}
System.out.println(Arrays.toString(data));
}
}
I am very new to java. I am trying to prompt the user to enter 4 integer numbers followed by a space and eventually print them out at the end. I am a little confused with the order of how I write things out and using the split(" ");
import java.util.Scanner;
public class calculations {
public static void main(String[] args) {
Scanner Keyboard = new Scanner(System.in);
System.out.println("Enter 4 integer numbers here: ");
int numbers = keyboard.nextInt();
// Need split(" "); here?
} // End main string args here
} // End class calculations here
Any help or advice is appreciated. I have looked at other ways on stackoverflow but somehow I keep getting errors.
Read it in one String with keyboard.nextLine
Use the split method of String for get an array of Strings
Convert every element of the array to int with Integer.parseInt
Print your ints.
import java.util.Scanner;
public class calculations {
public static void main(String[] args) {
Scanner Keyboard = new Scanner(System.in);
System.out.println("Enter 4 integer numbers here: ");
// Scan an entire line (containg 4 integers separated by spaces):
String lineWithNumbers = Keyboard.nextLine();
// Split the String by the spaces so that you get an array of size 4 with
// the numbers (in a String).
String[] numbers = lineWithNumbers.split(" ");
// For each String in the array, print them to the screen.
for(String numberString : numbers) {
System.out.println(numberString);
}
} // End main string args here
} // End class calculations here
This code will print all numbers, in case you actually want to do something with the Integers (for example mathematical operations) you can parse the String to an int, like so:
int myNumber = Integer.parseInt(numberString);
Hope this helps.
If would suggest to use the abilities of the Scanner class to retrieve numbers from the user input:
Scanner keyboard = new Scanner(System.in);
int[] numbers = new int[4];
System.out.println("Enter 4 integer numbers here: ");
for (int i = 0; i < 4 && keyboard.hasNextInt(); i++) {
numbers[i] = keyboard.nextInt();
}
System.out.println(Arrays.toString(numbers));
This code creates an array of size 4 and then loops over the user input reading the numbers from it. It will stop parsing the input if he has the four numbers, or if the user enters something different than a number. For example, if he enters 1 blub 3 4, then the array will be [1, 0, 0, 0].
This code has some advantages compared to the nextLine approaches of the over answers:
you don't have to care about the integer conversion (exception handling)
you can either write these number onto one line or each number on its own line
If you like to read an arbitrary amount of numbers, then use a List instead:
List<Integer> numbers = new ArrayList<>();
System.out.println("Enter some integer numbers here (enter something else than a number to stop): ");
while (keyboard.hasNextInt()) {
numbers.add(keyboard.nextInt());
}
System.out.println(numbers);
I am working on a program and I want to allow a user to enter multiple integers when prompted. I have tried to use a scanner but I found that it only stores the first integer entered by the user. For example:
Enter multiple integers: 1 3 5
The scanner will only get the first integer 1. Is it possible to get all 3 different integers from one line and be able to use them later? These integers are the positions of data in a linked list I need to manipulate based on the users input. I cannot post my source code, but I wanted to know if this is possible.
I use it all the time on hackerrank/leetcode
BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
String lines = br.readLine();
String[] strs = lines.trim().split("\\s+");
for (int i = 0; i < strs.length; i++) {
a[i] = Integer.parseInt(strs[i]);
}
Try this
public static void main(String[] args) {
Scanner in = new Scanner(System.in);
while (in.hasNext()) {
if (in.hasNextInt())
System.out.println(in.nextInt());
else
in.next();
}
}
By default, Scanner uses the delimiter pattern "\p{javaWhitespace}+" which matches at least one white space as delimiter. you don't have to do anything special.
If you want to match either whitespace(1 or more) or a comma, replace the Scanner invocation with this
Scanner in = new Scanner(System.in).useDelimiter("[,\\s+]");
You want to take the numbers in as a String and then use String.split(" ") to get the 3 numbers.
String input = scanner.nextLine(); // get the entire line after the prompt
String[] numbers = input.split(" "); // split by spaces
Each index of the array will hold a String representation of the numbers which can be made to be ints by Integer.parseInt()
Scanner has a method called hasNext():
Scanner scanner = new Scanner(System.in);
while(scanner.hasNext())
{
System.out.println(scanner.nextInt());
}
If you know how much integers you will get, then you can use nextInt() method
For example
Scanner sc = new Scanner(System.in);
int[] integers = new int[3];
for(int i = 0; i < 3; i++)
{
integers[i] = sc.nextInt();
}
Java 8
BufferedReader in = new BufferedReader(new InputStreamReader(System.in));
int arr[] = Arrays.stream(in.readLine().split(" ")).mapToInt(Integer::parseInt).toArray();
Here is how you would use the Scanner to process as many integers as the user would like to input and put all values into an array. However, you should only use this if you do not know how many integers the user will input. If you do know, you should simply use Scanner.nextInt() the number of times you would like to get an integer.
import java.util.Scanner; // imports class so we can use Scanner object
public class Test
{
public static void main( String[] args )
{
Scanner keyboard = new Scanner( System.in );
System.out.print("Enter numbers: ");
// This inputs the numbers and stores as one whole string value
// (e.g. if user entered 1 2 3, input = "1 2 3").
String input = keyboard.nextLine();
// This splits up the string every at every space and stores these
// values in an array called numbersStr. (e.g. if the input variable is
// "1 2 3", numbersStr would be {"1", "2", "3"} )
String[] numbersStr = input.split(" ");
// This makes an int[] array the same length as our string array
// called numbers. This is how we will store each number as an integer
// instead of a string when we have the values.
int[] numbers = new int[ numbersStr.length ];
// Starts a for loop which iterates through the whole array of the
// numbers as strings.
for ( int i = 0; i < numbersStr.length; i++ )
{
// Turns every value in the numbersStr array into an integer
// and puts it into the numbers array.
numbers[i] = Integer.parseInt( numbersStr[i] );
// OPTIONAL: Prints out each value in the numbers array.
System.out.print( numbers[i] + ", " );
}
System.out.println();
}
}
There is more than one way to do that but simple one is using String.split(" ")
this is a method of String class that separate words by a spacial character(s) like " " (space)
All we need to do is save this word in an Array of Strings.
Warning : you have to use scan.nextLine(); other ways its not going to work(Do not use scan.next();
String user_input = scan.nextLine();
String[] stringsArray = user_input.split(" ");
now we need to convert these strings to Integers. create a for loop and convert every single index of stringArray :
for (int i = 0; i < stringsArray.length; i++) {
int x = Integer.parseInt(stringsArray[i]);
// Do what you want to do with these int value here
}
Best way is converting the whole stringArray to an intArray :
int[] intArray = new int[stringsArray.length];
for (int i = 0; i < stringsArray.length; i++) {
intArray[i] = Integer.parseInt(stringsArray[i]);
}
now do any proses you want like print or sum or... on intArray
The whole code will be like this :
import java.util.Scanner;
public class Main {
public static void main(String[] args) {
Scanner scan = new Scanner(System.in);
String user_input = scan.nextLine();
String[] stringsArray = user_input.split(" ");
int[] intArray = new int[stringsArray.length];
for (int i = 0; i < stringsArray.length; i++) {
intArray[i] = Integer.parseInt(stringsArray[i]);
}
}
}
This works fine ....
int a = nextInt();
int b = nextInt();
int c = nextInt();
Or you can read them in a loop
Using this on many coding sites:
CASE 1: WHEN NUMBER OF INTEGERS IN EACH LINE IS GIVEN
Suppose you are given 3 test cases with each line of 4 integer inputs separated by spaces 1 2 3 4, 5 6 7 8 , 1 1 2 2
int t=3,i;
int a[]=new int[4];
Scanner scanner = new Scanner(System.in);
while(t>0)
{
for(i=0; i<4; i++){
a[i]=scanner.nextInt();
System.out.println(a[i]);
}
//USE THIS ARRAY A[] OF 4 Separated Integers Values for solving your problem
t--;
}
CASE 2: WHEN NUMBER OF INTEGERS in each line is NOT GIVEN
Scanner scanner = new Scanner(System.in);
String lines=scanner.nextLine();
String[] strs = lines.trim().split("\\s+");
Note that you need to trim() first: trim().split("\\s+") - otherwise, e.g. splitting a b c will emit two empty strings first
int n=strs.length; //Calculating length gives number of integers
int a[]=new int[n];
for (int i=0; i<n; i++)
{
a[i] = Integer.parseInt(strs[i]); //Converting String_Integer to Integer
System.out.println(a[i]);
}
created this code specially for the Hacker earth exam
Scanner values = new Scanner(System.in); //initialize scanner
int[] arr = new int[6]; //initialize array
for (int i = 0; i < arr.length; i++) {
arr[i] = (values.hasNext() == true ? values.nextInt():null);
// it will read the next input value
}
/* user enter = 1 2 3 4 5
arr[1]= 1
arr[2]= 2
and soo on
*/
It's working with this code:
Scanner input = new Scanner(System.in);
System.out.println("Enter Name : ");
String name = input.next().toString();
System.out.println("Enter Phone # : ");
String phone = input.next().toString();
A simple solution can be to consider the input as an array.
Scanner sc = new Scanner(System.in);
int n = sc.nextInt(); //declare number of integers you will take as input
int[] arr = new int[n]; //declare array
for(int i=0; i<arr.length; i++){
arr[i] = sc.nextInt(); //take values
}
You're probably looking for String.split(String regex). Use " " for your regex. This will give you an array of strings that you can parse individually into ints.
Better get the whole line as a string and then use StringTokenizer to get the numbers (using space as delimiter ) and then parse them as integers . This will work for n number of integers in a line .
Scanner sc = new Scanner(System.in);
List<Integer> l = new LinkedList<>(); // use linkedlist to save order of insertion
StringTokenizer st = new StringTokenizer(sc.nextLine(), " "); // whitespace is the delimiter to create tokens
while(st.hasMoreTokens()) // iterate until no more tokens
{
l.add(Integer.parseInt(st.nextToken())); // parse each token to integer and add to linkedlist
}
Using BufferedReader -
StringTokenizer st = new StringTokenizer(buf.readLine());
while(st.hasMoreTokens())
{
arr[i++] = Integer.parseInt(st.nextToken());
}
When we want to take Integer as inputs
For just 3 inputs as in your case:
import java.util.Scanner;
Scanner scan = new Scanner(System.in);
int a,b,c;
a = scan.nextInt();
b = scan.nextInt();
c = scan.nextInt();
For more number of inputs we can use a loop:
import java.util.Scanner;
Scanner scan = new Scanner(System.in);
int a[] = new int[n]; //where n is the number of inputs
for(int i=0;i<n;i++){
a[i] = scan.nextInt();
}
This method only requires users to enter the "return" key once after they have finished entering numbers:
It also skips special characters so that the final array will only contains integers
ArrayList<Integer> nums = new ArrayList<>();
// User input
Scanner sc = new Scanner(System.in);
String n = sc.nextLine();
if (!n.isEmpty()) {
String[] str = n.split(" ");
for (String s : str) {
try {
nums.add(Integer.valueOf(s));
} catch (NumberFormatException e) {
System.out.println(s + " cannot be converted to Integer, skipping...");
}
}
}
//Get user input as a 1 2 3 4 5 6 .... and then some of the even or odd number like as 2+4 = 6 for even number
Scanner sc = new Scanner(System.in);
int n = sc.nextInt();
int evenSum = 0;
int oddSum = 0;
while (n > 0) {
int last = n % 10;
if (last % 2 == 0) {
evenSum += last;
} else {
oddSum += last;
}
n = n / 10;
}
System.out.println(evenSum + " " + oddSum);
}
}
if ur getting nzec error, try this:
try{
//your code
}
catch(Exception e){
return;
}
i know it's old discuss :) i tested below code it's worked
`String day = "";
day = sc.next();
days[i] = Integer.parseInt(day);`
Essentially I have a file of lines of integers. Each line has 9 digits. And I want to read the file. And then input each line into its an array. I want the array to be the same one each time. As I am going to do some processing to the array created from the first line. And then process the same array using a different line.
My input file is as follows:
8 5 3 8 0 0 4 4 0
8 5 3 8 0 0 4 2 2
And the current code that I am using is:
BufferedReader br = new BufferedReader(new FileReader("c:/lol.txt"));
Scanner sc = new Scanner(new File("c:/lol.txt"));
String line;
while (sc.hasNextLine()){
line = sc.nextLine();
int k = Integer.parseInt(line);
Now clearly I should be doing something more, I am just not really sure how to go about it.
Any help is greatly appreciated.
Try:
import java.util.Scanner;
import java.io.File;
public class Test {
public static void main(String[] args) throws Exception {
Scanner sc = new Scanner(new File("c:/lol.txt"));
while (sc.hasNext()) {
String line = sc.nextLine();
// get String array from line
String[] strarr = line.split(" "); // attention: split expect regular expression, not just delimiter!
// initialize array
int[] intarr = new int[strarr.length];
// convert each element to integer
for (int i = 0; i < strarr.length; i++) {
intarr[i] = Integer.valueOf(strarr[i]); // <= update array from new line
}
}
}
}
Of course, you should handle exception instead to pass it.