如何使用 java 在数组中获取用户输入？

Question

how to take user input in Array using Java? ie we are not initializing it by ourself in our program but the user is going to give its value.. please guide!!

Answer 1

Here's a simple code that reads strings from stdin , adds them into List<String> , and then uses toArray to convert it to String[] (if you really need to work with arrays).

import java.util.*;

public class UserInput {
    public static void main(String[] args) {
        List<String> list = new ArrayList<String>();
        Scanner stdin = new Scanner(System.in);

        do {
            System.out.println("Current list is " + list);
            System.out.println("Add more? (y/n)");
            if (stdin.next().startsWith("y")) {
                System.out.println("Enter : ");
                list.add(stdin.next());
            } else {
                break;
            }
        } while (true);
        stdin.close();
        System.out.println("List is " + list);
        String[] arr = list.toArray(new String[0]);
        System.out.println("Array is " + Arrays.toString(arr));
    }
}

---

See also:

• Why is it preferred to use Lists instead of Arrays in Java?
• Fill a array with List data

Answer 2

package userinput;

import java.util.Scanner;

public class USERINPUT {

    public static void main(String[] args) {

        Scanner input = new Scanner(System.in);

        //allow user  input;
        System.out.println("How many numbers do you want to enter?");
        int num = input.nextInt();

        int array[] = new int[num];

        System.out.println("Enter the " + num + " numbers now.");

        for (int i = 0 ; i < array.length; i++ ) {
           array[i] = input.nextInt();
        }

        //you notice that now the elements have been stored in the array .. array[]

        System.out.println("These are the numbers you have entered.");
        printArray(array);

        input.close();

    }

    //this method prints the elements in an array......
    //if this case is true, then that's enough to prove to you that the user input has  //been stored in an array!!!!!!!
    public static void printArray(int arr[]){

        int n = arr.length;

        for (int i = 0; i < n; i++) {
            System.out.print(arr[i] + " ");
        }
    }

}

Answer 3

import java.util.Scanner;

class bigest {
    public static void main (String[] args) {
        Scanner input = new Scanner(System.in);

        System.out.println ("how many number you want to put in the pot?");
        int num = input.nextInt();
        int numbers[] = new int[num];

        for (int i = 0; i < num; i++) {
            System.out.println ("number" + i + ":");
            numbers[i] = input.nextInt();
        }

        for (int temp : numbers){
            System.out.print (temp + "\t");
        }

        input.close();
    }
}

Answer 4

You can do the following:

    import java.util.Scanner;

    public class Test {

            public static void main(String[] args) {

            int arr[];
            Scanner scan = new Scanner(System.in);
            // If you want to take 5 numbers for user and store it in an int array
            for(int i=0; i<5; i++) {
                System.out.print("Enter number " + (i+1) + ": ");
                arr[i] = scan.nextInt();    // Taking user input
            }

            // For printing those numbers
            for(int i=0; i<5; i++) 
                System.out.println("Number " + (i+1) + ": " + arr[i]);
        }
    }

Answer 5

It vastly depends on how you intend to take this input, ie how your program is intending to interact with the user.

The simplest example is if you're bundling an executable - in this case the user can just provide the array elements on the command-line and the corresponding array will be accessible from your application's main method.

Alternatively, if you're writing some kind of webapp, you'd want to accept values in the doGet / doPost method of your application, either by manually parsing query parameters, or by serving the user with an HTML form that submits to your parsing page.

If it's a Swing application you would probably want to pop up a text box for the user to enter input. And in other contexts you may read the values from a database/file, where they have previously been deposited by the user.

Basically, reading input as arrays is quite easy, once you have worked out a way to get input . You need to think about the context in which your application will run, and how your users would likely expect to interact with this type of application, then decide on an I/O architecture that makes sense.

Answer 6

import java.util.Scanner;

class Example{

//Checks to see if a string is consider an integer.

public static boolean isInteger(String s){

    if(s.isEmpty())return false;

    for (int i = 0; i <s.length();++i){

        char c = s.charAt(i);

        if(!Character.isDigit(c) && c !='-')

            return false;
    }

    return true;
}

//Get integer. Prints out a prompt and checks if the input is an integer, if not it will keep asking.

public static int getInteger(String prompt){
    Scanner input = new Scanner(System.in);
    String in = "";
    System.out.println(prompt);
    in = input.nextLine();
    while(!isInteger(in)){
        System.out.println(prompt);
        in = input.nextLine();
    }
    input.close();
    return Integer.parseInt(in);
}

public static void main(String[] args){
    int [] a = new int[6];
    for (int i = 0; i < a.length;++i){
        int tmp = getInteger("Enter integer for array_"+i+": ");//Force to read an int using the methods above.
        a[i] = tmp;
    }

}

}

Answer 7

**How to accept array by user Input

Answer:-

import java.io.*;

import java.lang.*;

class Reverse1  {

   public static void main(String args[]) throws IOException {

     int a[]=new int[25];

     int num=0,i=0;

     BufferedReader br=new BufferedReader(new InputStreamReader(System.in));

     System.out.println("Enter the Number of element");

     num=Integer.parseInt(br.readLine());

     System.out.println("Enter the array");

     for(i=1;i<=num;i++) {
        a[i]=Integer.parseInt(br.readLine());
     }

     for(i=num;i>=1;i--) {
        System.out.println(a[i]);    
     }

   }

}

Answer 8

int length;
    Scanner input = new Scanner(System.in);
    System.out.println("How many numbers you wanna enter?");
    length = input.nextInt();
    System.out.println("Enter " + length + " numbers, one by one...");
    int[] arr = new int[length];
    for (int i = 0; i < arr.length; i++) {
        System.out.println("Enter the number " + (i + 1) + ": ");
        //Below is the way to collect the element from the user
        arr[i] = input.nextInt();

        // auto generate the elements
        //arr[i] = (int)(Math.random()*100);
    }
    input.close();
    System.out.println(Arrays.toString(arr));

Answer 9

This is my solution if you want to input array in java and no. of input is unknown to you and you don't want to use List<> you can do this.

but be sure user input all those no. in one line seperated by space

 BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
 int[] arr = Arrays.stream(br.readLine().trim().split(" ")).mapToInt(Integer::parseInt).toArray();