如何使用 Java 在 Array 中获取用户输入?即我们不是在我们的程序中自己初始化它,但用户会给出它的值..请指导!!
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8 回答
7
这是一个简单的代码,它从 中读取字符串stdin
,将它们添加到List<String>
中,然后用于toArray
将其转换为String[]
(如果您确实需要使用数组)。
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));
}
}
也可以看看:
于 2010-04-12T15:39:49.420 回答
6
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] + " ");
}
}
}
于 2011-10-09T08:14:44.217 回答
2
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();
}
}
于 2015-02-21T15:53:00.467 回答
1
您可以执行以下操作:
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]);
}
}
于 2018-09-24T12:10:22.040 回答
0
这在很大程度上取决于您打算如何接受此输入,即您的程序打算如何与用户交互。
最简单的例子是,如果您正在捆绑一个可执行文件——在这种情况下,用户只需在命令行上提供数组元素,相应的数组就可以从您的应用程序的main
方法中访问。
或者,如果您正在编写某种 web 应用程序,您可能希望通过手动解析查询参数或通过向用户提供提交到解析页面的 HTML 表单来接受应用程序的doGet
/方法中的值。doPost
如果它是 Swing 应用程序,您可能希望弹出一个文本框供用户输入输入。在其他情况下,您可以从数据库/文件中读取值,这些值以前由用户存放在其中。
基本上,一旦您找到了获取input的方法,将 input 读取为数组非常容易。您需要考虑您的应用程序将在其中运行的上下文,以及您的用户可能希望如何与此类应用程序交互,然后决定一个有意义的 I/O 架构。
于 2010-04-12T14:21:13.867 回答
0
导入 java.util.Scanner;
类示例{
//检查一个字符串是否被认为是一个整数。
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;
}
}
}
于 2010-04-12T16:19:48.067 回答
0
**如何通过用户输入接受数组
回答:-
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]);
}
}
}
于 2010-10-02T14:34:46.337 回答
0
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));
于 2020-02-12T04:57:45.160 回答