Question

该程序将询问用户学生姓名和成绩，然后显示两者。用户输入的值存储在数组名称＆amp; 数组等级。我使用一个计数器控制的循环来将用户输入收集到数组中。如果我想为每个学生输入多个年级怎么办？编程相当新，任何输入或想法将非常感谢如何这样做...

public class Five {

public static void main(String[] args) {
    int students;

    Scanner input = new Scanner(System.in); //created input scanner


System.out.println("How many students are on your roster? If you wish to exit please type 00: ");// Initializing statement for program************
students = input.nextInt();

       if(students == 00) { //Exit program****************
       System.out.println("Maybe Next Time!!");
           System.exit(0);
         }

String[] names = new String[students];// Array names*******
String[]grade = new String[students]; //Array grade********

// Use Counter to go through Array**************************
        for(int counter =0; counter < students; counter++){
        System.out.println("Enter the name of a student: " +  (counter +1));
        names [counter] = input.next();
        System.out.println("Now enter that students grade A, B, C, D OR F: ");
        grade [counter] = input.next();
        }

input.close();// End Scanner object
//Use For loop for Printing names and grades entered by the user**************
System.out.println("Your students names and grades are as follows: ");
        for(int counter =0; counter < students; counter++){
        System.out.println("Name: " + names[counter]);
        System.out.println("Grade: " + grade[counter]);
        }
    }
}

Answer 1

您可以使用参差不齐的数组来输入多个等级你需要声明它，所以这是一种方法。

    String[] names = new String[students];// Array names*******
    String[][]grade = new String[names][]; //Array grade********

      for(int i=0; i<names[i];i++)
  {System.out.println("How many grades will you enter for "+names[i]+"?")
   int temp=input.nextInt();
  for(int j=0; j<names[i][temp-1];j++){
  grad[i][j]=input.next();         
           }}

Answer 2

您应该创建一个Student类和一个Grades类，并将它们存储为对象。使用数据结构，您最好的选择是hashmap。

HashMap<String, List<String>> grades = new HashMap<>();
grades.put("Jack", new ArrayList<>());
grades.get("Jack").add("A");

您可以找到有关HashMap here

的更多信息

在数组中存储多个用户输入

2 个答案: