使用Java向arraylist添加参数

时间:2013-10-01 03:16:23

标签: java arraylist java.util.scanner

我正在测试我的大学课程,当用户输入“添加”时,它会向大学班级的arraylist添加一名新学生。

   Scanner myInput=new Scanner (System.in);
   String information=myInput.next(); //I know this is incorrect not sure what else to do
    directory.addStudent(information);

这里的目录是包含学生的arraylist的对象。 addStudent方法在College类中,如下所示:

  public void addStudent (String studentName, long studentID, String address) {
        Student newStu= new Student( studentname, studentID, address);
        collegeList.add(newStu); //here collegeList is the name of the arraylist in College class
   }

无论如何我的问题似乎很简单,但无法弄清楚。我想知道如何拆分信息,以便它可以满足addStudent方法中的参数...你可以看到信息将是一个字符串,但我希望studentID是一个长而不是字符串我该怎么办?

3 个答案:

答案 0 :(得分:2)

无需在阅读后拆分字符串。您可以使用Scanner读取分隔的字符串。

我假设您的输入由空格分隔。如果是这样,您需要这样做:

    String name = myInput.next();
    long id = myInput.nextLong();
    String address = myInput.next();
    dictionary.add(name, id, address);

答案 1 :(得分:0)

试试这个:

import java.util.ArrayList;
import java.util.List;
import java.util.Scanner;
public class StudentMain {
    private static List<Student> collegeList = new ArrayList<Student> (); 
    public static void main(String...args)throws Throwable {
         String s= null; String name=null; long id=0L; String address = null; 
        System.out.print("What do you want to do today, press q to quit : ");
        Scanner sc = new Scanner(System.in);
        do{
        try{    
                s = sc.next();
                if(s.equalsIgnoreCase("add")){
                    System.out.print("Enter the name of student:  ");
                    name = sc.next();
                    System.out.print("Enter the Id of student : " );
                    id=sc.nextLong();
                    System.out.print("Enter address of student:  ");
                    address = sc.next();
                    addStudent(name,id,address);
                }
        }catch(NumberFormatException e){
            if(!s.equalsIgnoreCase("q"))
                System.out.println("Please enter q to quit else try again ==> ");
            }
        }while(!s.equalsIgnoreCase("q"));
    sc.close();
    }
    private static void addStudent(String name, long id, String address) {
        // TODO Auto-generated method stub
        Student newStu = new Student(name,id,address);
        collegeList.add(newStu);        
    }

}

class Student{
    String name;
    Long id;
    String address;

    Student(String name,Long id,String address){
    this.name= name;
    this.id = id;
    this.address=address;
    }
}

答案 2 :(得分:0)

您可以使用split()方法破解信息

你要做的是

String str = new String(information);
String[] s = str.split(" ", 3);

我在空间基础上分裂,第二个参数3意味着您必须分成3个字符串,即名称,ID,地址

在此之后你可以获得name via s[0] , id via s[1] and address via s[2] 所以你可以通过addStudent()方法轻松传递。但您需要根据addStudent()方法

的参数转换值