我想创建一个方法,显示我所创建的对象数组的所有数据。 我设法通过这里的人的帮助创建一个特定的搜索。 我能用他的代码吗?怎么样?我想要创建的方法是最后一部分。 如何在对象上的数组上显示我的所有数据
import java.util.Dictionary;
import java.util.Hashtable;
import java.util.Scanner;
public class TestMouse {
private static Dictionary<String, Mouse> mouseList = new Hashtable<String, Mouse>();
public static void main(String[] args) {
menu();
}
@SuppressWarnings("resource")
public static void menu() {
Scanner input = new Scanner(System.in);
int choice;
do {
System.out.println("CMPE 325 Student Record Holder System");
System.out.println("--------------------------------");
System.out.println("1.Add Student");
System.out.println("2.View Records");
System.out.println("3.Update Students");
System.out.println("4.Get Average");
System.out.println("5.Exit");
System.out.println();
System.out.println("--------------------------------");
System.out.print("Enter your choice: ");
choice = input.nextInt();
switch (choice) {
case 1:
record();
break;
case 2:
display();
// display();
break;
case 3:
menu();
// update();
// break();
case 4:
menu();
// average();
// break();
case 5:
break;
default:
continue;
}
} while (choice != 5);
System.out.println();
System.out.println("Good-Bye");
}
// end of menu(); method.
// ------------------------------------------------------------
@SuppressWarnings({ "resource", "unused" })
public static void record() {
Scanner Input = new Scanner(System.in);
int total;
System.out.println("How many students you are going to input? ");
total = Input.nextInt();
Mouse[] keyboard = new Mouse[total];
for (int index = 0; index < total; index++) {
String space;
keyboard[index] = new Mouse();
System.out.printf("Student[%d]", index + 1);
System.out.println();
System.out.println("");
space = Input.nextLine();
System.out.println("ID Number: ");
keyboard[index].setId(Input.nextLine());
System.out.println("First Name: ");
keyboard[index].setFirstName(Input.nextLine());
System.out.println("Middle Name: ");
keyboard[index].setMiddleName(Input.nextLine());
System.out.println("Last Name: ");
keyboard[index].setLastName(Input.nextLine());
System.out.println("Degree: ");
keyboard[index].setDegName(Input.nextLine());
System.out.println("Year Level: ");
keyboard[index].setYear(Input.nextInt());
//Save current object
mouseList.put(keyboard[index].getId(), keyboard[index]);
}
for (int index2 = 0; index2 < total; index2++) {
System.out.printf("Student[%d]", index2 + 1);
System.out.println();
System.out.println("ID Number:" + keyboard[index2].getId());
System.out.println("Name:" + keyboard[index2].getFirstName() + " "
+ keyboard[index2].getMiddleName() + " "
+ keyboard[index2].getLastName());
System.out.println("Degree:" + keyboard[index2].getDegName());
System.out.println("Year Level:" + keyboard[index2].getYear());
}
}
public static void specific() {
String id = "";
System.out.println("Enter an Id Number");
Scanner Input = new Scanner(System.in);
id = Input.nextLine();
Mouse mouse = mouseList.get(id);
if (mouse != null) {
System.out.println();
System.out.println("ID Number:" + mouse.getId());
System.out.println("Name:" + mouse.getFirstName() + " "
+ mouse.getMiddleName() + " "
+ mouse.getLastName());
System.out.println("Degree:" + mouse.getDegName());
System.out.println("Year Level:" + mouse.getYear());
}
}
@SuppressWarnings("resource")
public static void display(){
Scanner input = new Scanner(System.in);
int choice;
System.out.println("--------------------------------");
System.out.println("1.View List");
System.out.println("2.View Specific Record");
System.out.println("3.Exit");
System.out.println();
System.out.println("--------------------------------");
System.out.print("Enter your choice: ");
choice = input.nextInt();
switch (choice) {
case 1:
displayall();
break;
case 2:
specific();
break;
case 3:
menu();
break;
}
}
this part is the one im talking about. i hope u can help me
public static void displayall(){
String id = "";
System.out.println("Enter an Id Number");
Scanner Input = new Scanner(System.in);
id = Input.nextLine();
Mouse mouse = mouseList.get(id);
if (mouse != null) {
System.out.println();
System.out.println("ID Number:" + mouse.getId());
System.out.println("Name:" + mouse.getFirstName() + " "
+ mouse.getMiddleName() + " "
+ mouse.getLastName());
System.out.println("Degree:" + mouse.getDegName());
System.out.println("Year Level:" + mouse.getYear());
}
}
}
答案 0 :(得分:0)
因为你正在使用the legacy Dictionary
interface,这有点烦人,但是可以这样做。
你想要做的是创建一个Enumeration
元素,然后迭代它。该实例是通过调用Dictionary#elements()
或更具体地mouseList.elements()
来提供的。
然后迭代它as prescribed by Enumeration
.
如果Mouse
也覆盖toString()
,那实际上要容易得多,因为这是打印这些复杂对象的最佳方式。
最后,你在那个方法中做了很多不必要的调用 - 你不关心ID或类似的东西。您也无需检查null
,因为如果它不在元素集中,则不会在Enumeration
中。
如果你想更现代一点......那么请改用Map
。 Hashtable
已使用Map
接口,因此您无需更改具体对象(尽管我强烈推荐)。迭代Map
的元素会成为对Map#entrySet()
的调用,但需要更多的工作。