如何对所有对象调用一次函数？

Question

我很难理解如何只调用一次具有多个对象的函数（在for循环中）。

我试图写一个代码，其中对象的信息应该传递给另一个函数。

以下面的程序为例，函数被多次调用，而不是一次。

import java.util.Scanner;

class Attraction {

    String name;
    int open;

}

public class attractionGuide {

    public static void main(String[] args) {

        attractionDetails();
        System.exit(0);

    }

    public static Attraction createAttraction(String attractionName, int openingTime) {
        Attraction a = new Attraction();

        a.name = attractionName;
        a.open = openingTime;

        return a;
    }

    public static void attractionDetails() {

        Attraction TheEdenProject = createAttraction("The Eden Project", 9);
        Attraction LondonZoo = createAttraction("London Zoo", 10);
        Attraction TateModern = createAttraction("Tate Modern", 10);
        
        attractionInfo(TheEdenProject);
        attractionInfo(LondonZoo);
        attractionInfo(TateModern);

        // This is where the problem is^


   }

    public static Attraction attractionInfo(Attraction a) {

        Scanner scanner1 = new Scanner(System.in);
        System.out.print("Welcome. How many attractions do you need to know about? ");
        final int howMany = scanner1.nextInt();

        for (int i = 1; i <= howMany; i++) {

            System.out.print("\nName of attraction number number " + i + "?: ");
            Scanner scanner2 = new Scanner(System.in);
            String attraction_name = scanner2.nextLine();
            if (attraction_name.equalsIgnoreCase(a.name)) {
                System.out.println(a.name + " opens at " + a.open + "am.");
            } else {
                System.out.println("I have no information about that attraction.");
            }
        }
        return a;
    }
}

Welcome. How many attractions do you need to know about? 3     

Name of attraction number number 1?: the eden project
The Eden Project opens at 9am.

Name of attraction number number 2?: tate modern
I have no information about that attraction.

Name of attraction number number 3?: london zoo
I have no information about that attraction.
Welcome. How many attractions do you need to know about?

因为函数在循环中被调用了三次，所以一次只接受一个参数，而期望的输出应该是这样的：

Welcome. How many attractions do you need to know about? 3     

Name of attraction number number 1?: the eden project
The Eden Project opens at 9am.

Name of attraction number number 2?: tate modern
Tate Modern opens at 10am.

Name of attraction number number 3?: london zoo
London Zoo opens at 10am.

那么，我如何一次调用函数，让它一次接受所有参数呢？感谢您的帮助。谢谢

Answer 1

创建一个景点列表

List<Attraction> attractions = new ArrayList<>();

将每个景点添加到列表中：

attractions.add(TheEdenProject);
attractions.add(LondonZoo);
attractions.add(TateModern);

将列表传递给attractionInfo

attractionInfo(attractions);

将attractionInfo定义为

// Didn't see why you needed to return Attraction
public static void attractionInfo(List<Attraction> allKnown)

并按如下方式使用：

   Scanner scanner1 = new Scanner(System.in);
    System.out.print("Welcome. How many attractions do you need to know about? ");
    final int howMany = scanner1.nextInt();

    for (int i = 1; i <= howMany; i++) {

        System.out.print("\nName of attraction number number " + i + "?: ");
        Scanner scanner2 = new Scanner(System.in);
        String attraction_name = scanner2.nextLine();


        if (allKnown.contains(attraction_name)) {
            Attraction ofInterest = allKnown.get(attraction_name);
            System.out.println(ofInterest.name + " opens at " + ofInterest.open + "am.");
        } else {
            System.out.println("I have no information about that attraction.");
        }
    }

contains假设吸引力的等于，被实现为所需的名称比较。如果你需要帮助，尽管问。或者在SO中搜索[java]equals。