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我实现了一个地址簿和一个地址类！我想实现一个方法“searchForename”和“searchSurename”，这是我之前用“insert”添加的搜索元素。我查看了使用 ArrayList 搜索对象或元素的其他实现，但大多数情况下它们真的让我感到困惑，因为太多的代码没有任何解释。

第三次编辑新问题！：

第三次尝试：

import java.util.ArrayList;


public class Adressbook {

    private ArrayList<Adress> adresses = null;

    public Adressbook(){
        adresses = new ArrayList<Adress>();
    }

    public void insert(Adress adress){
        adresses.add(new Adress("Marvin","Wildman","Blastreet",9,12345,"Blatown"));
        adresses.add(new Adress("Anne","Wildman","Woodstreet",10,6789,"Exampletown"));
        adresses.add(new Adress("William","Wildman","Eaglestreet",11,73975,"Blubvalley"));
    }


    public void searchSurename(String surename){
        for(Adress s: adresses){
            if("Green".equals(surename)){
            System.out.println(s);
            }
        }

    }

    public void searchForename(String forename){
        for(Adress s: adresses){
            if("Anne".equals(forename)){
            System.out.println(s);
            }
        }
    }

    public String toString(){
        return null;

    }

}

我有几个问题：

1.Adressbook中的toString方法是怎样的？

2.Adress 类的 toString 是什么样子的？

3.Adress 类中的构造函数看起来正确吗？

4.我可以比这更容易/更有效地实现搜索方法吗？如果不正确，我该如何更改？

忘记我的班级地址：

public class Adress {
    public static String forename;
    public static String surename;
    public static String street;
    public static int houseno;
    public static int code;
    public static String state;



    public Adress(String forename, String surename, String street, int houseno, int code,String state){
        this.forename = forename;
        this.surename = surename;
        this.street = street;
        this.houseno = houseno;
        this.code = code;
        this.state = state;
    }



    public String toString(){
        return null;
    }

}
4

1 回答 1

4

Your method is a static method and that's why you can't use the this in a static context as there is no object instance in a call of a static method.

public static String searchSurename // static method

You need to remove the static from your method declaration.

In case you can't do the above change, then you'll have to make your ArrayList adresses as static.

Also, as a side note, use the equals() for String value comparisons. == is for object refernce comparison.

if(this.adresses.get(i).getSurename().equals(surename)) {
于 2013-10-26T14:30:34.647 回答