通过输入HashMap值不按预期工作来获取HashMap键

Question

问题出在getName()方法中

public class Phonebook implements PhonebookInterface {

    private Map<String, Set<String>> phonebook;
    private Map<String, Address> address;

    public Phonebook() {
        this.phonebook = new HashMap<String, Set<String>>();
        this.address = new HashMap<String, Address>();
    }

    @Override
    public void addNumber(String person, String number) {
        if (!this.phonebook.containsKey(person)) {
            this.phonebook.put(person, new HashSet<String>());
        }

        this.phonebook.get(person).add(number);
    }

    @Override
    public void getName(String number) {
        for (String person : this.phonebook.keySet()) {
            if (this.phonebook.get(person).contains(number)) {
                System.out.println(" " + person);
            } else {
                System.out.println("  not found");
            }
        }
    }

    @Override
    public void getNumber(String person) {
        try {
            for (String n : this.phonebook.get(person)) {
                if (this.phonebook.get(person).size() > 1) {
                    System.out.println(" " + n);
                } else {
                    System.out.println("number: " + n);
                }
            }
        } catch (Exception e) {
            System.out.println("  not found");
        }
    }

    public void addAddress(String person, String street, String city) {
        this.address.put(person, new Address(street, city));
    }

    public void getInfo(String person) {
        if (this.address.containsKey(person)) {
            System.out.println("  address: " + this.address.get(person));

            if (this.phonebook.containsKey(person)) {
                System.out.println("  phone numbers:");
                for (String n : this.phonebook.get(person)) {
                    System.out.println("  " + n);
                }
            } else {
                System.out.println("  phone number not found");
            }
        } else {
            System.out.println("  not found");
        }
    }

    public void removeInfo(String person) {
        if (this.address.containsKey(person)) {
            this.address.remove(person);
        } else {
            System.out.println(" not found");
        }
    }
}

public class Address {

    private String street;
    private String city;

    public Address() {
    }

    Address(String street, String city) {
        this.street = street;
        this.city = city;
    }

    public String getStreet() {
        return this.street;
    }

    public String getCity() {
        return this.city;
    }

    public String toString() {
        return this.street + " " + this.city;
    }

}

class Main {

    public static void main(String[] args) {
        Phonebook p = new Phonebook();

        System.out.println("Search for number: ");

        System.out.println("Pekka :");
        p.addNumber("Pekka", "014-1234");
        p.addNumber("Pekka", "015-5344");
        p.getNumber("Pekka");

        System.out.println("Matti :");
        p.addNumber("Matti", "013-4321");
        p.getNumber("Matti");

        System.out.println("\nsearch for a person by phone number: ");

        p.getName("013-4321");
        p.getName("1234567");
    }
}

预期的产出是：

搜索号码： 佩卡： 014-1234 015-5344 马蒂： 编号：013-4321

通过电话号码搜索某人： 马蒂 未找到

但输出是：

搜索号码： 佩卡： 014-1234 015-5344 马蒂： 编号：013-4321

按电话号码搜索某人： 未找到 马蒂 未找到 未找到

为什么输出打印“未找到”3次而不是一次？

Answer 1

迭代Map的所有键，并将相应的值与您要查找的值进行比较。如果没有匹配项，则打印“未找到”。

你的Map有两个键，所以如果你在Map中搜索一个值，你将打印匹配的键，但你也会打印另一个键的“not found”。

如果您要搜索Map中不存在的值，则会打印“未找到”两次（每个键一次）。

迭代所有键后，您应该只打印“未找到”：

public void getName(String number) {   
    for (String person : this.phonebook.keySet()) {   
        if (this.phonebook.get(person).contains(number)) {   
            System.out.println(" " + person);
            return;
        }    
    }
    System.out.println("  not found"); 
}