首页 文章

从HashMap打印值列表

提问于
浏览
2

我有两个类 Dog.javaDogSerach.java ,我想用HashMap打印狗的细节 . 我研究了这个问题的副本get string value from HashMap depending on key name并且还研究了Oracle doc http://docs.oracle.com/javase/tutorial/collections/interfaces/map.html,但仍然无法弄明白 .

我在DogSearch.java中尝试过

for (String key: dogs.keySet()) {
            System.out.println("Registration number : " + key);
           System.out.println("Detail : " +  dogs.get(key));
           }

但我明白了

Registration number : 1003
Detail : Dog [name=Luca, breed=Labrador, registrationNumber=1003]
Registration number : 1002
Detail : Dog [name=Gracie, breed=Rottweiler, registrationNumber=1002]
Registration number : 1001
Detail : Dog [name=Max, breed=German Shepherd, registrationNumber=1001]

我想像这样打印

Registration number: 1001
Name: Max
Breed: German Shepherd
... etc.

DogSearch.java

public class DogSearch {

    static Scanner scanner = new Scanner(System.in);
    public static void main(String[] args) {
        Map<String, Dog> dogs = new HashMap<String, Dog>();

        Dog max = new Dog("Max", "German Shepherd", "1001");
        Dog gracie = new Dog("Gracie", "Rottweiler", "1002");
        Dog luca = new Dog("Luca", "Labrador", "1003");


        dogs.put(max.getRegistrationNumber(), max);
        dogs.put(gracie.getRegistrationNumber(), gracie);
        dogs.put(luca.getRegistrationNumber(), luca);


        System.out.println("List of dogs by name: ");

        for (String key: dogs.keySet()) {
            System.out.println("Registration number : " + key);
            System.out.println("Breed : " +  dogs.get(key));

        }

    }
}

Dog.java

class Dog {
    private String name;
    private String breed;
    private String registrationNumber;

    public Dog(String name, String breed, String registrationNumber) {
        this.name = name;
        this.breed = breed;
        this.registrationNumber = registrationNumber;
    }


    public String getName() {
        return name;
    }

    public void setName(String name) {
        this.name = name;
    }

    public String getBreed() {
        return breed;
    }

    public void setBreed(String breed) {
        this.breed = breed;
    }

    public String getRegistrationNumber() {
        return registrationNumber;
    }

    public void setRegistrationNumber(String registrationNumber) {
        this.registrationNumber = registrationNumber;
    }

    @Override
    public int hashCode() {
        final int prime = 31;
        int result = 1;
        result = prime * result + ((breed == null) ? 0 : breed.hashCode());
        result = prime * result + ((name == null) ? 0 : name.hashCode());
        result = prime * result + ((registrationNumber == null) ? 0 : registrationNumber.hashCode());
        return result;
    }

    @Override
    public boolean equals(Object obj) {
        if (this == obj)
            return true;
        if (obj == null)
            return false;
        if (getClass() != obj.getClass())
            return false;
        Dog other = (Dog) obj;
        if (breed == null) {
            if (other.breed != null)
                return false;
        } else if (!breed.equals(other.breed))
            return false;
        if (name == null) {
            if (other.name != null)
                return false;
        } else if (!name.equals(other.name))
            return false;
        if (registrationNumber == null) {
            if (other.registrationNumber != null)
                return false;
        } else if (!registrationNumber.equals(other.registrationNumber))
            return false;
        return true;
    }

    @Override
    public String toString() {
        return "Dog [name=" + name + ", breed=" + breed + ", registrationNumber=" + registrationNumber + "]";
    }
}

3 回答

  • 1

    你可以这样做:

    for (String key: dogs.keySet()) {
            System.out.println("Registration number : " + key);
            System.out.println("Name: " +  dogs.get(key).getName());
            System.out.println("Breed: " +  dogs.get(key).getBreed());
           }
       }
    

    使用Java 8流,您可以根据您的狗的名称对 Map 进行排序 .

    Map<String, Dog> result = dogs.entrySet().stream() 
          .sorted(Map.Entry.comparingByValue(new MyComparator())) 
          .collect(Collectors.toMap(Map.Entry::getKey, Map.Entry::getValue, 
               (oldValue, newValue) -> oldValue, LinkedHashMap::new));
    

    然后遍历结果映射 .

    for (String key: result.keySet()) {
        System.out.println("Registration number : " + key);
         System.out.println("Name: " +  dogs.get(key).getName());
         System.out.println("Breed: " +  dogs.get(key).getBreed());
       }
    

    }

    而且你需要一个Comparator类

    public class MyComparator implements Comparator<Dog>{ 
    
         public int compare(Dog s1, Dog s2) { 
              return s1.getName().compareTo(s2.getName()); 
         } 
     }
    
  • 2

    有多种方法可以实现您的目标:

    第一个是更改 Dog.javatoString() 方法 . 使用 System.out.println() 时,java在将类作为参数传递时使用 toString() 方法 . 所以将你的_2915609改为:

    return "Name: " + name + "\n" + 
           "Breed: " + breed;
    

    应该做的伎俩 .

    第二种方法是更改在for循环中打印的内容 . 你可以做的一个例子是:

    for (String key: dogs.keySet()) {
        System.out.println("Registration number : " + key);
        Dog dog = dogs.get(key);
        System.out.println("Name : " +  dog.getName());
        System.out.println("Breed: " +  dog.getBreed());
    }
    
  • 0

    你为什么不在类 Dog 中创建一个方法并按照想要打印的方式编写实现!

    所以在这里你只需要

    for (String key: dogs.keySet()) {
        System.out.println(dogs.get(key).getPrintString());
    
    }
    

    这将在 Dog 课程中进行

    public String getPrintString() { 
     return "Registration number : "+ getRegistrationNumber() +"\nName: " +getName() +"\nBreed:"+getBreed(); // This can be optimized further using StringBuffer
    }
    

相关问题