java - 在文件中保存和加载列表

Question

我目前正在准备考试并正在完成以下任务:

如何将 ArrayList 传递给存储列表数据的“保存”方法以及将数据传回的另一个“加载”方法？

class Person {
    private String firstname;
    private String lastname;
    private String sortname;

public Person(String firstname, String lastname) {
        this.firstname = firstname;
        this.lastname = lastname;
        updateSortname();

//getter 和 setter..

根据任务我应该使用这些方法:

public static List<Person> load(String filename) throws IOException {
        return ??;
}


public static Person load(DataInputStream in) throws IOException {
        return ??;
}


public static void save(String filename, List<Person> list) throws IOException {
}


public static void save(DataOutputStream out, Person person) throws IOException {
}


public static List<Person> unserialize(String filename) throws IOException, ClassNotFoundException {
        return ??;
}


public static void serialize(String filename, List<Person> persons) throws IOException {
}

这是应产生以下输出的主要方法:

[Willy Wonka (WonkaWilly), Charlie Bucket (BucketCharlie), Grandpa Joe (JoeGrandpa)]

[Willy Wonka (WonkaWilly), Charlie Bucket (BucketCharlie), Grandpa Joe (JoeGrandpa)]

[Willy Wonka (WonkaWilly), Charlie Bucket (BucketCharlie), Grandpa Joe (JoeGrandpa)]

public class PersonTest {

public static void main(String[] args) throws IOException, ClassNotFoundException {
    List<Person> persons = new ArrayList<>();
    persons.add(new Person("Willy", "Wonka"));
    persons.add(new Person("Charlie", "Bucket"));
    persons.add(new Person("Grandpa", "Joe"));
    System.out.println(persons);

    Person.save("persons.sav", persons);
    persons = Person.load("persons.sav");
    System.out.println(persons);
    Person.serialize("persons.ser", persons);
    persons = Person.unserialize("persons.ser");
    System.out.println(persons);
}

}

它应该看起来像这样。但我不知道如何为 ArrayList 做到这一点。

public static void save(String filename , Graph graph ) throws IOException{

try (ObjectOutputStream out = new ObjectOutputStream(new BufferedOutputStream (new FileOutputStream (filename)))) {
out.writeObject (graph);
}
}

public static Graph load (String filename) throws IOException, ClassNotFoundException {

Graph graph = null;
try (ObjectInputStream in = new ObjectInputStream (new BufferedInputStream ( new FileInputStream (filename)))) {
graph = (Graph) in.readObject();
}
return graph;
}

Answer 1

由于您需要将 Person 对象输出为 as，我们需要重写 Person 类的 toString() 。

[Willy Wonka (WonkaWilly), Charlie Bucket (BucketCharlie), Grandpa Joe (JoeGrandpa)]

class Person {

//Respective Constructor, Getter & Setter methods


/* Returns the string representation of Person Class. 
 * The format of string is firstName lastName (lastNameFirstName)*/

  @Override
  public String toString() { 
    return String.format(firstName + " " + lastName + "("+ lastName + firstName + ")"); 
  } 
}

将对象写入文件的方法有很多。这是 PrintWriter

将对象保存到文件

public static void save(String filename, List<Person> list) throws IOException {
 PrintWriter pw = new PrintWriter(new FileOutputStream(fileName));
 for (Person person : list) {
    pw.println(person.toString());
   }
 pw.close();
}

或使用序列化

//您可以使用序列化机制。要使用它，您需要执行以下操作:

1. 将 Person 类声明为实现 Serialized:

   public class Person implements Serializable {
       ...
     @Override
     public String toString() { 
        return String.format(firstName + " " + lastName + "("+ lastName + firstName + ")"); 
     } 
   }
   

2. 将列表写入文件:

   public static void save(String filename, List<Person> list) throws IOException {
    FileOutputStream fos = new FileOutputStream(filename);
    ObjectOutputStream oos = new ObjectOutputStream(fos);
    oos.writeObject(list);
    oos.close();
   }
   

3. 从文件中读取列表:

   public static List<Person> load(String filename) throws IOException {
    FileInputStream fis = new FileInputStream(filename);
    ObjectInputStream ois = new ObjectInputStream(fis);
    List<Person> list = (List<Person>) ois.readObject();
    ois.close();
    return list;
   }