在文件中保存和加载列表

Question

我目前正在准备考试并正在执行以下任务：如何将 ArrayList 传递给存储列表数据的“保存”方法和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 {}这是应该产生以下输出的主要方法：[威利·旺卡（WonkaWilly）、查理·巴克特（BucketCharlie）、乔爷爷（JoeGrandpa）][威利·旺卡（WonkaWilly）、查理·巴克特（BucketCharlie）、乔爷爷（JoeGrandpa）][威利·旺卡（WonkaWilly）、查理·巴克特（BucketCharlie）、乔爷爷（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 ArrayLists 做这件事。public static void save(String filename , Graph graph ) throws IOException{try (ObjectOutputStream out = new ObjectOutputStream(new BufferedOutputStream (new FileOutputStream (filename)))) {out.writeObject (graph);}}

Answer 1

由于您需要Person对象的输出，因此我们需要重写toString()类Person。

[威利·旺卡（WonkaWilly）、查理·巴克特（BucketCharlie）、乔爷爷（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 + ")"); 

  } 

}

有许多方法可以将对象写入文件。这是与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();

}

或者使用序列化

// 你可以使用序列化机制。要使用它，您需要执行以下操作：

将Person类声明为实现Serializable：

public class Person implements Serializable {

    ...

  @Override

  public String toString() { 

     return String.format(firstName + " " + lastName + "("+ lastName + firstName + ")"); 

  } 

}

将您的列表写入文件：

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();

}

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;

}

Answer 2

你可以尝试这样的事情：

public static void save(String filename , ArrayList<Person> persons) throws IOException{

    try (ObjectOutputStream out = new ObjectOutputStream(new BufferedOutputStream (new FileOutputStream (filename)))) {

        for(int i = 0; i < persons.size; i++){

               out.writeObject(persons.get(i));

        }   

}

}