我目前正在準備考試并正在執(zhí)行以下任務：如何將 ArrayList 傳遞給存儲列表數(shù)據(jù)的“保存”方法和將數(shù)據(jù)傳回的另一個“加載”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..根據(jù)任務我應該使用這些方法：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 {}這是應該產(chǎn)生以下輸出的主要方法：[威利·旺卡（WonkaWilly）、查理·巴克特（BucketCharlie）、喬爺爺（JoeGrandpa）][威利·旺卡（WonkaWilly）、查理·巴克特（BucketCharlie）、喬爺爺（JoeGrandpa）][威利·旺卡（WonkaWilly）、查理·巴克特（BucketCharlie）、喬爺爺（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);}}它應該看起來像這樣。但我不知道如何為 ArrayLists 做這件事。public static void save(String filename , Graph graph ) throws IOException{try (ObjectOutputStream out = new ObjectOutputStream(new BufferedOutputStream (new FileOutputStream (filename)))) {out.writeObject (graph);}}
2 回答
Qyouu
TA貢獻1786條經(jīng)驗 獲得超11個贊
由于您需要Person對象的輸出，因此我們需要重寫toString()類Person。
[威利·旺卡（WonkaWilly）、查理·巴克特（BucketCharlie）、喬爺爺（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 + ")");
}
}
有許多方法可以將對象寫入文件。這是與PrintWriter
將對象保存到文件
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
// 你可以使用序列化機制。要使用它，您需要執(zhí)行以下操作：
將Person類聲明為實現(xiàn)Serializable：
public class Person implements Serializable {
...
@Override
public String toString() {
return String.format(firstName + " " + lastName + "("+ lastName + firstName + ")");
}
}
將您的列表寫入文件：
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();
}
從文件中讀取列表：
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;
}
斯蒂芬大帝
TA貢獻1827條經(jīng)驗 獲得超8個贊
你可以嘗試這樣的事情：
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));
}
}
}
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