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Function:
When serializing data, if you don’t want to serialize one of the attributes into a file, you need to use transient Modification, indicating that the attribute is a temporary attribute
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This is a student class:
public class Student implements Serializable {//注意:要想序列化,必须实现Serializable接口 private String name; private Integer age; private transient String address; //使用transient修饰 public Student() { } public Student(String name, Integer age, String address) { this.name = name; this.age = age; this.address = address; } //Getter/Setter }
When I serialize, I don’t plan to save the student’s address attribute. I only want to save the name and age attributes. I modify the adress attribute with the transient keyword and serialize it as follows:
public class TestStudent { public static void main(String[] args) throws IOException { List<Student> list = new ArrayList<>(); Student s1 = new Student("Jack", 20, "北京"); Student s2 = new Student("Rose", 21, "上海"); Student s3 = new Student("Hoke", 22, "深圳"); Student s4 = new Student("Mark", 23, "天津"); Student s5 = new Student("Json", 24, "成都"); list.add(s1); list.add(s2); list.add(s3); list.add(s4); list.add(s5); //将学生信息序列化到student.txt文件中 File file = new File("student.txt"); ObjectOutputStream oos = new ObjectOutputStream(new FileOutputStream(file)); oos.writeObject(list); } }
Proceed below Deserialize and verify the role of transient:
@Test public void test() throws IOException, ClassNotFoundException { ObjectInputStream ois = new ObjectInputStream(new FileInputStream(new File("student.txt"))); Object object = ois.readObject(); if (object instanceof List) { List<Student> list = (List<Student>) object; list.forEach(System.out::println); } }
Result:
You can see that the address attribute value in the output result is null. The value is not serialized;
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