我正在制作代码以将要序列化的对象列表传输到文件中并返回。问题是当我第一次序列化文件时,使用默认构造函数而不是第二个构造函数来实例化对象。
换句话说,输出带有默认值:
0 N/A N/A 01-01-1980 [UCLA]
0 N/A N/A 01-01-1980 [UCLA]
0 N/A N/A 01-01-1980 [UCLA]
但它应该是:
1234 Robert Smith 07-05-1980 [UCLA]
2345 Donald Trump 07-05-1980 [UCLA]
3456 Barack Obama 07-05-1980 [UCLA]
这是我的主要方法:
public static void main(String[] args) throws IOException, ClassNotFoundException {
// ArrayList list
ArrayList<Student> al = new ArrayList<Student>();
Date d = new Date(80, 5, 7);
Student s = new Student("Robert", "Smith", 1234, d, "UCLA");
Student s2 = new Student("Donald", "Trump", 2345, d, "UCLA");
Student s3 = new Student("Barack", "Obama", 3456, d, "UCLA");
al.add(s);
al.add(s2);
al.add(s3);
// serialization test
FileOutputStream fileOut = new FileOutputStream("StudentList.dat");
ObjectOutputStream out = new ObjectOutputStream(fileOut);
out.writeObject(al);
out.close();
fileOut.close();
// deserialization test
FileInputStream fileIn = new FileInputStream("StudentList.dat");
ObjectInputStream in = new ObjectInputStream(fileIn);
ArrayList<Student> a2 = (ArrayList<Student>) in.readObject();
in.close();
fileIn.close();
System.out.println(a2.size());
for (Student i : a2) {
System.out.println(i);
} // for
} // main
由于
编辑:添加学生班
package edu.uga.cs1302.gui;
import java.io.FileInputStream;
import java.io.FileOutputStream;
import java.io.IOException;
import java.io.ObjectInputStream;
import java.io.ObjectOutputStream;
import java.io.Serializable;
import java.util.ArrayList;
import java.util.Date;
@SuppressWarnings({ "serial", "rawtypes", "deprecation", "unchecked" })
public class Student extends Person implements Serializable {
private String collegeName;
/*
* public Student() { super(); collegeName = null; } // constructor
**/
public void setC(String c) {
collegeName = c;
} // set college
public String getC() {
return collegeName;
} // get college
public Student(String fName, String lName, int n, Date d, String college) {
super(fName, lName, n, d);
collegeName = college;
} // second constructor
public String toString() {
return super.toString() + " [" + collegeName + "]";
} // to string
} // class
答案 0 :(得分:1)
正在发生的事情是,只有正在编写和读取实现Serializable的类的字段。
这就是正确编写和读取collegeName
的原因,因为它在您继承的类中实现了Serializable。其他字段属于基类,不会这样做。
在Student
中声明要分别序列化的变量,或者使Person
实现Serializable。