我有一个Garden类,我在其中序列化和反序列化多个Plant类对象。序列化工作正在进行,但是如果想要在mein静态方法中将其分配给调用变量,则反序列化不起作用。
public void searilizePlant(ArrayList<Plant> _plants) {
try {
FileOutputStream fileOut = new FileOutputStream(fileName);
ObjectOutputStream out = new ObjectOutputStream(fileOut);
for (int i = 0; i < _plants.size(); i++) {
out.writeObject(_plants.get(i));
}
out.close();
fileOut.close();
} catch (IOException ex) {
}
}
反序列化代码:
public ArrayList<Plant> desearilizePlant() {
ArrayList<Plant> plants = new ArrayList<Plant>();
Plant _plant = null;
try {
ObjectInputStream in = new ObjectInputStream(new FileInputStream(fileName));
Object object = in.readObject();
// _plant = (Plant) object;
// TODO: ITERATE OVER THE WHOLE STREAM
while (object != null) {
plants.add((Plant) object);
object = in.readObject();
}
in.close();
} catch (IOException i) {
return null;
} catch (ClassNotFoundException c) {
System.out.println("Employee class not found");
return null;
}
return plants;
}
我的调用代码:
ArrayList<Plant> plants = new ArrayList<Plant>();
plants.add(plant1);
Garden garden = new Garden();
garden.searilizePlant(plants);
// THIS IS THE PROBLEM HERE
ArrayList<Plant> dp = new ArrayList<Plant>();
dp = garden.desearilizePlant();
修改
我得到一个空指针异常
@NilsH的解决方案工作正常,谢谢!
答案 0 :(得分:18)
如何序列化整个列表呢?无需序列化列表中的每个单独对象。
public void searilizePlant(ArrayList<Plant> _plants) {
try {
FileOutputStream fileOut = new FileOutputStream(fileName);
ObjectOutputStream out = new ObjectOutputStream(fileOut);
out.writeObject(_plants);
out.close();
fileOut.close();
} catch (IOException ex) {
}
}
public List<Plant> deserializePlant() {
List<Plants> plants = null;
try {
ObjectInputStream in = new ObjectInputStream(new FileInputStream(fileName));
plants = in.readObject();
in.close();
}
catch(Exception e) {}
return plants;
}
如果这不能解决您的问题,请发布有关您的错误的更多详细信息。
答案 1 :(得分:0)
反序列化整个对象列表(例如,由于内存问题)可能并不总是可行的。在这种情况下,尝试:
ObjectInputStream in = new ObjectInputStream(new FileInputStream(
filename));
while (true) {
try {
MyObject o = (MyObject) in.readObject();
// Do something with the object
} catch (EOFException e) {
break;
}
}
in.close();
或者使用Java SE 7 try-with-resources语句:
try (ObjectInputStream in = new ObjectInputStream(new FileInputStream(
filename))) {
while (true) {
MyObject o = (MyObject) in.readObject();
// Do something with the object
}
} catch (EOFException e) {
return;
}
答案 2 :(得分:0)
如果将序列化为数组线性列表,则可以在反序列化时将其强制转换为数组线性列表 - 所有其他方法对我来说都失败了:
import java.io.*;
import java.util.ArrayList;
import java.util.Arrays;
public class Program
{
public static void writeToFile(String fileName, Object obj, Boolean appendToFile) throws Exception
{
FileOutputStream fs = null;
ObjectOutputStream os = null;
try
{
fs = new FileOutputStream(fileName);
os = new ObjectOutputStream(fs);
//ObjectOutputStream.writeObject(object) inherently writes binary
os.writeObject(obj); //this does not use .toString() & if you did, the read in would fail
}
catch (FileNotFoundException e)
{
e.printStackTrace();
}
catch (IOException e)
{
e.printStackTrace();
}
catch(Exception e)
{
e.printStackTrace();
}
finally
{
try
{
os.close();
fs.close();
}
catch(Exception e)
{
//if this fails, it's probably open, so just do nothing
}
}
}
@SuppressWarnings("unchecked")
public static ArrayList<Person> readFromFile(String fileName)
{
FileInputStream fi = null;
ObjectInputStream os = null;
ArrayList<Person> peopleList = null;
try
{
fi = new FileInputStream(fileName);
os = new ObjectInputStream(fi);
peopleList = ((ArrayList<Person>)os.readObject());
}
catch (FileNotFoundException e)
{
e.printStackTrace();
}
catch(EOFException e)
{
e.printStackTrace();
}
catch(ClassNotFoundException e)
{
e.printStackTrace();
}
catch (IOException e)
{
e.printStackTrace();
}
catch(Exception e)
{
e.printStackTrace();
}
finally
{
try
{
os.close();
fi.close();
}
catch(Exception e)
{
//if this fails, it's probably open, so just do nothing
}
}
return peopleList;
}
public static void main(String[] args)
{
Person[] people = { new Person(1, 39, "Coleson"), new Person(2, 37, "May") };
ArrayList<Person> peopleList = new ArrayList<Person>(Arrays.asList(people));
System.out.println("Trying to write serializable object array: ");
for(Person p : people)
{
System.out.println(p);
}
System.out.println(" to binary file");
try
{
//writeToFile("output.bin", people, false); //serializes to file either way
writeToFile("output.bin", peopleList, false); //but only successfully read back in using single cast
} // peopleList = (ArrayList<Person>)os.readObject();
// Person[] people = (Person[])os.readObject(); did not work
// trying to read one at a time did not work either (not even the 1st object)
catch (Exception e)
{
e.printStackTrace();
}
System.out.println("\r\n");
System.out.println("Trying to read object from file. ");
ArrayList<Person> foundPeople = null;
try
{
foundPeople = readFromFile("input.bin");
}
catch (Exception e)
{
// TODO Auto-generated catch block
e.printStackTrace();
}
if (foundPeople == null)
{
System.out.println("got null, hummm...");
}
else
{
System.out.println("found: ");
for(int i = 0; i < foundPeople.size(); i++)
{
System.out.println(foundPeople.get(i));
}
//System.out.println(foundPeople); //implicitly calls .toString()
}
}
}