Serializing Java objects to XML with XStream

Question

The problem is that every time I execute the main method, the old content of a.xml is lost and is substituted with a new one. How to append content to the a.xml file without losing the previous information?

import java.io.FileNotFoundException;
import java.io.PrintWriter;

import com.thoughtworks.xstream.XStream;
import com.thoughtworks.xstream.io.xml.DomDriver;


public class Test {
    public static void main(String[] args) throws FileNotFoundException {
        XStream xs = new XStream(new DomDriver());
        Foo f = new Foo(1, "booo", new Bar(42));
        PrintWriter pw = new PrintWriter("a.xml");
        xs.toXML(f,pw);
    }
}


public class Bar {
    public int id;

    public Bar(int id) {
        this.id = id;
    }

}


public class Foo {
    public int a;
    public String b;
    public Bar boo;
    public Foo(int a, String b, Bar c) {
        this.a = a;
        this.b = b;
        this.boo = c;
    }
}

Martin Klinke · Accepted Answer · 2010-06-29 18:10:43Z

3

The question is, do you really want to append the serialized XML string to the file or do you want to add the new Foo instance to the XML structure.

Appending on a string basis would result in invalid XML about like this:

<foo>
  <a>1</a>
  <b>booo</b>
  <bar>
    <id>42</id>
  </bar>
</foo>
<foo>
  <a>1</a>
  <b>booo</b>
  <bar>
    <id>42</id>
  </bar>
</foo>

Instead you may want to preserve the data in a.xml by parsing it first, then add the new element and serialize the whole collection/array.

So something like this (assuming there is already a collection of Foos in a.xml):

List foos = xs.fromXml(...);
foos.add(new Foo(1, "booo", new Bar(42)));
xs.toXml(foos, pw);

... which gives you something along the lines of this:

<foos>
  <foo>
    <a>1</a>
    <b>booo</b>
    <bar>
      <id>42</id>
    </bar>
  </foo>
  <foo>
    <a>1</a>
    <b>booo</b>
    <bar>
      <id>42</id>
    </bar>
  </foo>
</foos>

HTH

answered Jun 29, 2010 at 18:10

Martin Klinke

7,3426 gold badges44 silver badges64 bronze badges

Sign up to request clarification or add additional context in comments.

3 Comments

brain_damage Over a year ago

Yes, that's just what I want to do. But what if the file is empty? Then List foos = xs.fromXML(...) won't be valid :?

Martin Klinke Over a year ago

You'll have to handle some special cases, that's true. But depending on XStream you will either get an exception or null, which you can then catch or check and proceed from there.

user238033 Over a year ago

Yes this is one way to do it. However, loading and then saving all for the goal of saving can be quite a tedious task.

chedine · Accepted Answer · 2010-06-29 17:54:17Z

2

Sample Code

public static void main(String a[]){
  //Other code omitted
  FileOutputStream fos = new FileOutputStream("c:\\yourfile",true); //true specifies append
  Foo f = new Foo(1, "booo", new Bar(42));
  xs.toXML(f,fos);
}

answered Jun 29, 2010 at 17:54

chedine

2,3743 gold badges19 silver badges24 bronze badges

Collectives™ on Stack Overflow

Serializing Java objects to XML with XStream

2 Answers 2

3 Comments

Comments

Your Answer

Linked

Hot Network Questions

Collectives™ on Stack Overflow

2 Answers 2

3 Comments

Comments

Your Answer

Sign up or log in

Post as a guest

Linked

Related