Xml Serialization in Java using Simple

So, I have to serialize some code in Java and I have used Simple (a java Xml Serialization library) before for Xml serialization with Android in Java, so I thought I would use it again for a regular java project.

Here is what I have done. Some examples have failed, some have succeed. Here are my results.

Example 1 – Serializing a simple Person object

Attempt 1 – Failed with exception

import org.simpleframework.xml.Element;
import org.simpleframework.xml.Root;

@Root
public class Person
{
	@Element
	public String FirstName;

	@Element
	public String LastName;
}

And here is the Run.java with the main method.

import java.io.File;
import org.simpleframework.xml.Serializer;
import org.simpleframework.xml.core.Persister;

public class Run
{
	public static void main(String[] args) throws Exception
	{
		Person p = new Person();

		Serializer serializer = new Persister();
		File file = new File("person.xml");

		serializer.write(p, file);
	}
}

Result

An Exception was thrown because FirstName is null. So maybe it cannot handle null values?

Attempt 2 – Succeeded

Lets try be setting the default values to an empty string.

import org.simpleframework.xml.Element;
import org.simpleframework.xml.Root;

@Root
public class Person
{
	@Element
	public String FirstName = "";

	@Element
	public String LastName = "";
}

It worked. Here is the xml file.

<person>
   <FirstName>
   <LastName>
</person>

Attempt 3 – Succeeded

Of course if we set the values for FirstName and LastName…

    Person p = new Person();
    p.FirstName = "Jared";
    p.LastName = "Barneck";

…they show in the Xml as well.

<person>
   <FirstName>Jared</FirstName>
   <LastName>Barneck</LastName>
</person>

Example 2 – Serializing a Person object with getters and setters

Attempt 1 – Succeeded

I changed the member variables to be private and created public getters and setters.

import org.simpleframework.xml.Element;
import org.simpleframework.xml.Root;

@Root
public class Person
{
	@Element
	private String FirstName = "";

	@Element
	private String LastName = "";

	public String getFirstName()
	{
		return FirstName;
	}

	public void setFirstName(String inFirstName)
	{
		FirstName = inFirstName;
	}

	public String getLastName()
	{
		return LastName;
	}

	public void setLastName(String inLastName)
	{
		LastName = inLastName;
	}
}

I now use the setters to set the values in the main method.

import java.io.File;
import org.simpleframework.xml.Serializer;
import org.simpleframework.xml.core.Persister;

public class Run
{
	public static void main(String[] args) throws Exception
	{
		Person p = new Person();
		p.setFirstName("Jared");
		p.setLastName("Barneck");

		Serializer serializer = new Persister();
		File file = new File("person.xml");

		serializer.write(p, file);
	}

}

That worked and output the desired Xml.

<person>
   <FirstName>Jared</FirstName>
   <LastName>Barneck</LastName>
</person>

Example 3 – Using a custom name

What if the name of the member variables were _FirstName and _LastName. We wouldn’t want the underscore “_” to show up in the Xml.

So what do we do? The documentation says to use this line:

@Element(name=”FirstName”)

Attempt 1 – Success

import org.simpleframework.xml.Element;
import org.simpleframework.xml.Root;

@Root
public class Person
{
	@Element(name="FirstName")
	private String _FirstName = "";

	@Element(name="LastName")
	private String _LastName = "";

	public String getFirstName()
	{
		return _FirstName;
	}

	public void setFirstName(String inFirstName)
	{
		_FirstName = inFirstName;
	}

	public String getLastName()
	{
		return _LastName;
	}

	public void setLastName(String inLastName)
	{
		_LastName = inLastName;
	}
}

This worked, and the Xml output was unchanged.

Example 3 – Serializing a List

Ok, now we want to serialize a list of Person objects.

Attempt 1 – Failed with exception

I tried to use an ArrayList<Person> and it didn’t work. The person object is the same as in example 2, but I changed the main method as follows:

import java.io.File;
import java.util.ArrayList;
import org.simpleframework.xml.Serializer;
import org.simpleframework.xml.core.Persister;

public class Run
{
	public static void main(String[] args) throws Exception
	{
		ArrayList<Person> people = new ArrayList<Person>();

		Person p1 = new Person();
		p1.setFirstName("Jared");
		p1.setLastName("Barneck");
		people.add(p1);

		Person p2 = new Person();
		p2.setFirstName("Mike");
		p2.setLastName("Michaels");
		people.add(p2);

		Serializer serializer = new Persister();
		File file = new File("people.xml");

		serializer.write(people, file);
	}
}

So that didn’t work.

Attempt 2 – Failed, no exception, just bad Xml

I created a People class that extends ArrayList<Person>.

import java.util.ArrayList;
import org.simpleframework.xml.Root;

@Root
public class People extends ArrayList<Person>
{
}

I then used this class in the main method.

	public static void main(String[] args) throws Exception
	{
		People people = new People();

		Person p1 = new Person();
		p1.setFirstName("Jared");
		p1.setLastName("Barneck");
		people.add(p1);

		Person p2 = new Person();
		p2.setFirstName("Mike");
		p2.setLastName("Michaels");
		people.add(p2);

		Serializer serializer = new Persister();
		File file = new File("people.xml");

		serializer.write(people, file);
	}

It didn’t throw and exception but the Xml was basically empty.

<people/>

That isn’t what we want.

Attempt 3 – Succeeded but not quite right

Ok, so how about a container object instead of an extending object. The documentation has an @ElementList attribute that can be used if the class contains a list. So lets use that.

import java.util.ArrayList;
import java.util.List;
import org.simpleframework.xml.ElementList;
import org.simpleframework.xml.Root;

@Root
public class People
{
	@ElementList
	List<Person> List = new ArrayList<Person>();
}

The main method changed slightly to accommodate the People class.

	public static void main(String[] args) throws Exception
	{
		People people = new People();

		Person p1 = new Person();
		p1.setFirstName("Jared");
		p1.setLastName("Barneck");
		people.List.add(p1);

		Person p2 = new Person();
		p2.setFirstName("Mike");
		p2.setLastName("Michaels");
		people.List.add(p2);

		Serializer serializer = new Persister();
		File file = new File("people.xml");

		serializer.write(people, file);
	}

The Xml was created and looks as follows:

<people>
   <List class="java.util.ArrayList">
      <person>
         <FirstName>Jared</FirstName>
         <LastName>Barneck</LastName>
      </person>
      <person>
         <FirstName>Mike</FirstName>
         <LastName>Michaels</LastName>
      </person>
   </List>
</people>

That is almost correct. However, we don’t need to the List line between People and person.

Attempt 4 – Succeeded

Ok, so I read the documentation and it said to use this to remove the useless List Xml element.

@ElementList(inline=true)

So I tried and sure enough, it worked.

<people>
   <person>
      <FirstName>Jared</FirstName>
      <LastName>Barneck</LastName>
   </person>
   <person>
      <FirstName>Mike</FirstName>
      <LastName>Michaels</LastName>
   </person>
</people>

Ok…lets continue this a bit later in another post.

Xml Serialization in Java using Simple – Inheritance

One Comment

  1. lesto says:

    null element are valid, you just have to specify (required=false), like "@Element(required=false)"

    another little trick is that you can use non-void constructor just specifying with variable they use:

    @Root
    public class OrderManager {
    @ElementList
    private final List orders;

    public OrderManager(@ElementList(name="orders") List orders) {
    this.orders = orders;
    }

    }

Leave a Reply

How to post code in comments?

*