JAXB does EXACTLY what you want. It's built into the JRE/JDK starting at 1.6
To expand on the "use JAXB" comments above,
In Windows
"%java_home%\bin\xjc" -p [your namespace] [xsd_file].xsd
e.g.,
"%java_home%\bin\xjc" -p com.mycompany.quickbooks.obj quickbooks.xsd
Wait a bit, and if you had a well-formed XSD file, you will get some well-formed Java classes
If you want to start coding Java to XML and XML to Java in less than 5 minutes, try Simple XML Serialization. Don't spend hours learning the JAXB API http://simple.sourceforge.net/download/stream/doc/tutorial/tutorial.php
However, if you are really keen on learning JAXB, here's an excellent tutorial http://blogs.oracle.com/teera/entry/jaxb_for_simple_java_xml
Contents of tutorial:
JAXB for simple Java-XML serialization
There're a number of way to do XML serialization in Java. If you want fine-grained control over parsing and serialization you can go for SAX, DOM, or Stax for better performance. Yet, what I often want to do is a simple mapping between POJOs and XML. However, creating Java classes to do XML event parsing manually is not trivial. I recently found JAXB to be a quick and convenient Java-XML mapping or serialization.
JAXB contains a lot of useful features, you can check out the reference implementation here. Kohsuke's Blog is also a good resource to learn more about JAXB. For this blog entry, I'll show you how to do a simple Java-XML serialization with JAXB.
POJO to XML
Let's say I have an Item Java object. I want to serialize an Item object to XML format. What I have to do first is to annotate this POJO with a few XML annotation from javax.xml.bind.annotation.* package. See code listing 1 for Item.java
From the code
@XmlRootElement(name="Item")
indicates that I want to be the root element.@XmlType(propOrder = {"name", "price"})
indicates the order that I want the element to be arranged in XML output.@XmlAttribute(name="id", ...)
indicates that id is an attribute to root element.@XmlElement(....)
indicates that I want price and name to be element within Item.My Item.java
is ready. I can then go ahead and create JAXB script for marshaling Item.
//creating Item data object
Item item = new Item();
item.setId(2);
item.setName("Foo");
item.setPrice(200);
.....
JAXBContext context = JAXBContext.newInstance(item.getClass());
Marshaller marshaller = context.createMarshaller();
//I want to save the output file to item.xml
marshaller.marshal(item, new FileWriter("item.xml"));
For complete code Listing please see Code Listing 2 main.java
. The output Code Listing 3 item.xml
file is created. It looks like this:
<?xml version="1.0" encoding="UTF-8" standalone="yes"?>
<ns1:item ns1:id="2" xmlns:ns1="http://blogs.sun.com/teera/ns/item">
<ns1:itemName>Foo</ns1:itemName>
<ns1:price>200</ns1:price>
</ns1:item>
Easy right? You can alternatively channel the output XML as text String, Stream, Writer, ContentHandler, etc by simply change the parameter of the marshal(...) method like
...
JAXBContext context = JAXBContext.newInstance(item.getClass());
Marshaller marshaller = context.createMarshaller();
// save xml output to the OutputStream instance
marshaller.marshal(item, <java.io.OutputStream instance>);
...
JAXBContext context = JAXBContext.newInstance(item.getClass());
Marshaller marshaller = context.createMarshaller();
StringWriter sw = new StringWriter();
//save to StringWriter, you can then call sw.toString() to get java.lang.String
marshaller.marshal(item, sw);
XML to POJO
Let's reverse the process. Assume that I now have a piece of XML string data and I want to turn it into Item.java object. XML data (Code listing 3) looks like
<?xml version="1.0" encoding="UTF-8" standalone="yes"?>
<ns1:item ns1:id="2" xmlns:ns1="http://blogs.sun.com/teera/ns/item">
<ns1:itemName>Bar</ns1:itemName>
<ns1:price>80</ns1:price>
</ns1:item>
I can then unmarshal this xml code to Item object by
...
ByteArrayInputStream xmlContentBytes = new ByteArrayInputStream (xmlContent.getBytes());
JAXBContext context = JAXBContext.newInstance(Item.getClass());
Unmarshaller unmarshaller = context.createUnmarshaller();
//note: setting schema to null will turn validator off
unmarshaller.setSchema(null);
Object xmlObject = Item.getClass().cast(unmarshaller.unmarshal(xmlContentBytes));
return xmlObject;
...
For complete code Listing please see Code Listing 2 (main.java). The XML source can come in many forms both from Stream and file. The only difference, again, is the method parameter:
...
unmarshaller.unmarshal(new File("Item.xml")); // reading from file
...
// inputStream is an instance of java.io.InputStream, reading from stream
unmarshaller.unmarshal(inputStream);
Validation with XML Schema
Last thing I want to mention here is validating input XML with schema before unmarshalling to Java object. I create an XML schema file called item.xsd. For complete code Listing please see Code Listing 4 (Item.xsd). Now what I have to do is register this schema for validation.
...
Schema schema = SchemaFactory.newInstance(XMLConstants.W3C_XML_SCHEMA_NS_URI)
.newSchema(new File("Item.xsd"));
unmarshaller.setSchema(schema); //register item.xsd shcema for validation
...
When I try to unmarshal XML data to POJO, if the input XML is not conformed to the schema, exception will be caught. For complete code Listing please see Code Listing 5 (invalid_item.xml).
javax.xml.bind.UnmarshalException
- with linked exception:
javax.xml.bind.JAXBException caught: null
[org.xml.sax.SAXParseException: cvc-datatype-valid.1.2.1: 'item1' is
not a valid value for 'integer'.]
Here I change the 'id' attribute to string instead of integer.
If XML input is valid against the schema, the XML data will be unmarshalled to Item.java object successfully.
Using Eclipse IDE:-
the easiest way is using command line. Just type in directory of your .xsd file:
xjc myFile.xsd.
So, the java will generate all Pojos.
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