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javaparsingjaxbxml-parsingmarshalling
John
John

Reputation: 135

Does JAXB always require XSD

http://www.oracle.com/technetwork/articles/javase/index-140168.html#binsch - This article says "JAXB requires that the XML document you want to access has a schema"

Here is a Hello World example , uses hello.xsd

http://jaxb.java.net/tutorial/section_1_3-Hello-World.html#Hello%20World -

And here is a Hello World example without using any xsd .

http://www.mkyong.com/java/jaxb-hello-world-example/

Does specification says that JAXB also works without xsd ,if that is true then how far the term "Binding" justifies. I believe binding happens through binding the schema.

Carification will be helpful . Thanks

Upvotes: 3

Views: 3955

Answers (3)

bdoughan
bdoughan

Reputation: 149047

JAXB 2 (JSR-222) implementations do not require an XML schema. You can start from an object model. You only need to add annotations where you want to override the default mapping behaviour.

Note: The link that you cited (http://www.oracle.com/technetwork/articles/javase/index-140168.html#binsch) was for JAXB 1 (JSR-31) which did require an XML schema. JAXB 2 which is what most people are referring to when they say JAXB does not require an XML schema.

Upvotes: 4

questzen
questzen

Reputation: 3297

xsd is just a formal representation of xml, analogous to 'Class' for an 'Object' . You can have a compliant xml without an xsd (in which case the xsd is hidden). While it is true that we start with xsd and generate java objects for the most part, you can do things the other way around as well. 'schemagen' actually lets you come up with a xsd for a given java class. This is just to say that there are multiple starting points.

Upvotes: 0

kkumar
kkumar

Reputation: 355

the Customer class in the tutorial is nothing but a binding object that will be generated from XSD using JAXB. The guy escaped that in his tutorial.

Upvotes: 0

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