Welcome to OGeek Q&A Community for programmer and developer-Open, Learning and Share
Welcome To Ask or Share your Answers For Others

Categories

0 votes
321 views
in Technique[技术] by (71.8m points)

java - Does JAXB support default schema values?

I have a schema that defines default values for elements and attributes. I am trying to parse a document using JAXB based on that schema but JAXB is not setting the default values. Any ideas on how to make JAXB honor the default values from the schema?

example.xsd:

<?xml version="1.0" encoding="UTF-8"?><xs:schemaxmlns:xs="http://www.w3.org/2001/XMLSchema" 
targetNamespace="http://www.example.org/example" 
xmlns:tns="http://www.example.org/example">

<xs:element name="root" type="tns:rootType"/>

<xs:complexType name="rootType">
    <xs:sequence>
        <xs:element name="child" type="tns:childType"/>
    </xs:sequence>
</xs:complexType>

<xs:complexType name="childType">
    <xs:sequence>
        <xs:element name="childVal" type="xs:string" default="defaultElVal"/>
    </xs:sequence>
    <xs:attribute name="attr" type="xs:string" default="defaultAttrVal"/>
</xs:complexType>

example1.xml

<?xml version="1.0" encoding="UTF-8"?>
<tns:root xmlns:tns="http://www.example.org/example" xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance" xsi:schemaLocation="http://www.example.org/example example.xsd ">
  <child>
    <childVal/>
  </child>
</tns:root>

TestParser.java

package test;  
import java.io.File;  
import javax.xml.XMLConstants;  
import javax.xml.bind.JAXBContext;  
import javax.xml.bind.Unmarshaller;  
import javax.xml.validation.Schema;  
import javax.xml.validation.SchemaFactory;  
public class TestParser {    
    public static void main(String[] pArgs) {  
        try {  
            JAXBContext context = JAXBContext.newInstance(RootElement.class);  
            Unmarshaller unmarshaller = context.createUnmarshaller();  

            SchemaFactory schemaFac = SchemaFactory.newInstance(XMLConstants.W3C_XML_SCHEMA_NS_URI);
            Schema sysConfigSchema = schemaFac.newSchema(
                    new File("example.xsd"));
            unmarshaller.setSchema(sysConfigSchema);
            RootElement root = (RootElement)unmarshaller.unmarshal(
                    new File("example1.xml"));
            System.out.println("Child Val: " + root.getChild().getChildVal());
            System.out.println("Child Attr: " + root.getChild().getAttr());
        } catch (Exception e) {
            System.out.println(e.getMessage());
            e.printStackTrace();
        }
    }
}

RootElement.java

package test;  
import javax.xml.bind.annotation.XmlRootElement;  

@XmlRootElement(name="root", namespace="http://www.example.org/example")  
public class RootElement {  

    private ChildEl child;  

    public RootElement() {}  

    public ChildEl getChild() {
        return child;
   }

    public void setChild(ChildEl pChild) {
        this.child = pChild;
    }
}

ChildEl.java

package test;

import javax.xml.bind.annotation.XmlAttribute;
import javax.xml.bind.annotation.XmlRootElement;

@XmlRootElement(name="child")
public class ChildEl {

    private String attr;
    private String childVal;

    public ChildEl() {};

    @XmlAttribute
    public String getAttr() {
        return attr;
    }
    public void setAttr(String pAttr) {
        this.attr = pAttr;
    }

    public String getChildVal() {
        return childVal;
    }
    public void setChildVal(String pVal) {
        this.childVal = pVal;
    }

}
See Question&Answers more detail:os

与恶龙缠斗过久,自身亦成为恶龙;凝视深渊过久,深渊将回以凝视…
Welcome To Ask or Share your Answers For Others

1 Reply

0 votes
by (71.8m points)

Element Default Value

To get the default value on the element property you need to annotate it as follows:

@XmlElement(defaultValue="defaultElVal")
public String getChildVal() {
    return childVal;
}

Attribute Default Value

If you use EclipseLink JAXB (MOXy) you will get the default attribute value using the code you supplied. There may be a bug in the Metro implementation of JAXB that is preventing this from working. Note I lead the MOXy implementation.


Alternate Approach

The following code should work with any JAXB implementation without requiring any code changes to your model. You could do the following and leverage SAXSource:

import java.io.File;  
import java.io.FileInputStream;

import javax.xml.XMLConstants;  
import javax.xml.bind.JAXBContext;  
import javax.xml.bind.Unmarshaller;  
import javax.xml.parsers.SAXParserFactory;
import javax.xml.transform.sax.SAXSource;
import javax.xml.validation.Schema;  
import javax.xml.validation.SchemaFactory;  

import org.xml.sax.InputSource;
import org.xml.sax.XMLReader;
public class TestParser {    
    public static void main(String[] pArgs) {  
        try {  
            JAXBContext context = JAXBContext.newInstance(RootElement.class);  
            Unmarshaller unmarshaller = context.createUnmarshaller();  

            SchemaFactory schemaFac = SchemaFactory.newInstance(XMLConstants.W3C_XML_SCHEMA_NS_URI);
            Schema sysConfigSchema = schemaFac.newSchema(
                    new File("example.xsd"));

            SAXParserFactory spf = SAXParserFactory.newInstance();
            spf.setNamespaceAware(true);
            spf.setSchema(sysConfigSchema);
            XMLReader xmlReader = spf.newSAXParser().getXMLReader();
            SAXSource source = new SAXSource(xmlReader, new InputSource(new FileInputStream("example1.xml")));
            RootElement root = (RootElement)unmarshaller.unmarshal(
                    source);
            System.out.println("Child Val: " + root.getChild().getChildVal());
            System.out.println("Child Attr: " + root.getChild().getAttr());
        } catch (Exception e) {
            System.out.println(e.getMessage());
            e.printStackTrace();
        }
    }
}

与恶龙缠斗过久,自身亦成为恶龙;凝视深渊过久,深渊将回以凝视…
OGeek|极客中国-欢迎来到极客的世界,一个免费开放的程序员编程交流平台!开放,进步,分享!让技术改变生活,让极客改变未来! Welcome to OGeek Q&A Community for programmer and developer-Open, Learning and Share
Click Here to Ask a Question

...