使用JAXB具有属性和内容的XML元素


78

如何使用JAXB生成以下XML?

<sport type="" gender="">
    sport description
</sport>

Answers:


129

使用注释类型和性别属性,@XmlAttribute使用描述属性@XmlValue

package org.example.sport;

import javax.xml.bind.annotation.*;

@XmlAccessorType(XmlAccessType.FIELD)
@XmlRootElement
public class Sport {

    @XmlAttribute
    protected String type;

    @XmlAttribute
    protected String gender;

    @XmlValue;
    protected String description;

}

想要查询更多的信息


10

正确的方案应为:

<?xml version="1.0" encoding="UTF-8"?>
<schema xmlns="http://www.w3.org/2001/XMLSchema" 
targetNamespace="http://www.example.org/Sport"
xmlns:tns="http://www.example.org/Sport" 
elementFormDefault="qualified"
xmlns:jaxb="http://java.sun.com/xml/ns/jaxb" 
jaxb:version="2.0">

<complexType name="sportType">
    <simpleContent>
        <extension base="string">
            <attribute name="type" type="string" />
            <attribute name="gender" type="string" />
        </extension>
    </simpleContent>
</complexType>

<element name="sports">
    <complexType>
        <sequence>
            <element name="sport" minOccurs="0" maxOccurs="unbounded"
                type="tns:sportType" />
        </sequence>
    </complexType>
</element>

为SportType生成的代码为:

package org.example.sport;

import javax.xml.bind.annotation.XmlAccessType;
import javax.xml.bind.annotation.XmlAccessorType;
import javax.xml.bind.annotation.XmlAttribute;
import javax.xml.bind.annotation.XmlType;


@XmlAccessorType(XmlAccessType.FIELD)
@XmlType(name = "sportType")
public class SportType {
    @XmlValue
    protected String value;
    @XmlAttribute
    protected String type;
    @XmlAttribute
    protected String gender;

    public String getValue() {
        return value;
    }

    public void setValue(String value) {
        this.value = value;
    }

    public String getType() {
    return type;
    }


    public void setType(String value) {
        this.type = value;
    }

    public String getGender() {
        return gender;
    }

    public void setGender(String value) {
        this.gender = value;
    }
}

8

这是有效的解决方案:

输出:

public class XmlTest {

    private static final Logger log = LoggerFactory.getLogger(XmlTest.class);

    @Test
    public void createDefaultBook() throws JAXBException {
        JAXBContext jaxbContext = JAXBContext.newInstance(Book.class);
        Marshaller marshaller = jaxbContext.createMarshaller();

        StringWriter writer = new StringWriter();
        marshaller.marshal(new Book(), writer);

        log.debug("Book xml:\n {}", writer.toString());
    }


    @XmlAccessorType(XmlAccessType.FIELD)
    @XmlRootElement(name = "book")
    public static class Book {

        @XmlElementRef(name = "price")
        private Price price = new Price();


    }

    @XmlAccessorType(XmlAccessType.FIELD)
    @XmlRootElement(name = "price")
    public static class Price {
        @XmlAttribute(name = "drawable")
        private Boolean drawable = true; //you may want to set default value here

        @XmlValue
        private int priceValue = 1234;

        public Boolean getDrawable() {
            return drawable;
        }

        public void setDrawable(Boolean drawable) {
            this.drawable = drawable;
        }

        public int getPriceValue() {
            return priceValue;
        }

        public void setPriceValue(int priceValue) {
            this.priceValue = priceValue;
        }
    }
}

输出:

22:00:18.471 [main]调试com.grebski.stack.XmlTest-图书xml:

<?xml version="1.0" encoding="UTF-8" standalone="yes"?>
<book>
    <price drawable="true">1234</price>
</book>

只是想添加指向类似答案的链接,以帮助那些将其视为解决方案的人。好答案,顺便说一句。stackoverflow.com/a/15429363/1686575
John Manko

你救我 !:DI使用@XmlValue进行解析
Angelo

4

更新的解决方案-使用我们正在讨论的架构解决方案。这使您得到答案:

样本架构:

<?xml version="1.0" encoding="UTF-8"?>
<schema xmlns="http://www.w3.org/2001/XMLSchema" 
targetNamespace="http://www.example.org/Sport"
xmlns:tns="http://www.example.org/Sport" 
elementFormDefault="qualified"
xmlns:jaxb="http://java.sun.com/xml/ns/jaxb" 
jaxb:version="2.0">

<complexType name="sportType">
    <attribute name="type" type="string" />
    <attribute name="gender" type="string" />
</complexType>

<element name="sports">
    <complexType>
        <sequence>
            <element name="sport" minOccurs="0" maxOccurs="unbounded"
                type="tns:sportType" />
        </sequence>
    </complexType>
</element>

生成代码

运动类型:

package org.example.sport; 

import javax.xml.bind.annotation.XmlAccessType;
import javax.xml.bind.annotation.XmlAccessorType;
import javax.xml.bind.annotation.XmlAttribute;
import javax.xml.bind.annotation.XmlType;


@XmlAccessorType(XmlAccessType.FIELD)
@XmlType(name = "sportType")
public class SportType {

    @XmlAttribute
    protected String type;
    @XmlAttribute
    protected String gender;

    public String getType() {
        return type;
    }


    public void setType(String value) {
        this.type = value;
    }

    public String getGender() {
        return gender;
    }

    public void setGender(String value) {
        this.gender = value;
    }

}

体育:

package org.example.sport;

import java.util.ArrayList;
import java.util.List;
import javax.xml.bind.annotation.XmlAccessType;
import javax.xml.bind.annotation.XmlAccessorType;
import javax.xml.bind.annotation.XmlRootElement;
import javax.xml.bind.annotation.XmlType;


@XmlAccessorType(XmlAccessType.FIELD)
@XmlType(name = "", propOrder = {
        "sport"
})
@XmlRootElement(name = "sports")
public class Sports {

    protected List<SportType> sport;

    public List<SportType> getSport() {
        if (sport == null) {
            sport = new ArrayList<SportType>();
        }
        return this.sport;
    }

}

通过在命令行上对模式运行xjc来生成输出类文件


1
不使用XSD。使用JAXB批注。
詹姆斯

当然可以,但是您可以生成一个非常简单的XSD来描述上述XML结构,以在带有注释的情况下为您生成Java代码文件。另外,您也可以手动构建它们,注释结构应该在文档中。使用XSD的优点是可以非常快速地更改结构并重新生成Java代码文件。这取决于您
Jim

1
感谢您的答复。但是我的问题是如何使用注释手动生成给定的XML。我如何定义实体类文件?
詹姆斯

请参阅上面的新解决方案,真的希望能对您
Jim
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