Java编组和UnMarshalling

Question

Employee.xsd

<?xml version="1.0" encoding="UTF-8" standalone="yes"?>
<xsd:schema xmlns:xsd="http://www.w3.org/2001/XMLSchema" xmlns:jxb="http://java.sun.com/xml/ns/jaxb" jxb:version="1.0" elementFormDefault="qualified">
<xsd:include schemaLocation="Family.xsd"/>
    <xsd:element name="NewFields">
        <xsd:complexType>
            <xsd:sequence>
                <xsd:element name="empFirstName" type="xsd:string" />
                <xsd:element name="empLastName" type="xsd:string" />
                <xsd:element name="family" type="FamilyFields" nillable="true" maxOccurs="unbounded" minOccurs="0"/>
            </xsd:sequence>
        </xsd:complexType>
    </xsd:element>
</xsd:schema>

Family.xsd

<?xml version="1.0" encoding="UTF-8"?>
<xsd:schema xmlns:xsd="http://www.w3.org/2001/XMLSchema" xmlns:jxb="http://java.sun.com/xml/ns/jaxb" jxb:version="1.0" elementFormDefault="qualified">
<xsd:complexType name="FamilyFields">
        <xsd:sequence>
            <xsd:element name="relation" type="xsd:string" />
            <xsd:element name="firstName" type="xsd:string" />
            <xsd:element name="lastName" type="xsd:string"/>
        </xsd:sequence>
</xsd:complexType>  
</xsd:schema>

使用这2种模式的第三方应用程序生成一个xml字符串，类似于-

<NewFields>
    <empFirstName>Kevin</empFirstName>
    <empLastName>Smith</empLastName>
    <family>
        <FamilyFields>
            <relation>self</relation>
            <firstName>New Kevin</firstName>
            <lastName>New Smith</lastName>
        </FamilyFields>
        <FamilyFields>
            <relation>wife</relation>
            <firstName>Jennifer</firstName>
            <lastName>Smith</lastName>
        </FamilyFields>
    </family>
</NewFields>

第二个模式始终是一个固定的模式。

我的要求是解析关系标记，如果关系是"self"，我需要用相应字段的empFirstName和lastName覆盖firstName和lastName。

我怎样才能做到这一点？

编辑1: Employee.xsd是动态的，可以是任何东西。但是Family.xsd是静态的，可以从任何其他xsd导入。

Answer 1

概要：(a)映射XML模式文件，(b)取消给定的XML，(c)修改内存中的XML对象，最后(d)将修改后的对象发送到任何您想要的位置。

映射XML架构文件

@XmlRootElement(name = "FamilyFields") public class FamilyFields {

  @XmlElement public String relation;
  @XmlElement public String firstName;
  @XmlElement public String lastName;

  public FamilyFields() {}
}

@XmlRootElement(name = "NewFields") public class NewFields {

  @XmlElement public String empFirstName;
  @XmlElement public String empLastName;
  @XmlElementWrapper(name = "family")
  @XmlElement(name = "FamilyFields")
  public List<FamilyFields> familyFields;

  public NewFields() {}
}

(如果你接受一个建议：，用手做，，__！使用XJC生成JAXB实体几乎从不输出您期望的内容，如果手头没有非常简单的模式文件，则可能会花费时间。)

收购JAXBContext，Unmarshaller和Marshaller

JAXBContext context = JAXBContext.newInstance(NewFields.class, FamilyFields.class);
Unmarshaller unmarshaller = context.createUnmarshaller();
Marshaller marshaller = context.createMarshaller();
// set optional properties
marshaller.setProperty(Marshaller.JAXB_FORMATTED_OUTPUT, true);

处理传入的XML

// the XML content you gave
String xml = ...

StringReader reader = new StringReader(xml);
NewFields newFields = (NewFields) unmarshaller.unmarshal(reader);

// modify the unmarshalled data to your heart's content
newFields.empLastName = ...

// marshal the modified data anywhere you want
marshaller.marshal(newFields, System.out);

Answer 2

如果您有一个XSD，那么使用XML数据绑定操作XML是最容易的。使用XSD，您可以创建表示XSD的类java。现在，您可以对java类进行getter和setter，然后输出XML。有许多XML数据绑定解决方案。试试XML bean：

http://xmlbeans.apache.org/