40

我们正在使用 JAXB 生成 Java 类，并且遇到了一些生成的复数方法名称不正确的情况。例如，我们期望getPhysicians得到getPhysicien. 我们将如何自定义 JAXB 如何使特定方法复数？

架构：

<xs:complexType name="physician">
    <xs:sequence>
       ...
    </xs:sequence>
</xs:complexType>

<xs:complexType name="physicianList">
    <xs:sequence>
        <xs:element name="Physician"
                    type="physician"
                    minOccurs="0"
                    maxOccurs="unbounded"/>
    </xs:sequence>
</xs:complexType>

生成的Java代码：

...
public class PhysicianList {
...

    @XmlElement(name = "Physician")
    protected List<Physician> physicien;
    ...

    public List<Physician> getPhysicien() {
        if (physicien == null) {
            physicien = new ArrayList<Physician>();
        }
        return this.physicien;
    }

更新

布莱斯已经回答了这个问题。但是，我不喜欢在 XML 模式中混合诸如 JAXB 自定义之类的问题。因此，对于那些有相同偏好的人，这里有一个 JAXB 绑定文件，它实现了与 Blaise 建议的相同的事情，将 JAXB 自定义排除在模式之外：

<jaxb:bindings xmlns:jaxb="http://java.sun.com/xml/ns/jaxb"
               xmlns:xs="http://www.w3.org/2001/XMLSchema"
               version="2.0">

    <jaxb:bindings schemaLocation="myschema.xsd">
        <jaxb:bindings node="//xs:complexType[@name='physicianList']//xs:element[@name='Physician']">
            <jaxb:property name="physicians"/>
        </jaxb:bindings>
    </jaxb:bindings>

</jaxb:bindings>
4

2 回答 2

31

默认情况下，将为您的架构片段生成以下内容：

    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.XmlElement;
    import javax.xml.bind.annotation.XmlType;

    @XmlAccessorType(XmlAccessType.FIELD)
    @XmlType(name = "physicianList", propOrder = {
        "physician"
    })
    public class PhysicianList {

        @XmlElement(name = "Physician")
        protected List<Physician> physician;

        public List<Physician> getPhysician() {
            if (physician == null) {
                physician = new ArrayList<Physician>();
            }
            return this.physician;
        }

    }

如果您注释您的 XML 模式：

    <xs:schema
        xmlns:jaxb="http://java.sun.com/xml/ns/jaxb"
        xmlns:xs="http://www.w3.org/2001/XMLSchema"
        jaxb:version="2.1">

        <xs:complexType name="physician">
            <xs:sequence>
            </xs:sequence>
        </xs:complexType>

        <xs:complexType name="physicianList">
            <xs:sequence>
                <xs:element name="Physician"
                            type="physician"
                            minOccurs="0"
                            maxOccurs="unbounded">
                      <xs:annotation>
                          <xs:appinfo>
                              <jaxb:property name="physicians"/>
                          </xs:appinfo>
                      </xs:annotation>
                 </xs:element>
            </xs:sequence>
        </xs:complexType>

    </xs:schema>

然后您可以生成所需的类：

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.XmlElement;
import javax.xml.bind.annotation.XmlType;

@XmlAccessorType(XmlAccessType.FIELD)
@XmlType(name = "physicianList", propOrder = {
    "physicians"
})
public class PhysicianList {

    @XmlElement(name = "Physician")
    protected List<Physician> physicians;

    public List<Physician> getPhysicians() {
        if (physicians == null) {
            physicians = new ArrayList<Physician>();
        }
        return this.physicians;
    }

}
于 2010-12-21T18:23:11.783 回答
18

也许现在回答有点晚了，但是还有另一种方法可以简单地生成复数名称，而无需混合XML Schema 和 JAXB 绑定。

通过使用带有“-extension”模式的 JAXB XJC 绑定编译器。需要添加一个自定义绑定文件，如下所示：

<?xml version="1.0"?>
<jxb:bindings version="1.0"
              xmlns:jxb="http://java.sun.com/xml/ns/jaxb"
              xmlns:xs="http://www.w3.org/2001/XMLSchema"
              xmlns:xjc="http://java.sun.com/xml/ns/jaxb/xjc"
              jxb:extensionBindingPrefixes="xjc">

  <jxb:globalBindings>              
    <xjc:simple/>
  </jxb:globalBindings>

</jxb:bindings>

参考 ：

于 2012-11-13T11:09:40.680 回答