如何自定义 JAXB 生成复数方法名称的方式?

2022-09-01 02:39:55

我们正在使用 JAXB 来生成 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,我们期望我们得到的地方。我们如何定制JAXB如何复数化特定方法?getPhysiciansgetPhysicien

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 模式中混合使用 JAXB 自定义等问题。因此,对于那些具有相同首选项的人来说,这里有一个JAXB绑定文件,它实现了与Blaise建议的相同的目标,使JAXB定制不在模式之外:

<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>

答案 1

默认情况下,将为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;
        }

    }

如果对 XML 架构进行批注:

    <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;
    }

}

答案 2

也许回答起来有点晚了,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模式和JAXB绑定。

通过将 JAXB XJC 绑定编译器与“扩展”方式结合使用。需要添加自定义绑定文件,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>

引用:


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