Java 如何自定义 JAXB 生成复数方法名称的方式?

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时间:2020-08-14 18:06:18  来源:igfitidea点击:

How do you customize how JAXB generates plural method names?

javaxmljaxbjaxb2

提问by SingleShot

We are using JAXB to generate Java classes and have encountered a few cases where generated plural method names are not correct. For example, where we expect getPhysicianswe are getting getPhysicien. How would we customize how JAXB pluralizes specific methods?

我们在使用JAXB生成Java类时遇到过几个生成的复数方法名不正确的情况。例如,我们期望getPhysicians得到的地方getPhysicien。我们将如何自定义 JAXB 如何使特定方法多元化?

The schema:

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>

The generated Java code:

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;
    }


Update

更新

This has been answered by Blaise. However, I prefer not mixing concerns such as JAXB customizations in an XML schema. So for those of you with the same preference, here is a JAXB binding file that achieves the same thing as what Blaise suggested, keeping JAXB customization out of the schema:

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 定制之类的问题。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 自定义保留在模式之外:

<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>

采纳答案by bdoughan

By default the following is generated for your schema fragment:

默认情况下,会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;
        }

    }

If you annotate your XML schema:

如果您注释您的 XML 架构:

    <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>

Then you can generate the desired class:

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;
    }

}

回答by Guillaume Husta

Maybe it's a little late to answer, but there's another way to generate plural names simply, without mixingXML Schema and JAXB Bindings.

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 模式和 JAXB 绑定。

By using JAXB XJC binding compiler with the "-extension" mode. A customization bindings file need to be added, like this one :

通过在“-extension”模式下使用 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>

References :

参考 :