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我正在尝试使用sbt-scalaxb为 生成绑定,FixRepository.xsd但它不喜欢 SUBJ。

作为最后的手段,我可​​以稍微更改架构,但是有没有办法调整sbt-scalaxb以理解原始文件?

sbt-scalaxb不喜欢的 XSD 片段是:

xml <xs:element name="component"> <xs:complexType> <xs:sequence> <xs:element ref="messageEntity" minOccurs="0" maxOccurs="unbounded"/> </xs:sequence> <xs:attributeGroup ref="entityAttribGrp"/> <xs:attribute name="id" type="id_t" use="required"/> <xs:attribute name="name" type="xs:string" use="required"/> <xs:attribute name="type" type="ComponentType_t" use="required"/> <xs:attribute name="repeating" type="BOOL_t" use="optional"/> <xs:attribute name="category" type="xs:string" use="optional"/> <xs:attribute name="abbrName" type="xs:string" use="optional"/> <xs:attribute name="notReqXML" type="BOOL_t" use="optional"/> <!-- would like to force a description of the component --> </xs:complexType> </xs:element>

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1 回答 1

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我自己想通了:scalaxbAttributePrefix设置为属性属性添加了一个前缀。

构建.sbt

scalaxbAttributePrefix in (Compile, scalaxb) := Some("attr")

生成的 FixRepository.scala

case class Fix(
    ...,
    components: ...fixrepo.Components,
    ...,
    attributes: Map[String, scalaxb.DataRecord[Any]] = Map()
) {
    ...
    lazy val attrComponents = attributes("@components").as[BOOL_t]
    ...
}
于 2018-08-26T13:50:25.250 回答