我有一个xsd文件a.xsd,它导入另一个xsd b.xsd。我想从a.xsd创建一个包含所有类的jar,但是忽略从b.xsd生成的类。我的pom文件如下。
<project xmlns="http://maven.apache.org/POM/4.0.0" xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance"
xsi:schemaLocation="http://maven.apache.org/POM/4.0.0 http://maven.apache.org/xsd/maven-4.0.0.xsd">
<modelVersion>4.0.0</modelVersion>
<groupId>com.personal.proj</groupId>
<artifactId>schema-model</artifactId>
<version>1.0.0</version>
<name>SchemaModel</name>
<build>
<plugins>
<plugin>
<groupId>org.jvnet.jaxb2.maven2</groupId>
<artifactId>maven-jaxb2-plugin</artifactId>
<version>0.12.3</version>
<executions>
<execution>
<id>id</id>
<goals>
<goal>generate</goal>
</goals>
<configuration>
<schemaDirectory>src/main/xsd</schemaDirectory>
<schemaIncludes>
<include>a.xsd</include>
</schemaIncludes>
<schemaExcludes>
<exclude>b.xsd</exclude>
</schemaExcludes>
<forceRegenerate>true</forceRegenerate>
<removeOldOutput>true</removeOldOutput>
<bindingDirectory>src/main/xsd</bindingDirectory>
<bindingIncludes>
<include>schema.xjb</include>
</bindingIncludes>
<generateDirectory>${project.build.directory}/generated-sources</generateDirectory>
</configuration>
</execution>
</executions>
</plugin>
</plugins>
</build>
</project>
我的schema.xjb文件是
<jxb:bindings xmlns:jxb="http://java.sun.com/xml/ns/jaxb"
jxb:version="2.0" xmlns:xs="http://www.w3.org/2001/XMLSchema">
<jxb:bindings schemaLocation="a.xsd" node="/xs:schema">
<jxb:schemaBindings>
<jxb:package name="com.a.model" />
</jxb:schemaBindings>
</jxb:bindings>
<jxb:bindings schemaLocation="b.xsd" node="/xs:schema">
<jxb:schemaBindings>
<jxb:package name="com.b.model" />
</jxb:schemaBindings>
</jxb:bindings>
这个link显示了如何做到这一点,但是我无法配置它。如果我需要添加剧集,我需要在同一个版本中生成它(b.episode)并将其用于生成of.jar
答案 0 :(得分:2)
在map="false"
绑定上使用b.xsd
:
<jxb:bindings schemaLocation="b.xsd" node="/xs:schema">
<jxb:schemaBindings map="false">
<jxb:package name="com.b.model" />
</jxb:schemaBindings>
</jxb:bindings>
仍然可能会生成一些类(XJC有一些问题)。我经常在构建中的后处理步骤中删除它们。
我通常建议使用episodes来做这些事情。这是一个test project for episodes。
免责声明:我是maven-jaxb2-plugin的作者。