我应该在哪里放置一个XSD文件用于JAXB代码生成和XML验证(Where should I place an XSD file to use for JAXB code gen and for XML validation)

 我在我的Java项目中创建了一个XSD文件，用于定义用户可编辑的输入文件（例如，假设XSD名为userinput.xsd，用户可编辑文件为userinput.xml）。 当程序运行时，它使用JAXB来验证用户是否在XML文件中没有出现任何错误，因为它将文件解组到DOM中。 
 
 我使用Maven标准目录布局构建了我的项目，并使用xjc生成了JAXB对象工厂和其他类，将它们放在名为/ src / main / java / my / name / space / generated / userinput的目录中（以匹配名称）的XSD）。 
 
 我已将XSD文件放在/ src / main / resources中。 
 
 当我打包我的jar文件时，该文件位于jar的根目录中，我可以在Java代码中引用它，如下所示（特别注意第4行中提到的资源）： 
 
JAXBContext jaxbContext = JAXBContext.newInstance("my.name.space.generated.userinput");
Unmarshaller unmarshaller = jaxbContext.createUnmarshaller();
SchemaFactory schemaFactory = SchemaFactory.newInstance(XMLConstants.W3C_XML_SCHEMA_NS_URI);
Schema schema = schemaFactory.newSchema(getClass().getResource("/userinput.xsd"));
unmarshaller.setSchema(schema);
JAXBElement<?> userinputType = (JAXBElement<?>) unmarshaller.unmarshal(new FileInputStream("userinput.xml"));
 
 这可行，但它似乎不正确，因为它意味着如果我想扩展到多个输入文件，我可能最终在我的源代码管理的资源目录和jar文件的根目录中有许多.xsd文件。 
 
 此外，当我从IDE而不是从打包的jar运行程序时，我必须将第四个语句更改为： 
 
Schema schema = schemaFactory.newSchema(getClass().getResource("src/main/resources/userinput.xsd"));
 
 我应该将XSD文件（a）放在我的源代码控制（即Maven结构）和（b）我的jar文件中？ [注意：通过我的IDE运行xjc将它放在... / generated / userinput目录中，但Maven在打包时会忽略它。 
 
 我正在寻找一个答案，表明存在某种共同的方法，所以如果可能的话，我想要一个参考。 如果这是当前未经指定的选择留给开发人员，那么请说明（最好引用），因为我理解stackoverflow不是收集意见的地方。 

I have created an XSD file in my Java project that defines a user-editable input file (for illustration, let's say the XSD is called userinput.xsd and the user-editable file is userinput.xml). When the program runs, it uses JAXB to validate that the user has not made any mistakes in the XML file as it unmarshalls the file into a DOM.
 
I have structured my project using the Maven Standard Directory Layout and generated the JAXB object factory and other classes using xjc, placing them in a directory called /src/main/java/my/name/space/generated/userinput (to match the name of the XSD).
 
I have placed the XSD file in /src/main/resources.
 
When I package my jar file, the file is in the root of the jar and I can refer to it in the Java code as follows (note in particular the resource mentioned in line 4):
 
JAXBContext jaxbContext = JAXBContext.newInstance("my.name.space.generated.userinput");
Unmarshaller unmarshaller = jaxbContext.createUnmarshaller();
SchemaFactory schemaFactory = SchemaFactory.newInstance(XMLConstants.W3C_XML_SCHEMA_NS_URI);
Schema schema = schemaFactory.newSchema(getClass().getResource("/userinput.xsd"));
unmarshaller.setSchema(schema);
JAXBElement<?> userinputType = (JAXBElement<?>) unmarshaller.unmarshal(new FileInputStream("userinput.xml"));
 
This works but it does not seem right as it means if I want to scale to multiple input files, I could end up with many .xsd files both in the resources directory in my source control and in the root of the jar file.
 
Also when I run the program from my IDE rather than from the packaged jar, I have to change the fourth statement to:
 
Schema schema = schemaFactory.newSchema(getClass().getResource("src/main/resources/userinput.xsd"));
 
Where should I put the XSD file (a) in my source control (i.e., the Maven structure) and (b) in my jar file? [Note: Running xjc through my IDE puts it in the .../generated/userinput directory but Maven then ignores it when packaging.]
 
I'm looking for an answer that indicates that there is some sort of common methodology, so would like a reference if possible. If this is currently an unspecified choice left to the developer, then please say so (preferably referenced) as I understand that stackoverflow is not the place to collect opinions.

原文：https://stackoverflow.com/questions/30983219