在 Python 中使用 lxml 将 XSD 解析为 XML 时遵循 xs:include [英] Following xs:include when parsing XSD as XML with lxml in Python

问题描述

所以，我的问题是我正在尝试做一些不正统的事情.我有一组复杂的 XSD 文件.但是我不想使用这些 XSD 文件来验证 XML 文件；我想将这些 XSD 解析为 XML 并像处理普通 XML 文件一样询问它们.这是可能的，因为 XSD 是有效的 XML.我在 Python3 中使用 lxml.

So, my problem is I'm trying to do something a little un-orthodox. I have a complicated set of XSD files. However I don't want to use these XSD files to verify an XML file; I want to parse these XSDs as XML and interrogate them just as I would a normal XML file. This is possible because XSDs are valid XML. I am using lxml with Python3.

我遇到的问题是语句:

<xs:include schemaLocation="sdm-extension.xsd"/>

如果我指示 lxml 创建一个 XSD 来进行这样的验证:

If I instruct lxml to create an XSD for verifying like this:

schema = etree.XMLSchema(schema_root)

这个依赖关系将被解决(文件与我刚刚加载的文件存在于同一目录中).但是，我将这些视为 XML，因此，正确地，lxml 只是将其视为具有属性的普通元素而不遵循它.

this dependency will be resolved (the file exists in the same directory as the one I've just loaded). HOWEVER, I am treating these as XML so, correctly, lxml just treats this as a normal element with an attribute and does not follow it.

是否有一种简单或正确的方法来扩展 lxml，以便我可能具有相同或相似的行为，例如

Is there an easy or correct way to extend lxml so that I may have the same or similar behaviour as, say

<xi:include href="metadata.xml" parse="xml" xpointer="title"/>

当然，我可以手动创建一个单独的 xml 文件，其中包含 XSD 架构中的所有依赖项.这也许是一个解决方案?

I could, of course, create a separate xml file manually that includes all the dependencies in the XSD schema. That is perhaps a solution?

推荐答案

试试这个:

def validate_xml(schema_file, xml_file):
    xsd_doc = etree.parse(schema_file)
    xsd = etree.XMLSchema(xsd_doc)
    xml = etree.parse(xml_file)
    return xsd.validate(xml)

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