针对架构的 XML (.xsd) 提要验证

问题描述

我有一个 XML 文件和一个 XML 架构.我想根据该架构验证文件并检查它是否符合该架构.我正在使用 python，但如果 python 中没有这样有用的库，我对任何语言都持开放态度.

I have a XML file and I have a XML schema. I want to validate the file against that schema and check if it adheres to that. I am using python but am open to any language for that matter if there is no such useful library in python.

我最好的选择是什么?我会担心我可以多快地启动和运行它.

What would be my best options here? I would worry about the how fast I can get this up and running.

推荐答案

绝对lxml.

使用预定义的模式定义一个 XMLParser，加载文件 fromstring() 并捕获任何 XML 架构错误:

Define an XMLParser with a predefined schema, load the the file fromstring() and catch any XML Schema errors:

from lxml import etree

def validate(xmlparser, xmlfilename):
    try:
        with open(xmlfilename, 'r') as f:
            etree.fromstring(f.read(), xmlparser)
        return True
    except etree.XMLSchemaError:
        return False

schema_file = 'schema.xsd'
with open(schema_file, 'r') as f:
    schema_root = etree.XML(f.read())

schema = etree.XMLSchema(schema_root)
xmlparser = etree.XMLParser(schema=schema)

filenames = ['input1.xml', 'input2.xml', 'input3.xml']
for filename in filenames:
    if validate(xmlparser, filename):
        print("%s validates" % filename)
    else:
        print("%s doesn't validate" % filename)

编码注意事项

如果架构文件包含带有编码的 xml 标记(例如 <?xml version="1.0" encoding="UTF-8"?>)，上面的代码将生成以下错误:

Note about encoding

If the schema file contains an xml tag with an encoding (e.g. <?xml version="1.0" encoding="UTF-8"?>), the code above will generate the following error:

Traceback (most recent call last):
  File "<input>", line 2, in <module>
    schema_root = etree.XML(f.read())
  File "src/lxml/etree.pyx", line 3192, in lxml.etree.XML
  File "src/lxml/parser.pxi", line 1872, in lxml.etree._parseMemoryDocument
ValueError: Unicode strings with encoding declaration are not supported. Please use bytes input or XML fragments without declaration.

一种解决方案是以字节模式打开文件:open(..., 'rb')

A solution is to open the files in byte mode: open(..., 'rb')

[...]
def validate(xmlparser, xmlfilename):
    try:
        with open(xmlfilename, 'rb') as f:
[...]
with open(schema_file, 'rb') as f:
[...]

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