针对模式的XML(.xsd)提要验证

Question

我有一个XML文件,我有一个XML模式.我想根据该模式验证该文件,并检查它是否符合该模式.我正在使用python,但如果在python中没有这样有用的库,我会对任何语言开放.

这里最好的选择是什么？我担心我能以多快的速度运行它.

Answer 1

当然lxml.

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

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)

注意编码

如果模式文件包含带编码的xml标记(例如<?xml version="1.0" encoding="UTF-8"?>),则上面的代码将生成以下错误:

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.

解决方案是以字节模式打开文件:open(..., 'rb')

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