绝对可以
lxml。
XMLParser使用预定义的架构定义，加载文件
fromstring()并捕获任何XML架构错误：
from lxml import etreedef validate(xmlparser, xmlfilename): try: with open(xmlfilename, 'r') as f: etree.fromstring(f.read(), xmlparser) return True except etree.XMLSchemaError: return Falseschema_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)关于编码的注意事项
如果模式文件包含带有编码（例如
<?xml version="1.0" encoding="UTF-8"?>）的xml标记，则上面的代码将产生以下错误：
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._parseMemorydocumentValueError: Unipre 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')
[...]def validate(xmlparser, xmlfilename): try: with open(xmlfilename, 'rb') as f:[...]with open(schema_file, 'rb') as f:[...]
