针对模式的XML（.xsd）提要验证

Question

我有一个XML文件，并且有一个XML模式。我想针对该架构验证文件，并检查其是否符合该架构。我正在使用python，但是如果python中没有这样有用的库，则可以使用任何语言。

我在这里最好的选择是什么？我会担心如何快速启动和运行它。

Answer 1

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 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)

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._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')

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