XML (.xsd) feed validation against a schema

为君一笑 提交于 2019-11-29 22:27:09
alecxe

Definitely lxml.

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)

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.

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:
[...]

The python snippet is good, but an alternative is to use xmllint:

xmllint -schema sample.xsd --noout sample.xml
易学教程内所有资源均来自网络或用户发布的内容,如有违反法律规定的内容欢迎反馈
该文章没有解决你所遇到的问题?点击提问,说说你的问题,让更多的人一起探讨吧!