Python Cookbook

Checking XML Well-Formedness

Credit: Paul Prescod

Problem

You need to check if an XML document is well-formed (not if it conforms to a DTD or schema), and you need to do this quickly.

Solution

SAX (presumably using a fast parser such as Expat underneath) is the fastest and simplest way to perform this task:

from xml.sax.handler import ContentHandler
from xml.sax import make_parser
from glob import glob
import sys

def parsefile(file):
    parser = make_parser(  )
    parser.setContentHandler(ContentHandler(  ))
    parser.parse(file)

for arg in sys.argv[1:]:
    for filename in glob(arg):
        try:
            parsefile(filename)
            print "%s is well-formed" % filename
        except Exception, e:
            print "%s is NOT well-formed! %s" % (filename, e)

Discussion

A text is a well-formed XML document if it adheres to all the basic syntax rules for XML documents. In other words, it has a correct XML declaration and a single root element, all tags are properly nested, tag attributes are quoted, and so on.

This recipe uses the SAX API with a dummy ContentHandler that does nothing. Generally, when we parse an XML document with SAX, we use a ContentHandler instance to process the document’s contents. But in this case, we only want to know if the document meets the most fundamental syntax constraints of XML; therefore, there is no processing that we need to do, and the do-nothing handler suffices.

The parsefile function parses the whole document and throws an exception if there is an error. The recipe’s main code catches any such exception and ...