Preserving original doctype and declaration of an lxml.etree parsed xml
Asked Answered
E

2

18

I'm using python's lxml and I'm trying to read an xml document, modify and write it back but the original doctype and xml declaration disappears. I'm wondering if there's an easy way of putting it back in whether through lxml or some other solution?

Ethelind answered 19/10, 2012 at 2:21 Comment(2)
Have you read the documentation for the tostring method? I thought it automatically preserved the DOCTYPE.Cuticle
you can use tostring to add in a doctype and declaration but i'll need to parse the info out first. lxml doesn't seem to preserve doctype or declaration to begin withEthelind
C
13

tl;dr

# adds declaration with version and encoding regardless of
# which attributes were present in the original declaration
# expects utf-8 encoding (encode/decode calls)
# depending on your needs you might want to improve that
from lxml import etree
from xml.dom.minidom import parseString
xml1 = '''\
<?xml version="1.0" encoding="UTF-8"?>
<!DOCTYPE root SYSTEM "example.dtd">
<root>...</root>
'''
xml2 = '''\
<root>...</root>
'''
def has_xml_declaration(xml):
    return parseString(xml).version
def process(xml):
    t = etree.fromstring(xml.encode()).getroottree()
    if has_xml_declaration(xml):
        print(etree.tostring(t, xml_declaration=True, encoding=t.docinfo.encoding).decode())
    else:
        print(etree.tostring(t).decode())
process(xml1)
process(xml2)

The following will include the DOCTYPE and the XML declaration:

from lxml import etree
from StringIO import StringIO

tree = etree.parse(StringIO('''<?xml version="1.0" encoding="iso-8859-1"?>
 <!DOCTYPE root SYSTEM "test" [ <!ENTITY tasty "eggs"> ]>
  <root>
   <a>&tasty;</a>
 </root>
'''))

docinfo = tree.docinfo
print etree.tostring(tree, xml_declaration=True, encoding=docinfo.encoding)

Note, tostring does not preserve the DOCTYPE if you create an Element (e.g. using fromstring), it only works when you process the XML using parse.

Update: as pointed out by J.F. Sebastian my assertion about fromstring is not true.

Here is some code to highlight the differences between Element and ElementTree serialization:

from lxml import etree
from StringIO import StringIO

xml_str = '''<?xml version="1.0" encoding="iso-8859-1"?>
 <!DOCTYPE root SYSTEM "test" [ <!ENTITY tasty "eggs"> ]>
  <root>
   <a>&tasty;</a>
 </root>
'''

# get the ElementTree using parse
parse_tree = etree.parse(StringIO(xml_str))
encoding = parse_tree.docinfo.encoding
result = etree.tostring(parse_tree, xml_declaration=True, encoding=encoding)
print "%s\nparse ElementTree:\n%s\n" % ('-'*20, result)

# get the ElementTree using fromstring
fromstring_tree = etree.fromstring(xml_str).getroottree()
encoding = fromstring_tree.docinfo.encoding
result = etree.tostring(fromstring_tree, xml_declaration=True, encoding=encoding)
print "%s\nfromstring ElementTree:\n%s\n" % ('-'*20, result)

# DOCTYPE is lost, and no access to encoding
fromstring_element = etree.fromstring(xml_str)
result = etree.tostring(fromstring_element, xml_declaration=True)
print "%s\nfromstring Element:\n%s\n" % ('-'*20, result)

and the output is:

--------------------
parse ElementTree:
<?xml version='1.0' encoding='iso-8859-1'?>
<!DOCTYPE root SYSTEM "test" [
<!ENTITY tasty "eggs">
]>
<root>
   <a>eggs</a>
 </root>

--------------------
fromstring ElementTree:
<?xml version='1.0' encoding='iso-8859-1'?>
<!DOCTYPE root SYSTEM "test" [
<!ENTITY tasty "eggs">
]>
<root>
   <a>eggs</a>
 </root>

--------------------
fromstring Element:
<?xml version='1.0' encoding='ASCII'?>
<root>
   <a>eggs</a>
 </root>
Cuticle answered 19/10, 2012 at 3:14 Comment(1)
Do note that doctype is not preserved if the root element doesn't match. Also, you might want to learn what was in XML declaration besides the version.U
S
7

You can also preserve DOCTYPE and the XML declaration with fromstring():

import sys
from StringIO import StringIO
from lxml import etree

xml = r'''<?xml version="1.0" encoding="UTF-8"?>
<!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Strict//EN"
  "http://www.w3.org/TR/xhtml1/DTD/xhtml1-strict.dtd">
<html xmlns="http://www.w3.org/1999/xhtml" xml:lang="en">
 <head>
 <title>example</title>
 </head>
 <body>
 <p>This is an example</p>
 </body>
</html>'''

tree = etree.fromstring(xml).getroottree() # or etree.parse(file)
tree.write(sys.stdout, xml_declaration=True, encoding=tree.docinfo.encoding)

Output

<?xml version='1.0' encoding='UTF-8'?>
<!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Strict//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-strict.dtd">
<html xmlns="http://www.w3.org/1999/xhtml" xml:lang="en" lang="en">
 <head><meta http-equiv="Content-Type" content="text/html; charset=UTF-8" />
 <title>example</title>
 </head>
 <body>
 <p>This is an example</p>
 </body>
</html>

Note the xml declaration (with correct encoding) and doctype are present. It even (possibly incorrectly) uses ' instead of " in the xml declaration and adds Content-Type to the <head>.

For the @John Keyes' example input it produces the same results as etree.tostring() in the answer.

Scooter answered 19/10, 2012 at 3:43 Comment(0)

© 2022 - 2024 — McMap. All rights reserved.