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

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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\

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.

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隐瞒了意图╮

2021-01-01 19:13

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>

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