I want to iterate on every "p" tags in a XML document and be able to get the current element's xpath but I don't find anything that does it.
The kind of code I tried:
from bs4 import BeautifulSoup
xml_file = open("./data.xml", "rb")
soup = BeautifulSoup(xml_file, "lxml")
for i in soup.find_all("p"):
print(i.xpath) # xpath doesn't work here (None)
print("\n")
Here is a sample XML file that I try to parse:
<?xml version="1.0" encoding="UTF-8"?>
<article>
<title>Sample document</title>
<body>
<p>This is a <b>sample document.</b></p>
<p>And there is another paragraph.</p>
</body>
</article>
I would like my code to output:
/article/body/p[0]
/article/body/p[1]
CodePudding user response:
You can use getpath() to get xpath from element:
result = root.xpath('//*[. = "XML"]')
for r in result:
print(tree.getpath(r))
you can try to use this function:
doc = etree.fromstring(xml)
btags = doc.xpath('//a/b')
for b in btags:
print b.text
def fast_iter(context, func, *args, **kwargs):
"""
fast_iter is useful if you need to free memory while iterating through a
very large XML file.
http://lxml.de/parsing.html#modifying-the-tree
Based on Liza Daly's fast_iter
http://www.ibm.com/developerworks/xml/library/x-hiperfparse/
See also http://effbot.org/zone/element-iterparse.htm
"""
for event, elem in context:
func(elem, *args, **kwargs)
# It's safe to call clear() here because no descendants will be
# accessed
elem.clear()
# Also eliminate now-empty references from the root node to elem
for ancestor in elem.xpath('ancestor-or-self::*'):
while ancestor.getprevious() is not None:
del ancestor.getparent()[0]
del context
def process_element(elt):
print(elt.text)
context=etree.iterparse(io.BytesIO(xml), events=('end',), tag='b')
fast_iter(context, process_element)
for more reference you can look here - https://newbedev.com/efficient-way-to-iterate-through-xml-elements