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I have encountered a very strange behaviour while working with BeautifulSoup today.
Let's have a look at a very simple html snippet:
<html><body><ix:nonfraction>lele</ix:nonfraction></body></html>
I am trying to get the content of the <ix:nonfraction> tag with BeautifulSoup.
Everything works fine when using the find method:
from bs4 import BeautifulSoup
html = "<html><body><ix:nonfraction>lele</ix:nonfraction></body></html>"
soup = BeautifulSoup(html, 'lxml') # The parser used here does not matter
soup.find('ix:nonfraction')
>>> <ix:nonfraction>lele</ix:nonfraction>
However, when trying to use the find_all method, I expect to have a list with this single element returned, which is not the case !
soup.find_all('ix:nonfraction')
>>> []
In fact, find_all seems to return an empty list everytime a colon is present in the tag I am searching for.
I have been able to reproduce the problem on two different computers.
Does anyone have an explanation, and more importantly, a workaround ?
I need to use the find_all method simply because my actual case requires me to get all these tags on a whole html page.
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The reason @yosemite_k's solution works is because in the source code of bs4, it's skipping a certain condition which causes this behavior. You can in fact do many variations which will produce this same result. Examples:
soup.find_all({"ix:nonfraction"})
soup.find_all('ix:nonfraction', limit=1)
soup.find_all('ix:nonfraction', text=True)
Below is a snippet from the source code of beautifulsoup that shows what happens when you call find or find_all. You will see that find just calls find_all with limit=1. In _find_all it checks for a condition:
if text is None and not limit and not attrs and not kwargs:
If it hits that condition, then the it might eventually make it down to this condition:
# Optimization to find all tags with a given name.
if name.count(':') == 1:
If it makes it there, then it does a reassignment of name:
# This is a name with a prefix.
prefix, name = name.split(':', 1)
This is where you're behavior is different. As long as find_all doesn't meet any of the prior conditions, then you'll find the element.
beautifulsoup4==4.6.0
def find(self, name=None, attrs={}, recursive=True, text=None,
**kwargs):
"""Return only the first child of this Tag matching the given
criteria."""
r = None
l = self.find_all(name, attrs, recursive, text, 1, **kwargs)
if l:
r = l[0]
return r
findChild = find
def find_all(self, name=None, attrs={}, recursive=True, text=None,
limit=None, **kwargs):
"""Extracts a list of Tag objects that match the given
criteria. You can specify the name of the Tag and any
attributes you want the Tag to have.
The value of a key-value pair in the 'attrs' map can be a
string, a list of strings, a regular expression object, or a
callable that takes a string and returns whether or not the
string matches for some custom definition of 'matches'. The
same is true of the tag name."""
generator = self.descendants
if not recursive:
generator = self.children
return self._find_all(name, attrs, text, limit, generator, **kwargs)
def _find_all(self, name, attrs, text, limit, generator, **kwargs):
"Iterates over a generator looking for things that match."
if text is None and 'string' in kwargs:
text = kwargs['string']
del kwargs['string']
if isinstance(name, SoupStrainer):
strainer = name
else:
strainer = SoupStrainer(name, attrs, text, **kwargs)
if text is None and not limit and not attrs and not kwargs:
if name is True or name is None:
# Optimization to find all tags.
result = (element for element in generator
if isinstance(element, Tag))
return ResultSet(strainer, result)
elif isinstance(name, str):
# Optimization to find all tags with a given name.
if name.count(':') == 1:
# This is a name with a prefix.
prefix, name = name.split(':', 1)
else:
prefix = None
result = (element for element in generator
if isinstance(element, Tag)
and element.name == name
and (prefix is None or element.prefix == prefix)
)
return ResultSet(strainer, result)
results = ResultSet(strainer)
while True:
try:
i = next(generator)
except StopIteration:
break
if i:
found = strainer.search(i)
if found:
results.append(found)
if limit and len(results) >= limit:
break
return results
leaving the tag name empty and using ix as attribute.
soup.find_all({"ix:nonfraction"})
works well
EDIT: 'ix:nonfraction' is not a tag name, so soup.find_all("ix:nonfraction") returned an empty list for a non existent tag.
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