Parsing HTML with entity ref - python

I am trying to parse some HTML which has as an example
<solids>
&sub2;
</solids>
The html file is read in as a string. I need to insert the HTML from a file that sub2 defines into the appropriate part of the string before then processing the whole string as XML.
I have tried HTMLParser and using its handlers with
class MyHTMLParser(HTMLParser):
def handle_entityref(self, name):
# This gets called when the entity is referenced
print "Entity reference : "+ name
print "Current Section : "+ self.get_starttag_text()
print self.getpos()
But getpos returns a line number and offset rather than position in the string. ( The insertion can be at any point in the file )
I found this link and this suggest to use lxml. I have looked at lxml but cannot see how it would solve the problem. Its scanner does not seem to have an entity handler and seems to be xml rather than html

Okay found that lxml will handle the ENTITY references for me.
Just had to setup parser with the option resolve_entities=True
parser = etree.XMLParser(resolve_entities=True)
root = etree.parse(filename, parser=parser)

Related

What is the meaning of "html.parser" when doing BeautifulSoup(source_code, 'html.parser')?

I am not getting the BeautifulSoup's syntax, especially the purpose of HTML parser inside the parenthesis.
BeautifulSoup(source_code, 'html.parser')
This seems to be the defining which library you want to use for parsing the source_code. Checkout the options in the docs and how they compare.
From what I understand, "html.parser" will use Python3 html module found here.
More reading on parsers:
Parser differences demo
A diagnostics method which shows you which packages are used
You can check out the BeautifulSoup source code to understand the constructor parameters and how they are used. Here is the code for the BeautifulSoup class __init__.py:
def __init__(self, markup="", features=None, builder=None,
parse_only=None, from_encoding=None, exclude_encodings=None,
**kwargs):
...
if builder is None:
original_features = features
if isinstance(features, basestring):
features = [features]
if features is None or len(features) == 0:
features = self.DEFAULT_BUILDER_FEATURES
builder_class = builder_registry.lookup(*features)
if builder_class is None:
raise FeatureNotFound(
"Couldn't find a tree builder with the features you "
"requested: %s. Do you need to install a parser library?"
% ",".join(features))
builder = builder_class()
if not (original_features == builder.NAME or
original_features in builder.ALTERNATE_NAMES):
if builder.is_xml:
markup_type = "XML"
else:
markup_type = "HTML"
The 1st argument is the markup code (ex. HTML code) and the 2nd argument specifies how to parse that markup, with the default being the built-in HTML parser but it can be overriden:
You can override this by specifying one of the following:
What type of markup you want to parse. Currently supported are “html”, “xml”, and “html5”.
The name of the parser library you want to use. Currently supported options are “lxml”, “html5lib”, and “html.parser” (Python’s built-in HTML parser).

Building a generic XML parser in Python?

I am a newbie and having 1 week experience writing python scripts.
I am trying to write a generic parser (Library for all my future jobs) which parses any input XML without any prior knowledge of tags.
Parse input XML.
Get the values from the XML and Set the values basing on the tags.
Use these values in the rest of the job.
I am using the "xml.etree.ElementTree" library and i am able to parse the XML in the below mentioned way.
#!/usr/bin/python
import os
import xml.etree.ElementTree as etree
import logging
logging.basicConfig(level=logging.DEBUG)
logger = logging.getLogger(__name__)
logger.info('start reading XML property file')
filename = "mood_ib_history_parameters_DEV.xml"
logger.info('getting the current location')
__currentlocation__ = os.getcwd()
__fullpath__ = os.path.join(__currentlocation__,filename)
logger.info('start parsing the XML property file')
tree = etree.parse(__fullpath__)
root = tree.getroot()
hive_db = root.find("hive_db").text
EDGE_HIVE_CONN = root.find("EDGE_HIVE_CONN").text
target_dir = root.find("target_dir").text
to_email_alias = root.find("to_email_alias").text
to_email_cc = root.find("to_email_cc").text
from_email_alias = root.find("from_email_alias").text
dburl = root.find("dburl").text
SQOOP_EDGE_CONN = root.find("SQOOP_EDGE_CONN").text
user_name = root.find("user_name").text
password = root.find("password").text
IB_log_table = root.find("IB_log_table").text
SR_DG_master_table = root.find("SR_DG_master_table").text
SR_DG_table = root.find("SR_DG_table").text
logger.info('Hive DB %s', hive_db)
logger.info('Hive DB %s', hive_db)
logger.info('Edge Hive Connection %s', EDGE_HIVE_CONN)
logger.info('Target Directory %s', target_dir)
logger.info('To Email address %s', to_email_alias)
logger.info('CC Email address %s', to_email_cc)
logger.info('From Email address %s', from_email_alias)
logger.info('DB URL %s',dburl)
logger.info('Sqoop Edge node connection %s',SQOOP_EDGE_CONN)
logger.info('Log table name %s',IB_log_table)
logger.info('Master table name %s',SR_DG_master_table)
logger.info('Data governance table name %s',SR_DG_table)
Now the question is if i want to parse an XML without any knowledge of the tags and elements and use the values how do i do it. I have gone through multiple tutorials but all of them help me with parsing the XML by using the tags like below
SQOOP_EDGE_CONN = root.find("SQOOP_EDGE_CONN").text
Can anybody point me to a right tutorial or library or a code snippet to parse the XML dynamically.
I think official documentation is pretty clear and contains some examples: https://docs.python.org/3/library/xml.etree.elementtree.html
The main part you need to implement is loop over the child nodes (potentially recursively):
for child in root:
# child.tag contains the tag name, child.attrib contains the attributes
print(child.tag, child.attrib)
Well parsing is easy as that - etree.parse(path)
Once you've got the root in hand using tree.getroot() you can just iterate over the tree using Python's "in":
for child_node in tree.getroot():
print child_node.text
Then, to see tags these child_nodes have, you do the same trick.
This lets you go over all tags in the XML without having to know the tag names at all.

Python: XML parsing using xml.dom.minidom - iterating over collection.getElementsByTagName

I have some xml I'm trying to parse using Python 2.7. The XML is in the format below. In the code (also included below), if I try to print using collection.getAttribute('ProjectId') I get the id number, but once I assign collection to tags and then run it through a loop, I don't get output(no error message). Any clues?
XML:
<SummaryReport ProjectId="37f8d135-1f1d-4e57-9b7d-b084770c6bf5" EntityId="016fbc07-69f0-407e-b5b5-0b0b6bba4307" Status="Failed">
<TotalCount>0</TotalCount>
<SuccessfulCount>0</SuccessfulCount>
<FailedCount>0</FailedCount>
<StartUtcTime>2015-09-09T16:43:11.810715Z</StartUtcTime>
<EndUtcTime>2015-09-09T16:43:44.5418427Z</EndUtcTime>
<IsIncremental>false</IsIncremental>
<OnDemand>true</OnDemand>
<TrackingId>c0972936-c8b6-4cdb-b089-d08c6f9702aa</TrackingId>
<Message>An error occurred during content building: Index was out of range. Must be non-negative and less than the size of the collection.
Parameter name: index</Message>
<LogEntries>
<LogEntry>
<Level>Info</Level>
<LogCodeType>System</LogCodeType>
<LogCode>PhaseSucceedInfo</LogCode>
<Name>Phase</Name>
<Message>'Load Metadata' succeeded in 00:00:00.1905878 seconds.</Message>
<Anchor>Info_7333babe-fc51-4b45-9167-bf263e7babcb</Anchor>
</LogEntry>
<LogEntry>
<Level>Info</Level>
<LogCodeType>System</LogCodeType>
<LogCode>PublishRequest</LogCode>
<Name>PublishTocAndArticleInit</Name>
<Message>'Load Metadata' succeeded in 00:00:01.1905878 seconds.</Message>
<Anchor>Info_51c10e71-d99a-49f9-b4aa-d83dc273426a</Anchor>
</LogEntry>
</LogEntries>
</SummaryReport>
Code:
#!/usr/bin/python
from xml.dom.minidom import parse
import xml.dom.minidom
# Open XML document using minidom parser
DOMTree = xml.dom.minidom.parse("file.xml")
collection = DOMTree.documentElement
#Get all the tags under summaryreport
tags = collection.getElementsByTagName("SummaryReport")
#print tag info
for tag in tags:
print '*******Tag Info************'
print 'Project Id: %s' % tag.getAttribute('ProjectId')
You get no output because collection.getElementsByTagName("SummaryReport") returns nothing :
>>> tags = collection.getElementsByTagName("SummaryReport")
>>> print(tags)
[]
That make sense since collection already reference SummaryReport element and it has no descendant element named the same.
UPDATE :
Simple for loop works fine to iterate through Level elements and print the value, for example :
>>> tags = collection.getElementsByTagName("Level")
>>> for tag in tags:
print(tag.firstChild.nodeValue)
Info
Info

Python 3.4 - XML Parse - IndexError: List Index Out of Range - How do I find range of XML?

Okay guys, I'm new to parsing XML and Python, and am trying to get this to work. If someone could help me with this it would be greatly appreciated. If you can help me (educate me) on how to figure it out for myself, that would be even better!
I am having trouble trying to figure out the range to reference for an XML document as I can't find any documentation on it. Here is my code and I'll include the entire Traceback after.
#import library to do http requests:
import urllib.request
#import easy to use xml parser called minidom:
from xml.dom.minidom import parseString
#all these imports are standard on most modern python implementations
#download the file:
file = urllib.request.urlopen('http://www.wizards.com/dndinsider/compendium/CompendiumSearch.asmx/KeywordSearch?Keywords=healing%20%word&nameOnly=True&tab=')
#convert to string:
data = file.read()
#close file because we dont need it anymore:
file.close()
#parse the xml you downloaded
dom = parseString(data)
#retrieve the first xml tag (<tag>data</tag>) that the parser finds with name tagName:
xmlTag = dom.getElementsByTagName('Data.Results.Power.ID')[0].toxml()
#strip off the tag (<tag>data</tag> ---> data):
xmlData=xmlTag.replace('<id>','').replace('</id>','')
#print out the xml tag and data in this format: <tag>data</tag>
print(xmlTag)
#just print the data
print(xmlData)
Traceback
/usr/bin/python3.4 /home/mint/PycharmProjects/DnD_Project/Power_Name.py
Traceback (most recent call last):
File "/home/mint/PycharmProjects/DnD_Project/Power_Name.py", line 14, in <module>
xmlTag = dom.getElementsByTagName('id')[0].toxml()
IndexError: list index out of range
Process finished with exit code 1
print len( dom.getElementsByTagName('id') )
EDIT:
ids = dom.getElementsByTagName('id')
if len( ids ) > 0 :
xmlTag = ids[0].toxml()
# rest of code
EDIT: I add example because I saw in other comment tha you don't know how to use it
BTW: I add some comment in code about file/connection
import urllib.request
from xml.dom.minidom import parseString
# create connection to data/file on server
connection = urllib.request.urlopen('http://www.wizards.com/dndinsider/compendium/CompendiumSearch.asmx/KeywordSearch?Keywords=healing%20%word&nameOnly=True&tab=')
# read from server as string (not "convert" to string):
data = connection.read()
#close connection because we dont need it anymore:
connection.close()
dom = parseString(data)
# get tags from dom
ids = dom.getElementsByTagName('Data.Results.Power.ID')
# check if there are any data
if len( ids ) > 0 :
xmlTag = ids[0].toxml()
xmlData=xmlTag.replace('<id>','').replace('</id>','')
print(xmlTag)
print(xmlData)
else:
print("Sorry, there was no data")
or you can use for loop if there is more tags
dom = parseString(data)
# get tags from dom
ids = dom.getElementsByTagName('Data.Results.Power.ID')
# get all tags - one by one
for one_tag in ids:
xmlTag = one_tag.toxml()
xmlData = xmlTag.replace('<id>','').replace('</id>','')
print(xmlTag)
print(xmlData)
BTW:
getElementsByTagName() expects tagname ID - not path Data.Results.Power.ID
tagname is ID so you have to replace <ID> not <id>
for this tag you can event use one_tag.firstChild.nodeValue in place of xmlTag.replace
.
dom = parseString(data)
# get tags from dom
ids = dom.getElementsByTagName('ID') # tagname
# get all tags - one by one
for one_tag in ids:
xmlTag = one_tag.toxml()
#xmlData = xmlTag.replace('<ID>','').replace('</ID>','')
xmlData = one_tag.firstChild.nodeValue
print(xmlTag)
print(xmlData)
I haven't used the built in xml library in a while, but it's covered in Mark Pilgrim's great Dive into Python book.
-- I see as I'm typing this that your question has already been answered but since you mention being new to Python I think you will find the text useful for xml parsing and as an excellent introduction to the language.
If you would like to try another approach to parsing xml and html, I highly recommend lxml.

Extracting text from HTML file using Python

I'd like to extract the text from an HTML file using Python. I want essentially the same output I would get if I copied the text from a browser and pasted it into notepad.
I'd like something more robust than using regular expressions that may fail on poorly formed HTML. I've seen many people recommend Beautiful Soup, but I've had a few problems using it. For one, it picked up unwanted text, such as JavaScript source. Also, it did not interpret HTML entities. For example, I would expect ' in HTML source to be converted to an apostrophe in text, just as if I'd pasted the browser content into notepad.
Update html2text looks promising. It handles HTML entities correctly and ignores JavaScript. However, it does not exactly produce plain text; it produces markdown that would then have to be turned into plain text. It comes with no examples or documentation, but the code looks clean.
Related questions:
Filter out HTML tags and resolve entities in python
Convert XML/HTML Entities into Unicode String in Python
The best piece of code I found for extracting text without getting javascript or not wanted things :
from urllib.request import urlopen
from bs4 import BeautifulSoup
url = "http://news.bbc.co.uk/2/hi/health/2284783.stm"
html = urlopen(url).read()
soup = BeautifulSoup(html, features="html.parser")
# kill all script and style elements
for script in soup(["script", "style"]):
script.extract() # rip it out
# get text
text = soup.get_text()
# break into lines and remove leading and trailing space on each
lines = (line.strip() for line in text.splitlines())
# break multi-headlines into a line each
chunks = (phrase.strip() for line in lines for phrase in line.split(" "))
# drop blank lines
text = '\n'.join(chunk for chunk in chunks if chunk)
print(text)
You just have to install BeautifulSoup before :
pip install beautifulsoup4
html2text is a Python program that does a pretty good job at this.
NOTE: NTLK no longer supports clean_html function
Original answer below, and an alternative in the comments sections.
Use NLTK
I wasted my 4-5 hours fixing the issues with html2text. Luckily i could encounter NLTK.
It works magically.
import nltk
from urllib import urlopen
url = "http://news.bbc.co.uk/2/hi/health/2284783.stm"
html = urlopen(url).read()
raw = nltk.clean_html(html)
print(raw)
Found myself facing just the same problem today. I wrote a very simple HTML parser to strip incoming content of all markups, returning the remaining text with only a minimum of formatting.
from HTMLParser import HTMLParser
from re import sub
from sys import stderr
from traceback import print_exc
class _DeHTMLParser(HTMLParser):
def __init__(self):
HTMLParser.__init__(self)
self.__text = []
def handle_data(self, data):
text = data.strip()
if len(text) > 0:
text = sub('[ \t\r\n]+', ' ', text)
self.__text.append(text + ' ')
def handle_starttag(self, tag, attrs):
if tag == 'p':
self.__text.append('\n\n')
elif tag == 'br':
self.__text.append('\n')
def handle_startendtag(self, tag, attrs):
if tag == 'br':
self.__text.append('\n\n')
def text(self):
return ''.join(self.__text).strip()
def dehtml(text):
try:
parser = _DeHTMLParser()
parser.feed(text)
parser.close()
return parser.text()
except:
print_exc(file=stderr)
return text
def main():
text = r'''
<html>
<body>
<b>Project:</b> DeHTML<br>
<b>Description</b>:<br>
This small script is intended to allow conversion from HTML markup to
plain text.
</body>
</html>
'''
print(dehtml(text))
if __name__ == '__main__':
main()
I know there are a lot of answers already, but the most elegent and pythonic solution I have found is described, in part, here.
from bs4 import BeautifulSoup
text = ' '.join(BeautifulSoup(some_html_string, "html.parser").findAll(text=True))
Update
Based on Fraser's comment, here is more elegant solution:
from bs4 import BeautifulSoup
clean_text = ' '.join(BeautifulSoup(some_html_string, "html.parser").stripped_strings)
Here is a version of xperroni's answer which is a bit more complete. It skips script and style sections and translates charrefs (e.g., ') and HTML entities (e.g., &).
It also includes a trivial plain-text-to-html inverse converter.
"""
HTML <-> text conversions.
"""
from HTMLParser import HTMLParser, HTMLParseError
from htmlentitydefs import name2codepoint
import re
class _HTMLToText(HTMLParser):
def __init__(self):
HTMLParser.__init__(self)
self._buf = []
self.hide_output = False
def handle_starttag(self, tag, attrs):
if tag in ('p', 'br') and not self.hide_output:
self._buf.append('\n')
elif tag in ('script', 'style'):
self.hide_output = True
def handle_startendtag(self, tag, attrs):
if tag == 'br':
self._buf.append('\n')
def handle_endtag(self, tag):
if tag == 'p':
self._buf.append('\n')
elif tag in ('script', 'style'):
self.hide_output = False
def handle_data(self, text):
if text and not self.hide_output:
self._buf.append(re.sub(r'\s+', ' ', text))
def handle_entityref(self, name):
if name in name2codepoint and not self.hide_output:
c = unichr(name2codepoint[name])
self._buf.append(c)
def handle_charref(self, name):
if not self.hide_output:
n = int(name[1:], 16) if name.startswith('x') else int(name)
self._buf.append(unichr(n))
def get_text(self):
return re.sub(r' +', ' ', ''.join(self._buf))
def html_to_text(html):
"""
Given a piece of HTML, return the plain text it contains.
This handles entities and char refs, but not javascript and stylesheets.
"""
parser = _HTMLToText()
try:
parser.feed(html)
parser.close()
except HTMLParseError:
pass
return parser.get_text()
def text_to_html(text):
"""
Convert the given text to html, wrapping what looks like URLs with <a> tags,
converting newlines to <br> tags and converting confusing chars into html
entities.
"""
def f(mo):
t = mo.group()
if len(t) == 1:
return {'&':'&', "'":''', '"':'"', '<':'<', '>':'>'}.get(t)
return '%s' % (t, t)
return re.sub(r'https?://[^] ()"\';]+|[&\'"<>]', f, text)
I know there's plenty of answers here already but I think newspaper3k also deserves a mention. I recently needed to complete a similar task of extracting the text from articles on the web and this library has done an excellent job of achieving this so far in my tests. It ignores the text found in menu items and side bars as well as any JavaScript that appears on the page as the OP requests.
from newspaper import Article
article = Article(url)
article.download()
article.parse()
article.text
If you already have the HTML files downloaded you can do something like this:
article = Article('')
article.set_html(html)
article.parse()
article.text
It even has a few NLP features for summarizing the topics of articles:
article.nlp()
article.summary
You can use html2text method in the stripogram library also.
from stripogram import html2text
text = html2text(your_html_string)
To install stripogram run sudo easy_install stripogram
There is Pattern library for data mining.
http://www.clips.ua.ac.be/pages/pattern-web
You can even decide what tags to keep:
s = URL('http://www.clips.ua.ac.be').download()
s = plaintext(s, keep={'h1':[], 'h2':[], 'strong':[], 'a':['href']})
print s
if you need more speed and less accuracy then you could use raw lxml.
import lxml.html as lh
from lxml.html.clean import clean_html
def lxml_to_text(html):
doc = lh.fromstring(html)
doc = clean_html(doc)
return doc.text_content()
PyParsing does a great job. The PyParsing wiki was killed so here is another location where there are examples of the use of PyParsing (example link). One reason for investing a little time with pyparsing is that he has also written a very brief very well organized O'Reilly Short Cut manual that is also inexpensive.
Having said that, I use BeautifulSoup a lot and it is not that hard to deal with the entities issues, you can convert them before you run BeautifulSoup.
Goodluck
This isn't exactly a Python solution, but it will convert text Javascript would generate into text, which I think is important (E.G. google.com). The browser Links (not Lynx) has a Javascript engine, and will convert source to text with the -dump option.
So you could do something like:
fname = os.tmpnam()
fname.write(html_source)
proc = subprocess.Popen(['links', '-dump', fname],
stdout=subprocess.PIPE,
stderr=open('/dev/null','w'))
text = proc.stdout.read()
Instead of the HTMLParser module, check out htmllib. It has a similar interface, but does more of the work for you. (It is pretty ancient, so it's not much help in terms of getting rid of javascript and css. You could make a derived class, but and add methods with names like start_script and end_style (see the python docs for details), but it's hard to do this reliably for malformed html.) Anyway, here's something simple that prints the plain text to the console
from htmllib import HTMLParser, HTMLParseError
from formatter import AbstractFormatter, DumbWriter
p = HTMLParser(AbstractFormatter(DumbWriter()))
try: p.feed('hello<br>there'); p.close() #calling close is not usually needed, but let's play it safe
except HTMLParseError: print ':(' #the html is badly malformed (or you found a bug)
I recommend a Python Package called goose-extractor
Goose will try to extract the following information:
Main text of an article
Main image of article
Any Youtube/Vimeo movies embedded in article
Meta Description
Meta tags
More :https://pypi.python.org/pypi/goose-extractor/
Anyone has tried bleach.clean(html,tags=[],strip=True) with bleach? it's working for me.
install html2text using
pip install html2text
then,
>>> import html2text
>>>
>>> h = html2text.HTML2Text()
>>> # Ignore converting links from HTML
>>> h.ignore_links = True
>>> print h.handle("<p>Hello, <a href='http://earth.google.com/'>world</a>!")
Hello, world!
Best worked for me is inscripts .
https://github.com/weblyzard/inscriptis
import urllib.request
from inscriptis import get_text
url = "http://www.informationscience.ch"
html = urllib.request.urlopen(url).read().decode('utf-8')
text = get_text(html)
print(text)
The results are really good
Beautiful soup does convert html entities. It's probably your best bet considering HTML is often buggy and filled with unicode and html encoding issues. This is the code I use to convert html to raw text:
import BeautifulSoup
def getsoup(data, to_unicode=False):
data = data.replace(" ", " ")
# Fixes for bad markup I've seen in the wild. Remove if not applicable.
masssage_bad_comments = [
(re.compile('<!-([^-])'), lambda match: '<!--' + match.group(1)),
(re.compile('<!WWWAnswer T[=\w\d\s]*>'), lambda match: '<!--' + match.group(0) + '-->'),
]
myNewMassage = copy.copy(BeautifulSoup.BeautifulSoup.MARKUP_MASSAGE)
myNewMassage.extend(masssage_bad_comments)
return BeautifulSoup.BeautifulSoup(data, markupMassage=myNewMassage,
convertEntities=BeautifulSoup.BeautifulSoup.ALL_ENTITIES
if to_unicode else None)
remove_html = lambda c: getsoup(c, to_unicode=True).getText(separator=u' ') if c else ""
Another non-python solution: Libre Office:
soffice --headless --invisible --convert-to txt input1.html
The reason I prefer this one over other alternatives is that every HTML paragraph gets converted into a single text line (no line breaks), which is what I was looking for. Other methods require post-processing. Lynx does produce nice output, but not exactly what I was looking for. Besides, Libre Office can be used to convert from all sorts of formats...
I had a similar question and actually used one of the answers with BeautifulSoup.
The problem was it was really slow. I ended up using library called selectolax.
It's pretty limited but it works for this task.
The only issue was that I had manually remove unnecessary white spaces.
But it seems to be working much faster that BeautifulSoup solution.
from selectolax.parser import HTMLParser
def get_text_selectolax(html):
tree = HTMLParser(html)
if tree.body is None:
return None
for tag in tree.css('script'):
tag.decompose()
for tag in tree.css('style'):
tag.decompose()
text = tree.body.text(separator='')
text = " ".join(text.split()) # this will remove all the whitespaces
return text
Another option is to run the html through a text based web browser and dump it. For example (using Lynx):
lynx -dump html_to_convert.html > converted_html.txt
This can be done within a python script as follows:
import subprocess
with open('converted_html.txt', 'w') as outputFile:
subprocess.call(['lynx', '-dump', 'html_to_convert.html'], stdout=testFile)
It won't give you exactly just the text from the HTML file, but depending on your use case it may be preferable to the output of html2text.
#PeYoTIL's answer using BeautifulSoup and eliminating style and script content didn't work for me. I tried it using decompose instead of extract but it still didn't work. So I created my own which also formats the text using the <p> tags and replaces <a> tags with the href link. Also copes with links inside text. Available at this gist with a test doc embedded.
from bs4 import BeautifulSoup, NavigableString
def html_to_text(html):
"Creates a formatted text email message as a string from a rendered html template (page)"
soup = BeautifulSoup(html, 'html.parser')
# Ignore anything in head
body, text = soup.body, []
for element in body.descendants:
# We use type and not isinstance since comments, cdata, etc are subclasses that we don't want
if type(element) == NavigableString:
# We use the assumption that other tags can't be inside a script or style
if element.parent.name in ('script', 'style'):
continue
# remove any multiple and leading/trailing whitespace
string = ' '.join(element.string.split())
if string:
if element.parent.name == 'a':
a_tag = element.parent
# replace link text with the link
string = a_tag['href']
# concatenate with any non-empty immediately previous string
if ( type(a_tag.previous_sibling) == NavigableString and
a_tag.previous_sibling.string.strip() ):
text[-1] = text[-1] + ' ' + string
continue
elif element.previous_sibling and element.previous_sibling.name == 'a':
text[-1] = text[-1] + ' ' + string
continue
elif element.parent.name == 'p':
# Add extra paragraph formatting newline
string = '\n' + string
text += [string]
doc = '\n'.join(text)
return doc
I've had good results with Apache Tika. Its purpose is the extraction of metadata and text from content, hence the underlying parser is tuned accordingly out of the box.
Tika can be run as a server, is trivial to run / deploy in a Docker container, and from there can be accessed via Python bindings.
While alot of people mentioned using regex to strip html tags, there are a lot of downsides.
for example:
<p>hello world</p>I love you
Should be parsed to:
Hello world
I love you
Here's a snippet I came up with, you can cusomize it to your specific needs, and it works like a charm
import re
import html
def html2text(htm):
ret = html.unescape(htm)
ret = ret.translate({
8209: ord('-'),
8220: ord('"'),
8221: ord('"'),
160: ord(' '),
})
ret = re.sub(r"\s", " ", ret, flags = re.MULTILINE)
ret = re.sub("<br>|<br />|</p>|</div>|</h\d>", "\n", ret, flags = re.IGNORECASE)
ret = re.sub('<.*?>', ' ', ret, flags=re.DOTALL)
ret = re.sub(r" +", " ", ret)
return ret
in a simple way
import re
html_text = open('html_file.html').read()
text_filtered = re.sub(r'<(.*?)>', '', html_text)
this code finds all parts of the html_text started with '<' and ending with '>' and replace all found by an empty string
In Python 3.x you can do it in a very easy way by importing 'imaplib' and 'email' packages. Although this is an older post but maybe my answer can help new comers on this post.
status, data = self.imap.fetch(num, '(RFC822)')
email_msg = email.message_from_bytes(data[0][1])
#email.message_from_string(data[0][1])
#If message is multi part we only want the text version of the body, this walks the message and gets the body.
if email_msg.is_multipart():
for part in email_msg.walk():
if part.get_content_type() == "text/plain":
body = part.get_payload(decode=True) #to control automatic email-style MIME decoding (e.g., Base64, uuencode, quoted-printable)
body = body.decode()
elif part.get_content_type() == "text/html":
continue
Now you can print body variable and it will be in plaintext format :) If it is good enough for you then it would be nice to select it as accepted answer.
Here's the code I use on a regular basis.
from bs4 import BeautifulSoup
import urllib.request
def processText(webpage):
# EMPTY LIST TO STORE PROCESSED TEXT
proc_text = []
try:
news_open = urllib.request.urlopen(webpage.group())
news_soup = BeautifulSoup(news_open, "lxml")
news_para = news_soup.find_all("p", text = True)
for item in news_para:
# SPLIT WORDS, JOIN WORDS TO REMOVE EXTRA SPACES
para_text = (' ').join((item.text).split())
# COMBINE LINES/PARAGRAPHS INTO A LIST
proc_text.append(para_text)
except urllib.error.HTTPError:
pass
return proc_text
I hope that helps.
you can extract only text from HTML with BeautifulSoup
url = "https://www.geeksforgeeks.org/extracting-email-addresses-using-regular-expressions-python/"
con = urlopen(url).read()
soup = BeautifulSoup(con,'html.parser')
texts = soup.get_text()
print(texts)
Another example using BeautifulSoup4 in Python 2.7.9+
includes:
import urllib2
from bs4 import BeautifulSoup
Code:
def read_website_to_text(url):
page = urllib2.urlopen(url)
soup = BeautifulSoup(page, 'html.parser')
for script in soup(["script", "style"]):
script.extract()
text = soup.get_text()
lines = (line.strip() for line in text.splitlines())
chunks = (phrase.strip() for line in lines for phrase in line.split(" "))
text = '\n'.join(chunk for chunk in chunks if chunk)
return str(text.encode('utf-8'))
Explained:
Read in the url data as html (using BeautifulSoup), remove all script and style elements, and also get just the text using .get_text(). Break into lines and remove leading and trailing space on each, then break multi-headlines into a line each chunks = (phrase.strip() for line in lines for phrase in line.split(" ")). Then using text = '\n'.join, drop blank lines, finally return as sanctioned utf-8.
Notes:
Some systems this is run on will fail with https:// connections because of SSL issue, you can turn off the verify to fix that issue. Example fix: http://blog.pengyifan.com/how-to-fix-python-ssl-certificate_verify_failed/
Python < 2.7.9 may have some issue running this
text.encode('utf-8') can leave weird encoding, may want to just return str(text) instead.
Answer using Pandas to get table data from HTML.
If you want to extract table data quickly from HTML. You can use the read_HTML function, docs are here. Before using this function you should read the gotchas/issues surrounding the BeautifulSoup4/html5lib/lxml parsers HTML parsing libraries.
import pandas as pd
http = r'https://www.ibm.com/docs/en/cmofz/10.1.0?topic=SSQHWE_10.1.0/com.ibm.ondemand.mp.doc/arsa0257.htm'
table = pd.read_html(http)
df = table[0]
df
output
There are a number of option that can be played with see here and here.

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