.. image:: https://badge.fury.io/py/htmlement.svg :target: https://pypi.python.org/pypi/htmlement
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.. image:: https://codecov.io/gh/willforde/python-htmlement/branch/master/graph/badge.svg?token=D5EKKLIVBP :target: https://codecov.io/gh/willforde/python-htmlement
.. image:: https://api.codeclimate.com/v1/badges/7d593426acc83cba5ef7/maintainability :target: https://codeclimate.com/github/willforde/python-htmlement/maintainability :alt: Maintainability
HTMLement is a pure Python HTML Parser.
The object of this project is to be a "pure-python HTML parser" which is also "faster" than "beautifulsoup". And like "beautifulsoup", will also parse invalid html.
The most simple way to do this is to use ElementTree XPath expressions__.
Python does support a simple (read limited) XPath engine inside its "ElementTree" module.
A benefit of using "ElementTree" is that it can use a "C implementation" whenever available.
This "HTML Parser" extends html.parser.HTMLParser_ to build a tree of ElementTree.Element_ instances.
Run ::
pip install htmlement
-or- ::
pip install git+https://github.com/willforde/python-htmlement.git
Here I’ll be using a sample "HTML document" that will be "parsed" using "htmlement": ::
html = """
<html>
<head>
<title>GitHub</title>
</head>
<body>
<a href="https://github.com/marmelo">GitHub</a>
<a href="https://github.com/marmelo/python-htmlparser">GitHub Project</a>
</body>
</html>
"""
# Parse the document
import htmlement
root = htmlement.fromstring(html)
Root is an ElementTree.Element_ and supports the ElementTree API with XPath expressions. With this I'm easily able to get both the title and all anchors in the document. ::
# Get title
title = root.find("head/title").text
print("Parsing: %s" % title)
# Get all anchors
for a in root.iterfind(".//a"):
print(a.get("href"))
And the output is as follows: ::
Parsing: GitHub
https://github.com/willforde
https://github.com/willforde/python-htmlement
Here I’ll be using a slightly more complex "HTML document" that will be "parsed" using "htmlement with a filter" to fetch only the menu items. This can be very useful when dealing with large "HTML documents" since it can be a lot faster to only "parse the required section" and to ignore everything else. ::
html = """
<html>
<head>
<title>Coffee shop</title>
</head>
<body>
<ul class="menu">
<li>Coffee</li>
<li>Tea</li>
<li>Milk</li>
</ul>
<ul class="extras">
<li>Sugar</li>
<li>Cream</li>
</ul>
</body>
</html>
"""
# Parse the document
import htmlement
root = htmlement.fromstring(html, "ul", attrs={"class": "menu"})
In this case I'm not unable to get the title, since all elements outside the filter were ignored. But this allows me to be able to extract all "list_item elements" within the menu list and nothing else. ::
# Get all listitems
for item in root.iterfind(".//li"):
# Get text from listitem
print(item.text)
And the output is as follows: ::
Coffee
Tea
Milk
.. _html.parser.HTMLParser: https://docs.python.org/3.6/library/html.parser.html#html.parser.HTMLParser .. _ElementTree.Element: https://docs.python.org/3.6/library/xml.etree.elementtree.html#xml.etree.ElementTree.Element .. Xpath: https://docs.python.org/3.6/library/xml.etree.elementtree.html#xpath-support __ XPath
FAQs
Pure-Python HTML parser with ElementTree support.
We found that htmlement demonstrated a healthy version release cadence and project activity because the last version was released less than a year ago. It has 1 open source maintainer collaborating on the project.
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