I'm trying to match something like this with beautifulsoup.
<a href="./SlimLineUSB3/SlimLine1BayUSB3.php">
<b>1 Bay SlimLine with both eSATA and USB 3.0</b>
</a>
In a regexp, it would look something like this. I want to capture the url.
<a href="\.(.*)">
<b>.*</b>
</a>
How do I go about doing something like this with BeautifulSoup? I need to use the b tags inside of the 'a' tags I want, since that's the only thing that differentiates these 'a's from any other link on the page. It seems like I can only write regexps to match the tag name or specific attributes?
If you just want to get the href from all a tags which contain one b tag:
>>> from BeautifulSoup import BeautifulSoup
>>> html = """
... <html><head><title>Title</title></head><body>
... <b>first</b>
... <a><b>no-href</b></a>
... <div><b>second</b></div>
... <div><b>third</b></div>
... no-bold-tag
... <b>text</b><p>other-stuff</p>
... </body></html>
... ... """
>>> soup = BeautifulSoup(html)
>>> [a['href'] for a in soup('a', href=True) if a.b and len(a) == 1]
[u'first/index.php', u'second/index.php', u'third/index.php']
This can be done quite elegantly using an XPath expression if you don't mind using lxml.
import lxml.html as lh
html = '''
<a href="./SlimLineUSB3/SlimLine1BayUSB3.php">
<b>1 Bay SlimLine with both eSATA and USB 3.0</b>
</a>
<a href="./Another/URL.php">
<b>foo</b>
<p>bar</p>
</a>
'''
tree = lh.fromstring(html)
for link in tree.xpath('a[count(b) = 1 and count(*) = 1]'):
print lh.tostring(link)
Result:
<a href="./SlimLineUSB3/SlimLine1BayUSB3.php">
<b>1 Bay SlimLine with both eSATA and USB 3.0</b>
</a>
Or if you wanted to use a method more similar to #ekhumoro's with lxml you could do:
[a for a in tree.xpath('a[#href]') if a.find('b') != None and len(a) == 1]
Related
I'm trying to scrap a forum but I can't deal with the comments, because the users use emoticons, and bold font, and cite previous messages, and and and...
For example, here's one of the comments that I have a problem with:
<div class="content">
<blockquote>
<div>
<cite>User write:</cite>
I DO NOT WANT THIS <img class="smilies" alt=":116:" title="116">
</div>
</blockquote>
<br/>
THIS IS THE COMMENT THAT I NEED!
</div>
I searching for help for the last 4 days and I couldn't find anything, so I decided to ask here.
This is the code that I'm using:
def make_soup(url):
html = urlopen(url).read()
return BeautifulSoup(html, "lxml")
def get_messages(url):
soup = make_soup(url)
msg = soup.find("div", {"class" : "content"})
# I get in msg the hole message, exactly as I wrote previously
print msg
# Here I get:
# 1. <blockquote> ... </blockquote>
# 2. <br/>
# 3. THIS IS THE COMMENT THAT I NEED!
for item in msg.children:
print item
I'm looking for a way to deal with messages in a general way, no matter how they are. Sometimes they put emoticons between the text and I need to remove them and get the hole message (in this situation, bsp will put each part of the message (first part, emoticon, second part) in different items).
Thanks in advance!
Use decompose http://www.crummy.com/software/BeautifulSoup/bs4/doc/#decompose
Decompose extracts tags that you don't want. In your case:
soup.blockquote.decompose()
or all unwanted tags:
for tag in ['blockquote', 'img', ... ]:
soup.find(tag).decompose()
Your example:
>>> from bs4 import BeautifulSoup
>>> html = """<div class="content">
... <blockquote>
... <div>
... <cite>User write:</cite>
... I DO NOT WANT THIS <img class="smilies" alt=":116:" title="116">
... </div>
... </blockquote>
... <br/>
... THIS IS THE COMMENT THAT I NEED!
... </div>"""
>>> soup = BeautifulSoup(html, 'html.parser')
>>> soup.find('blockquote').decompose()
>>> soup.find("div", {"class" : "content"}).text.strip()
u'THIS IS THE COMMENT THAT I NEED!'
Update
Sometimes all you have is a tag starting point but you are actually interested in the content before or after that starting point. You can use .next_sibling and .previous_sibling to navigate between page elements that are on the same level of the parse tree:
>>> html = """<div>No<blockquote>No</blockquote>Yes.<em>Yes!</em>Yes?</div>No!"""
>>> soup = BeautifulSoup(html, 'html.parser')
>>> elm = soup.blockquote.next_sibling
>>> txt = ""
>>> while elm:
... txt += elm.string
... elm = elm.next_sibling
...
>>> print(txt)
u'Yes.Yes!Yes?'
BeautifulSoup has a get_text method. Maybe this is what you want.
From their documentation:
markup = '\nI linked to <i>example.com</i>\n'
soup = BeautifulSoup(markup)
soup.get_text()
u'\nI linked to example.com\n'
soup.i.get_text()
u'example.com'
If the text you want is never within any additional tags, as in your example, you can use extract() to get rid of all the tags and their contents:
html = '<div class="content">\
<blockquote>\
<div>\
<cite>User write:</cite>\
I DO NOT WANT THIS <img class="smilies" alt=":116:" title="116">\
</div>\
</blockquote>\
<br/>\
THIS IS THE COMMENT THAT I NEED!\
</div>'
from bs4 import BeautifulSoup
soup = BeautifulSoup(html, 'lxml')
div = soup.find('div', class_='content')
tags = div.findAll(recursive=False)
for tag in tags:
tag.extract()
text = div.get_text(strip=True)
print(text)
This gives:
THIS IS THE COMMENT THAT I NEED!
To deal with emoticons, you'll have to do something more complicated. You'll probably have to define a list of emoticons to recognize yourself, and then parse the text to look for them.
I'm not very familliar with BeautifulSoup.
I have the html code like (it's only part of it):
<div class="central-featured-lang lang1" lang="en">
<a class="link-box" href="//en.wikibooks.org/">
<strong>English</strong><br>
<em>Open-content textbooks</em><br>
<small>51 000+ pages</small></a>
</div>
On the output I should get (and for other languages):
English: 51 000+ pages.
I tried something like:
for item in soup.find_all('div'):
print item.get('class')
But this does not work. Can you help me, or at least lead to solution?
item.get() returns attribute values, not text contained under an element.
You can get the text directly contained in elements with the Element.string attribute, or all contained text (recursively) with the Element.get_text() method.
Here, I'd search for div elements with a lang attribute, then use the contained elements to find strings:
for item in soup.find_all('div', lang=True):
if not (item.strong and item.small):
continue
language = item.strong.string
pages = item.small.string
print '{}: {}'.format(language, pages)
Demo:
>>> from bs4 import BeautifulSoup
>>> sample = '''\
... <div class="central-featured-lang lang1" lang="en">
... <a class="link-box" href="//en.wikibooks.org/">
... <strong>English</strong><br>
... <em>Open-content textbooks</em><br>
... <small>51 000+ pages</small></a>
... </div>
... '''
>>> soup = BeautifulSoup(sample)
>>> for item in soup.find_all('div', lang=True):
... if not (item.strong and item.small):
... continue
... language = item.strong.string
... pages = item.small.string
... print '{}: {}'.format(language, pages)
...
English: 51 000+ pages
I am trying to parse all elements under div using beautifulsoup the issue is that I don't know all the elements underneath the div prior to parsing. For example a div can have text data in paragraph mode and bullet format along with some href elements. Each url that I open can have different elements underneath the specific div class that I am looking at:
example:
url a can have following:
<div class='content'>
<p> Hello I have a link </p>
<li> I have a bullet point
foo
</div>
but url b
can have
<div class='content'>
<p> I only have paragraph </p>
</div>
I started as doing something like this:
content = souping_page.body.find('div', attrs={'class': 'content})
but how to go beyond this is little confuse. I was hoping to create one string from all the parse data as a end result.
At the end I want the following string to be obtain from each example:
Example 1: Final Output
parse_data = Hello I have a link I have a bullet point
parse_links = foo.com
Example 2: Final Output
parse_data = I only have paragraph
You can get just the text of a text with element.get_text():
>>> from bs4 import BeautifulSoup
>>> sample1 = BeautifulSoup('''\
... <div class='content'>
... <p> Hello I have a link </p>
...
... <li> I have a bullet point
...
... foo
... </div>
... ''').find('div')
>>> sample2 = BeautifulSoup('''\
... <div class='content'>
... <p> I only have paragraph </p>
...
... </div>
... ''').find('div')
>>> sample1.get_text()
u'\n Hello I have a link \n I have a bullet point\n\nfoo\n'
>>> sample2.get_text()
u'\n I only have paragraph \n'
or you can strip it down a little using element.stripped_strings:
>>> ' '.join(sample1.stripped_strings)
u'Hello I have a link I have a bullet point foo'
>>> ' '.join(sample2.stripped_strings)
u'I only have paragraph'
To get all links, look for all a elements with href attributes and gather these in a list:
>>> [a['href'] for a in sample1.find_all('a', href=True)]
['foo.com']
>>> [a['href'] for a in sample2.find_all('a', href=True)]
[]
The href=True argument limits the search to <a> tags that have a href attribute defined.
Per the Beautiful Soup docs, to iterate over the children of a tag use either .contents to get them as a list or .children (a generator).
for child in title_tag.children:
print(child)
So, in your case, for example, you grab the .text of each tag and concatenate it together. I'm not clear on whether you want the link location or simply the label, if the former, refer to this SO question.
These are my first steps with python, please bear with me.
Basically I want to parse a Table of Contents from a single Dokuwiki page with Beautiful Soup. The TOC looks like this:
<div id="dw__toc">
<h3 class="toggle">Table of Contents</h3>
<div>
<ul class="toc">
<li class="level1"><div class="li">#</div>
<ul class="toc">
<li class="level2"><div class="li">One</div></li>
<li class="level2"><div class="li">Two</div></li>
<li class="level2"><div class="li">Three</div></li>
I would like to be able to search in the content of the a-tags and if a result is found return its content and also return the href-link. So if I search for "one" the result should be
One
#link1
What I have done so far:
#!/usr/bin/python2
from BeautifulSoup import BeautifulSoup
import urllib2
#Grab and open URL, create BeatifulSoup object
url = "http://www.somewiki.at/wiki/doku.php"
page = urllib2.urlopen(url)
soup = BeautifulSoup(page.read())
#Grab Table of Contents
grab_toc = soup.find('div', {"id":"dw__toc"})
#Look for all divs with class: li
ftext = grab_toc.findAll('div', {"class":"li"})
#Look for links
links = grab_toc.findAll('a',href=True)
#Iterate
for everytext in ftext:
text = ''.join(everytext.findAll(text=True))
data = text.strip()
print data
for everylink in links:
print everylink['href']
This prints out the data I want but I'm kind of lost to rewrite it to be able to search within the result and only return the searchterm. Tried something like
if data == 'searchtearm':
print data
break
else:
print 'Nothing found'
But this is kind of a weak search. Is there a nicer way to do this? In my example the Beatiful Soup resultset is changed into a list. Is it better to search in the result set in the first place, if so then how to do this?
Instead of searching through the links one-by-one, have BeautifulSoup search for you, using a regular expression:
import re
matching_link = grab_toc.find('a', text=re.compile('one', re.IGNORECASE))
This would find the first a link in the table of contents with the 3 characters one in the text somewhere. Then just print the link and text:
print matching_link.string
print matching_link['href']
Short demo based on your sample:
>>> from bs4 import BeautifulSoup
>>> import re
>>> soup = BeautifulSoup('''\
... <div id="dw__toc">
... <h3 class="toggle">Table of Contents</h3>
... <div>
...
... <ul class="toc">
... <li class="level1"><div class="li">#</div>
... <ul class="toc">
... <li class="level2"><div class="li">One</div></li>
... <li class="level2"><div class="li">Two</div></li>
... <li class="level2"><div class="li">Three</div></li>
... </ul></ul>''')
>>> matching_link = soup.find('a', text=re.compile('one', re.IGNORECASE))
>>> print matching_link.string
One
>>> print matching_link['href']
#link1
In BeautifulSoup version 3, the above .find() call returns the contained NavigableString object instead. To get back to the parent a element, use the .parent attribute:
matching_link = grab_toc.find('a', text=re.compile('one', re.IGNORECASE)).parent
print matching_link.string
print matching_link['href']
Here's a piece of HTML code (from delicious):
<h4>
<a rel="nofollow" class="taggedlink " href="http://imfy.us/" >Generate Secure Links with Anonymous Referers & Anti-Bot Protection</a>
<span class="saverem">
<em class="bookmark-actions">
<strong><a class="inlinesave action" href="/save?url=http%3A%2F%2Fimfy.us%2F&title=Generate%20Secure%20Links%20with%20Anonymous%20Referers%20%26%20Anti-Bot%20Protection&jump=%2Fdux&key=fFS4QzJW2lBf4gAtcrbuekRQfTY-&original_user=dux©user=dux©tags=web+apps+url+security+generator+shortener+anonymous+links">SAVE</a></strong>
</em>
</span>
</h4>
I'm trying to find all the links where class="inlinesave action". Here's the code:
sock = urllib2.urlopen('http://delicious.com/theuser')
html = sock.read()
soup = BeautifulSoup(html)
tags = soup.findAll('a', attrs={'class':'inlinesave action'})
print len(tags)
But it doesn't find anything!
Any thoughts?
Thanks
If you want to look for an anchor with exactly those two classes you'd, have to use a regexp, I think:
tags = soup.findAll('a', attrs={'class': re.compile(r'\binlinesave\b.*\baction\b')})
Keep in mind that this regexp won't work if the ordering of the class names is reversed (class="action inlinesave").
The following statement should work for all cases (even though it looks ugly imo.):
soup.findAll('a',
attrs={'class':
re.compile(r'\baction\b.*\binlinesave\b|\binlinesave\b.*\baction\b')
})
Python string methods
html=open("file").read()
for item in html.split("<strong>"):
if "class" in item and "inlinesave action" in item:
url_with_junk = item.split('href="')[1]
m = url_with_junk.index('">')
print url_with_junk[:m]
May be that issue is fixed in verion 3.1.0, I could parse yours,
>>> html="""<h4>
... <a rel="nofollow" class="taggedlink " href="http://imfy.us/" >Generate Secure Links with Anony
... <span class="saverem">
... <em class="bookmark-actions">
... <strong><a class="inlinesave action" href="/save?url=http%3A%2F%2Fimfy.us%2F&title=Gen
... </em>
... </span>
... </h4>"""
>>>
>>> from BeautifulSoup import BeautifulSoup
>>> soup = BeautifulSoup(html)
>>> tags = soup.findAll('a', attrs={'class':'inlinesave action'})
>>> print len(tags)
1
>>> tags
[<a class="inlinesave action" href="/save?url=http%3A%2F%2Fimfy.us%2F&title=Generate%20Secure%
>>>
I have tried with BeautifulSoup 2.1.1 also, its does not work at all.
You might make some forward progress using pyparsing:
from pyparsing import makeHTMLTags, withAttribute
htmlsrc="""<h4>... etc."""
atag = makeHTMLTags("a")[0]
atag.setParseAction(withAttribute(("class","inlinesave action")))
for result in atag.searchString(htmlsrc):
print result.href
Gives (long result output snipped at '...'):
/save?url=http%3A%2F%2Fimfy.us%2F&title=Genera...+anonymous+links