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Extract content with BeautifulSoup and Python

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.

how to grab alternating child tags in python beautifulsoup

I am trying to get a series of data from alternating tags in a html page.
The html looks like this:
<div>
<h3>title</h3>
<div>text</div>
<h3>title</h3>
<div>text</div>
...
</div>
Since I can't grab each h3/div pair in a "for each pair in div", how to I grab them efficiently?

Find all headers, and grab the next sibling from there:
for header in soup.select('div h3'):
next_div = header.find_next_sibling('div')
element.find_next_sibling() returns an element or None if no such sibling can be found.
Demo:
>>> from bs4 import BeautifulSoup
>>> soup = BeautifulSoup('''\
... <div>
... <h3>First header</h3>
... <div>First div to go with a header</div>
... <h3>Second header</h3>
... <div>Second div to go with a header</div>
... </div>
... ''')
>>> for header in soup.select('div h3'):
... next_div = header.find_next_sibling('div')
... print(header.text, next_div.text)
...
First header First div to go with a header
Second header Second div to go with a header

There are lots of ways to do this, but the easiest for me would be to select all the h3 tags and then walk the DOM to get their next sibling.

Beautiful Soup: searching for a nested pattern?

soup.find_all will search a BeautifulSoup document for all occurrences of a single tag. Is there a way to search for particular patterns of nested tags?
For example, I would like to search for all occurrences of this pattern:
<div class="separator">
<a>
<img />
</a>
</div>

There are multiple ways to find the pattern, but the easiest one would be to use a CSS selector:
for img in soup.select('div.separator > a > img'):
print img # or img.parent.parent to get the "div"
Demo:
>>> from bs4 import BeautifulSoup
>>> data = """
... <div>
... <div class="separator">
... <a>
... <img src="test1"/>
... </a>
... </div>
...
... <div class="separator">
... <a>
... <img src="test2"/>
... </a>
... </div>
...
... <div>test3</div>
...
... <div>
... <a>test4</a>
... </div>
... </div>
... """
>>> soup = BeautifulSoup(data)
>>>
>>> for img in soup.select('div.separator > a > img'):
... print img.get('src')
...
test1
test2
I do understand that, strictly speaking, the solution would not work if the div has more than just one a child, or inside the a tag there is smth else except the img tag. If this is the case the solution can be improved with additional checks (will edit the answer if needed).

Check out this part of the docs. You probably want a function like this:
def nested_img(div):
child = div.contents[0]
return child.name == "a" and child.contents[0].name == "img"
soup.find_all("div", nested_img)
P.S.: This is untested.

Python: Parse all elements under a div

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.

Matching HTML subtree with BeautifulSoup

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]