Beautiful Soup - how to get href - beautifulsoup

I can't seem to be able to extract the href (there is only one <strong>Website:</strong> on the page) from the following soup of html:
<div id='id_Website'>
<strong>Website:</strong>
<a href='http://google.com' target='_blank' rel='nofollow'>www.google.com</a>
</div></div><div>
This is what I thought should work
href = soup.find("strong" ,text=re.compile(r'Website')).next["href"]

.next in this case is a NavigableString containing the whitespace between the <strong> tag and the <a> tag. Also, the text= attribute is for matching NavigableStrings, rather than elements.
The following does what you want, I think:
import re
from BeautifulSoup import BeautifulSoup
html = '''<div id='id_Website'>
<strong>Website:</strong>
<a href='http://google.com' target='_blank' rel='nofollow'>www.google.com</a>
</div></div><div>'''
soup = BeautifulSoup(html)
for t in soup.findAll(text=re.compile(r'Website:')):
# Find the parent of the NavigableString, and see
# whether that's a <strong>:
s = t.parent
if s.name == 'strong':
print s.nextSibling.nextSibling['href']
... but that isn't very robust. If the enclosing div has a predictable ID, then it would better to find that, and then find the first <a> element within it.

Related

BS4 - Replacing text content, preserving tags

I have an HTML document that uses the text-styling style attribute to change case. When I see that style, I'd like to change all text for which that tag applies, retaining the HTML tags.
I have a partial solution that replaces the tag entirely. The approach that seems like it ought to be correct gives me AttributeError: 'NoneType' object has no attribute 'next_element'
Example:
from bs4 import BeautifulSoup, NavigableString, Tag
import re
html = '''
<div style="text-transform: uppercase;">
Foo0
<font>Foo0</font>
<div>Foo1
<div>Foo2</div>
</div>
</div>
'''
upper_patt = re.compile('(?i)text-transform:\s*uppercase')
# works, but replaces all text, removing the HTML tags
for node in soup.find_all(attrs={'style': upper_patt}):
node.replace_with(node.text.upper())
# does not work, throws AttributeError error
soup = BeautifulSoup(html, "html.parser")
for node in soup.find_all(attrs={'style': upper_patt}):
for txt in node.strings:
txt.replace_with(txt.upper())
Seems like you want to change the inner text to uppercase for all the children of an element with text-transform: uppercase.
Instead of altering the result of find_all, loop over the children text with node.findChildren(text=True) of the result, and use replace_with() to change the text:
from bs4 import BeautifulSoup, NavigableString, Tag
import re
html = '''
<div style="text-transform: uppercase;">
Foo0
<font>Foo0</font>
<div>Foo1
<div>Foo2</div>
</div>
</div>
'''
upper_patt = re.compile('(?i)text-transform:\s*uppercase')
soup = BeautifulSoup(html, "html.parser")
for node in soup.find_all(attrs={'style': upper_patt}):
for child in node.findChildren(recursive=True, text=True):
child.replace_with(child.text.upper())
print(soup)
Prints:
<div style="text-transform: uppercase;">
FOO0
<font>FOO0</font>
<div>FOO1
<div>FOO2</div>
</div>
</div>

Pandas web scraping(Beautiful soup) find in tag with class, another tag with a link. Then following the link inside href

I tried fins 'td' tag with specific attribute, and then find 'a' tag inside of the 'td' tag
for row in bs4.find_all('<td class="series-column"'):
for link in bs4.find_all('a'):
if link.has_attr('href') and (link.has_attr('class') == 'formatted-title external-link result-url'):
print(link.attrs['href'])
On the screenshot you see html for this page
Your bs4.find_all('<td class="series-column"') is wrong. You have to supply tag name and attributes you want to find, for example bs4.find_all('td', class_='series-column'). Or use CSS selector:
from bs4 import BeautifulSoup
txt = '''
<td class="series-column">
<a class="formatted-title external-link result-url" href="//knoema.com/...">link text</a>
</td>'''
soup = BeautifulSoup(txt, 'html.parser')
for link in soup.select('td.series-column a.formatted-title.external-link.result-url'):
print(link['href'])
Prints:
//knoema.com/...

extract text from html string with Scrapy

Here is the html string in question.
<div class="def ddef_d db">a <a class="query" href="https://dictionary.cambridge.org/us/dictionary/english/book" title="book">book</a> of grammar <a class="query" href="https://dictionary.cambridge.org/us/dictionary/english/rule" title="rules">rules</a>: </div>
With BeautifulSoup, this code
from bs4 import BeautifulSoup
soup = BeautifulSoup(htmltxt, 'lxml')
soup.text
gets me
a book of grammar rules:
which is exactly what I want.
With scrapy, how do I get the same result?
from scrapy import Selector
sel = Selector(text=htmltxt)
sel.css('.ddef_d::text').getall()
this code gets me
['a ', ' of grammar ', ': ']
How should I fix it?
aYou can use this code to get all text inside div and its child:
text = ''.join(sel.css('.ddef_d ::text').getall())
print(text)
your selector returns text only from the div, but part of text located inside child elements (a), that's why you have to add space before ::text to include child text into result.

How to use BeautifulSoup to get content inside over-line tags

I would like to extract the content("_The_important_content_") from an HTML snippet as follows:
<div
class="
a:2
c:gray
m:da
"
>
_The_important_content_
</div>
My code is just:
for i in soup.findAll('div', class_="a:2 c:gray m:da"):
print(i.text)
But because the "class" field contains new line symbols and is expanded to multiple line so that BeautifulSoup cannot match, the code returns nothing. How can I specify the correct class field and get the content?
There are many tags with the same "class" value and other "class" value but I want to extract the contents from the tags with that specific "class" value.
Try this:
html='''
<div
class="
a:2
c:gray
m:da
"
>
_The_important_content_
</div>
'''
from bs4 import BeautifulSoup
soup = BeautifulSoup(html,"lxml")
item = soup.select("[class^=]")[0].text
print(item.strip())
Result:
_The_important_content_

Getting a specific part of a website with Beautiful Soup 4

I got the basics down of finding stuff with Beautiful Soup 4. However right now I am stuck with a specific problem.I want to scrape the "2DKT94P" from the data-oid of the below code:
<div class="js-object listitem_wrap " data-estateid="45784882" data-oid="2DKT94P">
<div class="listitem relative js-listitem ">
Any pointers on how I might do this? I would also appreciate a pointer for an advanced tutorial that covers this, and/or a link on where I would have been able to find this in the official documentation because I failed to recognize the correct part...
Thanks in advance!
you should locate the div tag using class attribute then get it's data-oid attribute
div = soup.find("div", class_="js-object")
oid = div['data-oid']
If your data is well formated you can do this via this way:
from bs4 import BeautifulSoup
example = """
<div class="js-object listitem_wrap " data-estateid="45784882" data-
oid="2DKT94P">
<div class="listitem relative js-listitem ">2DKT94P DIV</div>
</div>
<div>other div</div>"""
soup = BeautifulSoup(example, "html.parser")
RandomDIV = soup.find(attrs= {"data-oid":"2DKT94P"})
print (RandomDIV.get_text().strip())
Outputs:
2DKT94P DIV
Find more info about find or find_all with attributes here.
Or via select:
RandomDIV = soup.select("div[data-oid='2DKT94P']")
print (RandomDIV[0].get_text().strip())
Find more about select.
EDIT:
Totally misunderstood the question. If you want to search only for data-oid you can do like this:
soup = BeautifulSoup(example, "html.parser")
RandomDIV = soup.find_all(lambda tag: [t for t in tag.attrs if
t == 'data-oid'])
for div in RandomDIV:
#data-oid
print(div["data-oid"])
#text
print (div.text.strip())
Learn more here.