from bs4 import BeautifulSoup
import requests
url ="http://www.basketnews.lt/lygos/59-nacionaline-krepsinio-asociacija/2013/naujienos.html"
r = requests.get(url)
soup = BeautifulSoup(r.text)
naujienos = soup.findAll('a', {'class':'title'})
print naujienos
Here is important part of HTML:
<div class="title">
<span class="feedbacks"></span>
</div>
I get empty list. Where is my mistake?
EDIT:
Thanks it worked. Now I want to print news titles. This is how I am trying to do it:
nba = soup.select('div.title > a')
for i in nba:
print ""+i.string+"\n"
I get max 5 titles and error occurs: cannot concatenate 'str' and 'NoneType' objects
soup.findAll('a', {'class':'title'})
This says, give me all a tags that also have class="title". That's obviously not what you're trying to do.
I think you want a tags that are the direct descendant of a tag with class="title". You can try using a css selector:
soup.select('div.title > a')
Out[58]:
[Blatche'as: âGarantuoju, kad laimÄsimeâ,
<a href="/news-73147-rockets-veikiausiai-pasiliks-mchalea.html">âRocketsâ veikiausiai pasiliks McHaleâÄ
</a>,
# snip lots of other links
]
Related
How does one access a link from HTML divs?
Here is the HTML I am trying to scrape, I want to get the href value:
<div class="item-info-wrap">
<div class="news-feed_item-meta icon-font-before icon-espnplus-before"> <span class="timestamp">5d</span><span class="author">Field Yates</span> </div>
<h1> <a name="&lpos=nfl:feed:5:news" href="/nfl/insider/story/_/id/31949666/six-preseason-nfl-trades-teams-make-imagining-deals-nick-foles-xavien-howard-more" class=" realStory" data-sport="nfl" data-mptype="story">
Six NFL trades we'd love to see in August: Here's where Foles could help, but it's not the Colts</a></h1>
<p>Nick Foles is running the third team in Chicago. Xavien Howard wants out of Miami. Let's project six logical deals.</p></div>
Here is the code I have been trying to use to access the href value:
from bs4 import BeautifulSoup
import requests
source = requests.get('https://www.espn.com/nfl/team/_/name/phi/philadelphia-eagles').text
soup = BeautifulSoup(source, 'lxml')
for article in soup.find_all('div', class_='item-info-wrap'):
headline = article.h1.a.text
print(headline)
summary = article.p.text
print(summary)
try:
link_src = article.h1.a.href # Having difficulty getting href value
print(link_src)
link = f'https://espn.com/{link_src}'
except Exception as e:
link = None
print(link)
The output I am getting is https://espn.com/None for every ESPN article. Appreciate any help and feedback!
If you change the code in line 12 like the code below, it should work.
link_src = article.h1.a["href"]
FYI https://www.crummy.com/software/BeautifulSoup/bs4/doc/#attributes
I just started using beautifulsoup and am stuck on an issue regarding getting attributes of tags inside other tags. I am using the whitehouse.gov/briefing-room/ for practice. What I'm trying to do right now is just get all the links on this page and append them to an empty list. This is my code right now:
result = requests.get("https://www.whitehouse.gov/briefing-room/")
src = result.content
soup = BeautifulSoup(src, 'lxml')
urls = []
for h2_tags in soup.find_all('h2'):
a_tag = h2_tags.find('a')
urls.append(a_tag.attr['href']) # This is where I get the NoneType error
This code returns the <a tags, but the first and last 3 tags it returns are 'None' and because of this, get a type error when trying to access the attributes to get the href for these <a tags
The problem is, that some <h2> tags don't contain <a> tags. So you have to check for that alternative. Or just select all <a> tags that are under <h2> using CSS selector:
import requests
from bs4 import BeautifulSoup
result = requests.get("https://www.whitehouse.gov/briefing-room/")
src = result.content
soup = BeautifulSoup(src, 'lxml')
urls = []
for a_tag in soup.select('h2 a'): # <-- select <A> tags that are under <H2> tags
urls.append(a_tag.attrs['href'])
print(*urls, sep='\n')
Prints:
https://www.whitehouse.gov/briefing-room/statements-releases/2021/03/10/statement-by-nsc-spokesperson-emily-horne-on-national-security-advisor-jake-sullivan-leading-the-first-virtual-meeting-of-the-u-s-israel-strategic-consultative-group/
https://www.whitehouse.gov/briefing-room/press-briefings/2021/03/09/press-briefing-by-press-secretary-jen-psaki-and-deputy-director-of-the-national-economic-council-bharat-ramamurti-march-9-2021/
https://www.whitehouse.gov/briefing-room/statements-releases/2021/03/09/readout-of-the-white-houses-meeting-with-climate-finance-leaders/
https://www.whitehouse.gov/briefing-room/statements-releases/2021/03/09/readout-of-vice-president-kamala-harris-call-with-prime-minister-erna-solberg-of-norway/
https://www.whitehouse.gov/briefing-room/statements-releases/2021/03/09/nomination-sent-to-the-senate-3/
https://www.whitehouse.gov/briefing-room/statements-releases/2021/03/09/president-biden-announces-key-hire-for-the-office-of-management-and-budget/
https://www.whitehouse.gov/briefing-room/speeches-remarks/2021/03/09/remarks-by-president-biden-during-tour-of-w-s-jenks-son/
https://www.whitehouse.gov/briefing-room/statements-releases/2021/03/09/president-joseph-r-biden-jr-approves-louisiana-disaster-declaration/
https://www.whitehouse.gov/briefing-room/statements-releases/2021/03/09/statement-by-president-joe-biden-on-the-house-taking-up-the-pro-act/
https://www.whitehouse.gov/briefing-room/statements-releases/2021/03/09/white-house-announces-additional-staff/
I have the following:
html =
'''<div class=“file-one”>
<a href=“/file-one/additional” class=“file-link">
<h3 class=“file-name”>File One</h3>
</a>
<div class=“location”>
Down
</div>
</div>'''
And would like to get just the text of href which is /file-one/additional. So I did:
from bs4 import BeautifulSoup
soup = BeautifulSoup(html, 'html.parser')
link_text = “”
for a in soup.find_all(‘a’, href=True, text=True):
link_text = a[‘href’]
print “Link: “ + link_text
But it just prints a blank, nothing. Just Link:. So I tested it out on another site but with a different HTML, and it worked.
What could I be doing wrong? Or is there a possibility that the site intentionally programmed to not return the href?
Thank you in advance and will be sure to upvote/accept answer!
The 'a' tag in your html does not have any text directly, but it contains a 'h3' tag that has text. This means that text is None, and .find_all() fails to select the tag. Generally do not use the text parameter if a tag contains any other html elements except text content.
You can resolve this issue if you use only the tag's name (and the href keyword argument) to select elements. Then add a condition in the loop to check if they contain text.
soup = BeautifulSoup(html, 'html.parser')
links_with_text = []
for a in soup.find_all('a', href=True):
if a.text:
links_with_text.append(a['href'])
Or you could use a list comprehension, if you prefer one-liners.
links_with_text = [a['href'] for a in soup.find_all('a', href=True) if a.text]
Or you could pass a lambda to .find_all().
tags = soup.find_all(lambda tag: tag.name == 'a' and tag.get('href') and tag.text)
If you want to collect all links whether they have text or not, just select all 'a' tags that have a 'href' attribute. Anchor tags usually have links but that's not a requirement, so I think it's best to use the href argument.
Using .find_all().
links = [a['href'] for a in soup.find_all('a', href=True)]
Using .select() with CSS selectors.
links = [a['href'] for a in soup.select('a[href]')]
You can also use attrs to get the href tag with regex search
soup.find('a', href = re.compile(r'[/]([a-z]|[A-Z])\w+')).attrs['href']
First of all, use a different text editor that doesn't use curly quotes.
Second, remove the text=True flag from the soup.find_all
You could solve this with just a couple lines of gazpacho:
from gazpacho import Soup
html = """\
<div class="file-one">
<a href="/file-one/additional" class="file-link">
<h3 class="file-name">File One</h3>
</a>
<div class="location">
Down
</div>
</div>
"""
soup = Soup(html)
soup.find("a", {"class": "file-link"}).attrs['href']
Which would output:
'/file-one/additional'
A bit late to the party but I had the same issue recently scraping some recipes and got mine printing clean by doing this:
from bs4 import BeautifulSoup
import requests
source = requests.get('url for website')
soup = BeautifulSoup(source, 'lxml')
for article in soup.find_all('article'):
link = article.find('a', href=True)['href'}
print(link)
I am a bit confused in getting content of the paragraph tag.
<div class="SomeID">
<p>What a voice! </p>
</div>
I reached at this point
list = soup.find_all("div","SomeID")
But how to get the paragraph content.(What a voice!)
The basic problem is to get the content of all paragraph tags from
import urllib
from bs4 import BeautifulSoup
html = urllib.urlopen('http://www.dawn.com/news/1267272/democracys-woes').read()
soup = BeautifulSoup(html, 'html.parser')
list = soup.find_all("div","comment__body cf")
print list
You can actually do it one go with a CSS selector:
for p in soup.select("div.SomeID > p"):
print(p.get_text(strip=True))
Or, if you need a single p element:
soup.select_one("div.SomeID > p").get_text(strip=True)
Note that > here means the direct parent-child relationship.
I'm working on a python script to do some web scraping. I want to find the base URL of a given section on a web page that looks like this:
<div class='pagination'>
<a href='webpage-category/page/1'>1</a>
<a href='webpage-category/page/2'>2</a>
...
</div>
So, I just need to get everything from the first href besides the number('webpage-category/page/') and I have the following working code:
pages = [l['href'] for link in soup.find_all('div', class_='pagination')
for l in link.find_all('a') if not re.search('pageSub', l['href'])]
s = pages[0]
f = ''.join([i for i in s if not i.isdigit()])
The problem is, generating this list is a waste, since I just need the first href. I think a Generator would be the answer but I couldn't pull this off. Maybe you guys could help me to make this code more concise?
What about this:
from bs4 import BeautifulSoup
html = """ <div class='pagination'>
<a href='webpage-category/page/1'>1</a>
<a href='webpage-category/page/2'>2</a>
</div>"""
soup = BeautifulSoup(html)
link = soup.find('div', {'class': 'pagination'}).find('a')['href']
print '/'.join(link.split('/')[:-1])
prints:
webpage-category/page
Just FYI, speaking about the code you've provided - you can use [next()][-1] instead of a list comprehension:
s = next(l['href'] for link in soup.find_all('div', class_='pagination')
for l in link.find_all('a') if not re.search('pageSub', l['href']))
UPD (using the website link provided):
import urllib2
from bs4 import BeautifulSoup
url = "http://www.hdwallpapers.in/cars-desktop-wallpapers/page/2"
soup = BeautifulSoup(urllib2.urlopen(url))
links = soup.find_all('div', {'class': 'pagination'})[1].find_all('a')
print next('/'.join(link['href'].split('/')[:-1]) for link in links
if link.text.isdigit() and link.text != "1")