I am trying to webscrape this website. To do so,
I run the following code:
from bs4 import BeautifulSoup
import requests
url = "https://www.ecb.europa.eu/press/pressconf/2022/html/index_include.en.html"
soup = BeautifulSoup(requests.get(url).content)
data = []
for link in soup.select('div.title > a'):
soup = BeautifulSoup(requests.get(f"https://www.ecb.europa.eu{link['href']}").content)
data.append({
'text':' '.join([p.text for p in soup.select('main .section p:not([class])')])
})
print(data)
This works fine. What is the issue? The problem is that the webscraped text comes without information on paragraphs split and on bold character. This is a problem since I would then need to make some calls on the basis of that.
Can anyone suggest how to maintain meta-information in the text?
Thanks a lot!
A solution is to determine in the website code source what are the markers for paragraphs split and bold characters.
Then, the "soup" variable, you can localize what interests you using the markers as a string to be searched in "soup".
Looking briefly at the source code of your website, I think the answer relies in following markers (I needed to add ' otherwise the markers are hidden by stackoverflow):
"<'/a><'/div><'div class="subtitle">"
Related
I can't figure out why BS4 is not seeing the text inside of the span in the following scenario:
Page: https://pypi.org/project/requests/
Text I'm looking for - number of stars on the left hand side (around 43,000 at the time of writing)
My code:
stars = soup.find('span', {'class': 'github-repo-info__item', 'data-key': 'stargazers_count'}).text
also tried:
stars = soup.find('span', {'class': 'github-repo-info__item', 'data-key': 'stargazers_count'}).get_text()
Both return an empty string ''. The element itself seems to be located correctly (I can browse through parents / siblings in PyCharm debugger without a problem. Fetching text in other parts of the website also works perfectly fine. It's just the github-related stats that fail to fetch.
Any ideas?
Because this page use Javascript to load the page dynamically.So you couldn't get it directly by response.text
The source code of the page:
You could crawl the API directly:
import requests
r = requests.get('https://api.github.com/repos/psf/requests')
print(r.json()["stargazers_count"])
Result:
43010
Using bs4, we can't scrape stars rate.
After inspecting the site, please check response html.
There, there is class information named "github-repo-info__item", but there is no text information.
in this case, use selenium.
I am downloading the verb conjugations to aid my learning. However one thing I can't seem to get from this web page is the english translation near the top of the page.
The code I have is below. When I print results_eng it prints the section I want but there is no english translation, what am I missing?
import requests
from bs4 import BeautifulSoup
URL = 'https://conjugator.reverso.net/conjugation-portuguese-verb-ser.html'
page = requests.get(URL)
soup = BeautifulSoup(page.content, 'html.parser')
results_eng = soup.find(id='list-translations')
eng = results_eng.find_all('p', class_='context_term')
In a normal website, you should be able to find the text in a paragraph witht the function get_text(), but in this case this is a search, wich means it's probably pulling the data from a database and its not in the paragraph itself. At least that's what I can come up with, since I tried to use that function and I got an empty string in return. Can't you try another website and see what happens?
p.d: I'm a beginner, sorry if I'm guessing wrong
Hello there stack community!
I'm having an issue that I can't seem to resolve since it looks like most of the help out there is for Python 2.7.
I want to pull a table from a webpage and then just get the linktext and not the whole anchor.
Here is the code:
from urllib.request import urlopen
from bs4 import BeautifulSoup
import re
url = 'http://www.craftcount.com/category.php?cat=5'
html = urlopen(url).read()
soup = BeautifulSoup(html)
alltables = soup.findAll("table")
## This bit captures the input from the previous sequence
results=[]
for link in alltables:
rows = link.findAll('a')
## Find just the names
top100 = re.findall(r">(.*?)<\/a>",rows)
print(top100)
When I run it, I get: "TypeError: expected string or buffer"
Up to the second to the last line, it does everything correctly (when I swap out 'print(top100)' for 'print(rows)').
As an example of the response I get:
michellechangjewelry
And I just need to get:
michellechangjewelry
According to pythex.org, my (ir)regular expression should work, so I wanted to see if anyone out there knew how to do that. As an additional issue, it looks like most people like to go the other way, that is, from having the full text and only wanting the URL part.
Finally, I'm using BeautifulSoup out of "convenience", but I'm not beholden to it if you can suggest a better package to narrow down the parsing to the linktext.
Many thanks in advance!!
BeautifulSoup results are not strings; they are Tag objects, mostly.
Look for the text of the <a> tags, use the .string attribute:
for table in alltables:
link = table.find('a')
top100 = link.string
print(top100)
This finds the first <a> link in a table. To find all text of all links:
for table in alltables:
links = table.find_all('a')
top100 = [link.string for link in links]
print(top100)
I'm trying to use BeautifulSoup to parse some HTML in Python. Specifically, I'm trying to create two arrays of soup objects: one for the dates of postings on a website, and one for the postings themselves. However, when I use findAll on the div class that matches the postings, only the initial tag is returned, not the text inside the tag. On the other hand, my code works just fine for the dates. What is going on??
# store all texts of posts
texts = soup.findAll("div", {"class":"quote"})
# store all dates of posts
dates = soup.findAll("div", {"class":"datetab"})
The first line above returns only
<div class="quote">
which is not what I want. The second line returns
<div class="datetab">Feb<span>2</span></div>
which IS what I want (pre-refining).
I have no idea what I'm doing wrong. Here is the website I'm trying to parse. This is for homework, and I'm really really desperate.
Which version of BeautifulSoup are you using? Version 3.1.0 performs significantly worse with real-world HTML (read: invalid HTML) than 3.0.8. This code works with 3.0.8:
import urllib2
from BeautifulSoup import BeautifulSoup
page = urllib2.urlopen("http://harvardfml.com/")
soup = BeautifulSoup(page)
for incident in soup.findAll('span', { "class" : "quote" }):
print incident.contents
That site is powered by Tumblr. Tumblr has an API.
There's a python port of Tumblr that you can use to read documents.
from tumblr import Api
api = Api('harvardfml.com')
freq = {}
posts = api.read()
for post in posts:
#do something here
for your bogus findAll, without the actual source code of your program it is hard to see what is wrong.
Given an HTML link like
texttxt
how can I isolate the url and the text?
Updates
I'm using Beautiful Soup, and am unable to figure out how to do that.
I did
soup = BeautifulSoup.BeautifulSoup(urllib.urlopen(url))
links = soup.findAll('a')
for link in links:
print "link content:", link.content," and attr:",link.attrs
i get
*link content: None and attr: [(u'href', u'_redirectGeneric.asp?genericURL=/root /support.asp')]* ...
...
Why am i missing the content?
edit: elaborated on 'stuck' as advised :)
Use Beautiful Soup. Doing it yourself is harder than it looks, you'll be better off using a tried and tested module.
EDIT:
I think you want:
soup = BeautifulSoup.BeautifulSoup(urllib.urlopen(url).read())
By the way, it's a bad idea to try opening the URL there, as if it goes wrong it could get ugly.
EDIT 2:
This should show you all the links in a page:
import urlparse, urllib
from BeautifulSoup import BeautifulSoup
url = "http://www.example.com/index.html"
source = urllib.urlopen(url).read()
soup = BeautifulSoup(source)
for item in soup.fetchall('a'):
try:
link = urlparse.urlparse(item['href'].lower())
except:
# Not a valid link
pass
else:
print link
Here's a code example, showing getting the attributes and contents of the links:
soup = BeautifulSoup.BeautifulSoup(urllib.urlopen(url))
for link in soup.findAll('a'):
print link.attrs, link.contents
Looks like you have two issues there:
link.contents, not link.content
attrs is a dictionary, not a string. It holds key value pairs for each attribute in an HTML element. link.attrs['href'] will get you what you appear to be looking for, but you'd want to wrap that in a check in case you come across an a tag without an href attribute.
Though I suppose the others might be correct in pointing you to using Beautiful Soup, they might not, and using an external library might be massively over-the-top for your purposes. Here's a regex which will do what you ask.
/<a\s+[^>]*?href="([^"]*)".*?>(.*?)<\/a>/
Here's what it matches:
'text'
// Parts: "url", "text"
'text<span>something</span>'
// Parts: "url", "text<span>something</span>"
If you wanted to get just the text (eg: "textsomething" in the second example above), I'd just run another regex over it to strip anything between pointed brackets.