Hi I am still a beginner at python and I was experimenting.
I am looking for a way to request a url and get the data of the webpage so the page does not need to open.
Once I get the data, I need to search the data for a tag, for example, if it has 'hello' somewhere on the home page that is requested.
Here is an example:
import urllib.request
fp = urllib.request.urlopen("http://www.python.org")
mybytes = fp.read()
mystr = mybytes.decode("utf8")
fp.close()
x = mystr.find('testing word tag');
print(x)
Please bear with me as I am still a rookie and can't find an example of what I am looking for.
^ found this code on here but it does not seem to work to find a string.
Anyone knows the best way to do it?
Thank you guys :)
Here are the most used libraries for this kind of work:
Requests to get the HTML of the page.
BeautifulSoup to find elements (and much more)
$ pip install requests bs4
And in your favorite IDE:
import requests
from bs4 import BeautifulSoup
r = requests.get("http://www.python.org")
soup = BeautifulSoup(r.content, "html.parser")
sometag = soup.find("sometag")
print(sometag)
Try this.
import requests
url = "https://stackoverflow.com/questions/63577634/extract-html-and-search-in-python"
res = requests.get(url)
print(res.text)
Another method.
from simplified_scrapy import SimplifiedDoc,req
html = req.get('https://www.python.org')
doc = SimplifiedDoc(html)
title = doc.getElement('title').text
print (title)
title = doc.getElementByText('Welcome to', tag='title').text
print (title)
Result:
Welcome to Python.org
Welcome to Python.org
Here are more examples: https://github.com/yiyedata/simplified-scrapy-demo/tree/master/doc_examples
I'm new to Python, and for my second attempt at a project, I wanted to extract a substring – specifically, an identifying number – from a hyper-reference on a url.
For example, this url is the result of my search query, giving the hyper-reference http://www.chessgames.com/perl/chessgame?gid=1012809. From this I want to extract the identifying number "1012809" and append it to navigate to the url http://www.chessgames.com/perl/chessgame?gid=1012809, after which I plan to download the file at the url http://www.chessgames.com/pgn/alekhine_naegeli_1932.pgn?gid=1012809 . But I am currently stuck a few steps behind this because I can't figure out a way to extract the identifier.
Here is my MWE:
from bs4 import BeautifulSoup
url = 'http://www.chessgames.com/perl/chess.pl?yearcomp=exactly&year=1932&playercomp=white&pid=&player=Alekhine&pid2=&player2=Naegeli&movescomp=exactly&moves=&opening=&eco=&result=1%2F2-1%2F2'
page = urllib2.urlopen(url)
soup = BeautifulSoup(page, 'html.parser')
import re
y = str(soup)
x = re.findall("gid=[0-9]+",y)
print x
z = re.sub("gid=", "", x(1)) #At this point, things have completely broken down...
As Albin Paul commented, re.findall return a list, you need to extract elements from it. By the way, you don't need BeautifulSoup here, use urllib2.urlopen(url).read() to get the string of the content, and the re.sub is also not needed here, one regex pattern (?:gid=)([0-9]+) is enough.
import re
import urllib2
url = 'http://www.chessgames.com/perl/chess.pl?yearcomp=exactly&year=1932&playercomp=white&pid=&player=Alekhine&pid2=&player2=Naegeli&movescomp=exactly&moves=&opening=&eco=&result=1%2F2-1%2F2'
page = urllib2.urlopen(url).read()
result = re.findall(r"(?:gid=)([0-9]+)",page)
print(result[0])
#'1012809'
You don't need regex here at all. Css selector along with string manipulation will lead you to the right direction. Try the below script:
import requests
from bs4 import BeautifulSoup
page_link = 'http://www.chessgames.com/perl/chess.pl?yearcomp=exactly&year=1932&playercomp=white&pid=&player=Alekhine&pid2=&player2=Naegeli&movescomp=exactly&moves=&opening=&eco=&result=1%2F2-1%2F2'
soup = BeautifulSoup(requests.get(page_link).text, 'lxml')
item_num = soup.select_one("[href*='gid=']")['href'].split("gid=")[1]
print(item_num)
Output:
1012809
This is my first Python project so it is very basic and rudimentary.
I often have to clean off viruses for friends and the free programs that I use are updated often. Instead of manually downloading each program, I was trying to create a simple way to automate the process. Since I am also trying to learn python I thought it would be a good opportunity to practice.
Questions:
I have to find the .exe file with some of the links. I can find the correct URL, but I get an error when it tries to download.
Is there a way to add all of the links into a list, and then create a function to go through the list and run the function on each url? I've Google'd quite a bit and I just cannot seem to make it work. Maybe I am not thinking in the right direction?
import urllib, urllib2, re, os
from BeautifulSoup import BeautifulSoup
# Website List
sas = 'http://cdn.superantispyware.com/SUPERAntiSpyware.exe'
tds = 'http://support.kaspersky.com/downloads/utils/tdsskiller.exe'
mbam = 'http://www.bleepingcomputer.com/download/malwarebytes-anti-malware/dl/7/?1'
tr = 'http://www.simplysup.com/tremover/download.html'
urllist = [sas, tr, tds, tr]
urrllist2 = []
# Find exe files to download
match = re.compile('\.exe')
data = urllib2.urlopen(urllist)
page = BeautifulSoup(data)
# Check links
#def findexe():
for link in page.findAll('a'):
try:
href = link['href']
if re.search(match, href):
urllist2.append(href)
except KeyError:
pass
os.chdir(r"C:\_VirusFixes")
urllib.urlretrieve(urllist2, os.path.basename(urllist2))
As you can see, I have left the function commented out as I cannot get it to work correctly.
Should I abandon the list and just download them individually? I was trying to be efficient.
Any suggestions or if you could point me in the right direction, it would be most appreciated.
In addition to mikez302's answer, here's a slightly more readable way to write your code:
import os
import re
import urllib
import urllib2
from BeautifulSoup import BeautifulSoup
websites = [
'http://cdn.superantispyware.com/SUPERAntiSpyware.exe'
'http://support.kaspersky.com/downloads/utils/tdsskiller.exe'
'http://www.bleepingcomputer.com/download/malwarebytes-anti-malware/dl/7/?1'
'http://www.simplysup.com/tremover/download.html'
]
download_links = []
for url in websites:
connection = urllib2.urlopen(url)
soup = BeautifulSoup(connection)
connection.close()
for link in soup.findAll('a', {href: re.compile(r'\.exe$')}):
download_links.append(link['href'])
for url in download_links:
urllib.urlretrieve(url, r'C:\_VirusFixes', os.path.basename(url))
urllib2.urlopen is a function for accessing a single URL. If you want to access multiple ones, you should loop over the list. You should do something like this:
for url in urllist:
data = urllib2.urlopen(url)
page = BeautifulSoup(data)
# Check links
for link in page.findAll('a'):
try:
href = link['href']
if re.search(match, href):
urllist2.append(href)
except KeyError:
pass
os.chdir(r"C:\_VirusFixes")
urllib.urlretrieve(urllist2, os.path.basename(urllist2))
The code above didn't work for me, in my case it was because the pages assemble their links through a script instead of including it in the code. When I ran into that problem I used the following code which is just a scraper:
import os
import re
import urllib
import urllib2
from bs4 import BeautifulSoup
url = ''
connection = urllib2.urlopen(url)
soup = BeautifulSoup(connection) #Everything the same up to here
regex = '(.+?).zip' #Here we insert the pattern we are looking for
pattern = re.compile(regex)
link = re.findall(pattern,str(soup)) #This finds all the .zip (.exe) in the text
x=0
for i in link:
link[x]=i.split(' ')[len(i.split(' '))-1]
# When it finds all the .zip, it usually comes back with a lot of undesirable
# text, luckily the file name is almost always separated by a space from the
# rest of the text which is why we do the split
x+=1
os.chdir("F:\Documents")
# This is the filepath where I want to save everything I download
for i in link:
urllib.urlretrieve(url,filename=i+".zip") # Remember that the text we found doesn't include the .zip (or .exe in your case) so we want to reestablish that.
This is not as efficient as the codes in the previous answers but it will work for most almost any site.
This is my code to get a web page's image's URLs
for some webpage, it works very well, while it' dosen't work for some web page
this is my code:
#!/usr/bin/python
import urllib2
import re
#bufOne = urllib2.urlopen(r"http://vgirl.weibo.com/5show/user.php?fid=17262", timeout=4).read()
bufTwo = urllib2.urlopen(r"http://541626.com/pages/38307", timeout=4).read()
jpgRule = re.findall(r'http://[\w/]*?jpg', bufOne, re.IGNORECASE)
jpgRule = re.findall(r'http://[\w/]*?jpg', bufTwo, re.IGNORECASE)
print jpgRule
bufOne work well, but bufTwodidn't work. so how to write a ruler for it to make bufTwo work well?
Don't use regex to parse HTML. Rather use Beautiful Soup to find all img tags and then get the src attributes.
from BeautifullSoup import BeautifullSoup
#...
soup = BeautifulSoup(bufTwo)
imgTags = soup.findAll('img')
img = [tag['src'] for tag in imgTags]
I'll take this chance ddk gave to show you an easier way of getting all the images.
Using Beautiful Soup like that:
from BeautifulSoup import BeautifulSoup
all_imgs = soup.findAll("img", { "src" : re.compile(r'http://[\w/]*?jpg') })
That will already give you a list with all the images you want.
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How can I retrieve the links of a webpage and copy the url address of the links using Python?
Here's a short snippet using the SoupStrainer class in BeautifulSoup:
import httplib2
from bs4 import BeautifulSoup, SoupStrainer
http = httplib2.Http()
status, response = http.request('http://www.nytimes.com')
for link in BeautifulSoup(response, parse_only=SoupStrainer('a')):
if link.has_attr('href'):
print(link['href'])
The BeautifulSoup documentation is actually quite good, and covers a number of typical scenarios:
https://www.crummy.com/software/BeautifulSoup/bs4/doc/
Edit: Note that I used the SoupStrainer class because it's a bit more efficient (memory and speed wise), if you know what you're parsing in advance.
For completeness sake, the BeautifulSoup 4 version, making use of the encoding supplied by the server as well:
from bs4 import BeautifulSoup
import urllib.request
parser = 'html.parser' # or 'lxml' (preferred) or 'html5lib', if installed
resp = urllib.request.urlopen("http://www.gpsbasecamp.com/national-parks")
soup = BeautifulSoup(resp, parser, from_encoding=resp.info().get_param('charset'))
for link in soup.find_all('a', href=True):
print(link['href'])
or the Python 2 version:
from bs4 import BeautifulSoup
import urllib2
parser = 'html.parser' # or 'lxml' (preferred) or 'html5lib', if installed
resp = urllib2.urlopen("http://www.gpsbasecamp.com/national-parks")
soup = BeautifulSoup(resp, parser, from_encoding=resp.info().getparam('charset'))
for link in soup.find_all('a', href=True):
print link['href']
and a version using the requests library, which as written will work in both Python 2 and 3:
from bs4 import BeautifulSoup
from bs4.dammit import EncodingDetector
import requests
parser = 'html.parser' # or 'lxml' (preferred) or 'html5lib', if installed
resp = requests.get("http://www.gpsbasecamp.com/national-parks")
http_encoding = resp.encoding if 'charset' in resp.headers.get('content-type', '').lower() else None
html_encoding = EncodingDetector.find_declared_encoding(resp.content, is_html=True)
encoding = html_encoding or http_encoding
soup = BeautifulSoup(resp.content, parser, from_encoding=encoding)
for link in soup.find_all('a', href=True):
print(link['href'])
The soup.find_all('a', href=True) call finds all <a> elements that have an href attribute; elements without the attribute are skipped.
BeautifulSoup 3 stopped development in March 2012; new projects really should use BeautifulSoup 4, always.
Note that you should leave decoding the HTML from bytes to BeautifulSoup. You can inform BeautifulSoup of the characterset found in the HTTP response headers to assist in decoding, but this can be wrong and conflicting with a <meta> header info found in the HTML itself, which is why the above uses the BeautifulSoup internal class method EncodingDetector.find_declared_encoding() to make sure that such embedded encoding hints win over a misconfigured server.
With requests, the response.encoding attribute defaults to Latin-1 if the response has a text/* mimetype, even if no characterset was returned. This is consistent with the HTTP RFCs but painful when used with HTML parsing, so you should ignore that attribute when no charset is set in the Content-Type header.
Others have recommended BeautifulSoup, but it's much better to use lxml. Despite its name, it is also for parsing and scraping HTML. It's much, much faster than BeautifulSoup, and it even handles "broken" HTML better than BeautifulSoup (their claim to fame). It has a compatibility API for BeautifulSoup too if you don't want to learn the lxml API.
Ian Blicking agrees.
There's no reason to use BeautifulSoup anymore, unless you're on Google App Engine or something where anything not purely Python isn't allowed.
lxml.html also supports CSS3 selectors so this sort of thing is trivial.
An example with lxml and xpath would look like this:
import urllib
import lxml.html
connection = urllib.urlopen('http://www.nytimes.com')
dom = lxml.html.fromstring(connection.read())
for link in dom.xpath('//a/#href'): # select the url in href for all a tags(links)
print link
import urllib2
import BeautifulSoup
request = urllib2.Request("http://www.gpsbasecamp.com/national-parks")
response = urllib2.urlopen(request)
soup = BeautifulSoup.BeautifulSoup(response)
for a in soup.findAll('a'):
if 'national-park' in a['href']:
print 'found a url with national-park in the link'
The following code is to retrieve all the links available in a webpage using urllib2 and BeautifulSoup4:
import urllib2
from bs4 import BeautifulSoup
url = urllib2.urlopen("http://www.espncricinfo.com/").read()
soup = BeautifulSoup(url)
for line in soup.find_all('a'):
print(line.get('href'))
Links can be within a variety of attributes so you could pass a list of those attributes to select.
For example, with src and href attributes (here I am using the starts with ^ operator to specify that either of these attributes values starts with http):
from bs4 import BeautifulSoup as bs
import requests
r = requests.get('https://stackoverflow.com/')
soup = bs(r.content, 'lxml')
links = [item['href'] if item.get('href') is not None else item['src'] for item in soup.select('[href^="http"], [src^="http"]') ]
print(links)
Attribute = value selectors
[attr^=value]
Represents elements with an attribute name of attr whose value is prefixed (preceded) by value.
There are also the commonly used $ (ends with) and * (contains) operators. For a full syntax list see the link above.
Under the hood BeautifulSoup now uses lxml. Requests, lxml & list comprehensions makes a killer combo.
import requests
import lxml.html
dom = lxml.html.fromstring(requests.get('http://www.nytimes.com').content)
[x for x in dom.xpath('//a/#href') if '//' in x and 'nytimes.com' not in x]
In the list comp, the "if '//' and 'url.com' not in x" is a simple method to scrub the url list of the sites 'internal' navigation urls, etc.
just for getting the links, without B.soup and regex:
import urllib2
url="http://www.somewhere.com"
page=urllib2.urlopen(url)
data=page.read().split("</a>")
tag="<a href=\""
endtag="\">"
for item in data:
if "<a href" in item:
try:
ind = item.index(tag)
item=item[ind+len(tag):]
end=item.index(endtag)
except: pass
else:
print item[:end]
for more complex operations, of course BSoup is still preferred.
This script does what your looking for, But also resolves the relative links to absolute links.
import urllib
import lxml.html
import urlparse
def get_dom(url):
connection = urllib.urlopen(url)
return lxml.html.fromstring(connection.read())
def get_links(url):
return resolve_links((link for link in get_dom(url).xpath('//a/#href')))
def guess_root(links):
for link in links:
if link.startswith('http'):
parsed_link = urlparse.urlparse(link)
scheme = parsed_link.scheme + '://'
netloc = parsed_link.netloc
return scheme + netloc
def resolve_links(links):
root = guess_root(links)
for link in links:
if not link.startswith('http'):
link = urlparse.urljoin(root, link)
yield link
for link in get_links('http://www.google.com'):
print link
To find all the links, we will in this example use the urllib2 module together
with the re.module
*One of the most powerful function in the re module is "re.findall()".
While re.search() is used to find the first match for a pattern, re.findall() finds all
the matches and returns them as a list of strings, with each string representing one match*
import urllib2
import re
#connect to a URL
website = urllib2.urlopen(url)
#read html code
html = website.read()
#use re.findall to get all the links
links = re.findall('"((http|ftp)s?://.*?)"', html)
print links
Why not use regular expressions:
import urllib2
import re
url = "http://www.somewhere.com"
page = urllib2.urlopen(url)
page = page.read()
links = re.findall(r"<a.*?\s*href=\"(.*?)\".*?>(.*?)</a>", page)
for link in links:
print('href: %s, HTML text: %s' % (link[0], link[1]))
Here's an example using #ars accepted answer and the BeautifulSoup4, requests, and wget modules to handle the downloads.
import requests
import wget
import os
from bs4 import BeautifulSoup, SoupStrainer
url = 'https://archive.ics.uci.edu/ml/machine-learning-databases/eeg-mld/eeg_full/'
file_type = '.tar.gz'
response = requests.get(url)
for link in BeautifulSoup(response.content, 'html.parser', parse_only=SoupStrainer('a')):
if link.has_attr('href'):
if file_type in link['href']:
full_path = url + link['href']
wget.download(full_path)
I found the answer by #Blairg23 working , after the following correction (covering the scenario where it failed to work correctly):
for link in BeautifulSoup(response.content, 'html.parser', parse_only=SoupStrainer('a')):
if link.has_attr('href'):
if file_type in link['href']:
full_path =urlparse.urljoin(url , link['href']) #module urlparse need to be imported
wget.download(full_path)
For Python 3:
urllib.parse.urljoin has to be used in order to obtain the full URL instead.
BeatifulSoup's own parser can be slow. It might be more feasible to use lxml which is capable of parsing directly from a URL (with some limitations mentioned below).
import lxml.html
doc = lxml.html.parse(url)
links = doc.xpath('//a[#href]')
for link in links:
print link.attrib['href']
The code above will return the links as is, and in most cases they would be relative links or absolute from the site root. Since my use case was to only extract a certain type of links, below is a version that converts the links to full URLs and which optionally accepts a glob pattern like *.mp3. It won't handle single and double dots in the relative paths though, but so far I didn't have the need for it. If you need to parse URL fragments containing ../ or ./ then urlparse.urljoin might come in handy.
NOTE: Direct lxml url parsing doesn't handle loading from https and doesn't do redirects, so for this reason the version below is using urllib2 + lxml.
#!/usr/bin/env python
import sys
import urllib2
import urlparse
import lxml.html
import fnmatch
try:
import urltools as urltools
except ImportError:
sys.stderr.write('To normalize URLs run: `pip install urltools --user`')
urltools = None
def get_host(url):
p = urlparse.urlparse(url)
return "{}://{}".format(p.scheme, p.netloc)
if __name__ == '__main__':
url = sys.argv[1]
host = get_host(url)
glob_patt = len(sys.argv) > 2 and sys.argv[2] or '*'
doc = lxml.html.parse(urllib2.urlopen(url))
links = doc.xpath('//a[#href]')
for link in links:
href = link.attrib['href']
if fnmatch.fnmatch(href, glob_patt):
if not href.startswith(('http://', 'https://' 'ftp://')):
if href.startswith('/'):
href = host + href
else:
parent_url = url.rsplit('/', 1)[0]
href = urlparse.urljoin(parent_url, href)
if urltools:
href = urltools.normalize(href)
print href
The usage is as follows:
getlinks.py http://stackoverflow.com/a/37758066/191246
getlinks.py http://stackoverflow.com/a/37758066/191246 "*users*"
getlinks.py http://fakedomain.mu/somepage.html "*.mp3"
There can be many duplicate links together with both external and internal links. To differentiate between the two and just get unique links using sets:
# Python 3.
import urllib
from bs4 import BeautifulSoup
url = "http://www.espncricinfo.com/"
resp = urllib.request.urlopen(url)
# Get server encoding per recommendation of Martijn Pieters.
soup = BeautifulSoup(resp, from_encoding=resp.info().get_param('charset'))
external_links = set()
internal_links = set()
for line in soup.find_all('a'):
link = line.get('href')
if not link:
continue
if link.startswith('http'):
external_links.add(link)
else:
internal_links.add(link)
# Depending on usage, full internal links may be preferred.
full_internal_links = {
urllib.parse.urljoin(url, internal_link)
for internal_link in internal_links
}
# Print all unique external and full internal links.
for link in external_links.union(full_internal_links):
print(link)
import urllib2
from bs4 import BeautifulSoup
a=urllib2.urlopen('http://dir.yahoo.com')
code=a.read()
soup=BeautifulSoup(code)
links=soup.findAll("a")
#To get href part alone
print links[0].attrs['href']