I've read quite a few posts here about this, but I'm very new to Python in general so I was hoping for some more info.
Essentially, I'm trying to write something that will pull word definitions from a site and write them to a file. I've been using BeautifulSoup, and I've made quite some progress, but here's my issue -
from __future__ import print_function
import requests
import urllib2, urllib
from BeautifulSoup import BeautifulSoup
wordlist = open('test.txt', 'a')
word = raw_input('Paste your word ')
url = 'http://services.aonaware.com/DictService/Default.aspx?action=define&dict=wn&query=%s' % word
# print url
html = urllib.urlopen(url).read()
# print html
soup = BeautifulSoup(html)
visible_text = soup.find('pre')(text=True)
print(visible_text, file=wordlist)
this seems to pull what I need, but puts it in this format
[u'passable\n adj 1: able to be passed or traversed or crossed; "the road is\n passable"
but I need it to be in plaintext. I've tried using a sanitizer (I was running it through bleach, but that didn't work. I've read some of the other answers here, but they don't explain HOW the code works, and I don't want to add something if I don't understand how it works.
Is there any way to just pull the plaintext?
edit: I ended up doing
from __future__ import print_function
import requests
import urllib2, urllib
from bs4 import BeautifulSoup
wordlist = open('test.txt', 'a')
word = raw_input('Paste your word ')
url = 'http://services.aonaware.com/DictService/Default.aspx?action=define&dict=wn&query=%s' % word
# print url
html = urllib.urlopen(url).read()
# print html
soup = BeautifulSoup(html)
visible_text = soup.find('pre')(text=True)[0]
print(visible_text, file=wordlist)
The code is already giving you plaintext, it just happens to have some characters encoded as entity references. In this case, special characters, which form part of the XML/HTML syntax are encoded to prevent them from breaking the structure of the text.
To decode them, use the HTMLParser module:
import HTMLParser
h = HTMLParser.HTMLParser()
h.unescape('"the road is passable"')
>>> u'"the road is passable"'
Related
I am trying to scrape a website, which so far I am able to scrape but I want to output the file to a text file then from there I want to delete some strings in it.
from urllib.request import urlopen
from bs4 import BeautifulSoup
delete = ['https://', 'http://', 'b\'http://', 'b\'https://']
url = urlopen('https://openphish.com/feed.txt')
bs = BeautifulSoup(url.read(), 'html.parser' )
print(bs.encode('utf_8'))
The result is a lot of links, I can show a sample.
"b'https://certain-wrench.000webhostapp.com/auth/signin/details.html\nhttps://sweer-adherence.000webhostapp.com/auth/signin/details.html\n"
UPDATED
import requests
from bs4 import BeautifulSoup
url = "https://openphish.com/feed.txt"
url_get = requests.get(url)
soup = BeautifulSoup(url_get.content, 'lxml')
with open('url.txt', 'w', encoding='utf-8') as f_out:
f_out.write(soup.prettify())
delete = ["</p>", "</body>", "</html>", "<body>", "<p>", "<html>", "www.",
"https://", "http://", " ", " ", " "]
with open(r'C:\Users\v-morisv\Desktop\scripts\url.txt', 'r') as file:
with open(r'C:\Users\v-morisv\Desktop\scripts\url1.txt', 'w') as
file1:
for line in file:
for word in delete:
line = line.replace(word, "")
print(line, end='')
file1.write(line)
This code above works, but I have a problem because I am not getting only the domain I am getting everything after the forwarddash so it looks like this
bofawebplus.webcindario.com/index4.html and I want to remove "/" and everything after it.
This seems like a proper situation using Regular Expression.
from urllib.request import urlopen
from bs4 import BeautifulSoup
url = urlopen('https://openphish.com/feed.txt')
bs = BeautifulSoup(url.read(), 'html.parser' )
import re
domain_list = re.findall(re.compile('http[s]?://([^/]*)/'), bs.text)
print('\n'.join(domain_list))
There's no reason to use BeautifulSoup here, it is used for parsing HTML, but the URL being opened is plain text.
Here's a solution that should do what you need. It uses the Python urlparse as an easier and more reliable way of extracting the domain name.
This also uses a python set to remove duplicate entries, since there were quite a few.
from urllib.request import urlopen
from urllib.parse import urlparse
feed_list = urlopen('https://openphish.com/feed.txt')
domains = set()
for line in feed_list:
url = urlparse(line)
domain = url.netloc.decode('utf-8') # decode from utf-8 to string
domains.add(domain) # Keep all the domains in the set to remove duplicates
for domain in domains:
print(domains)
I'm using lxml.html module
from lxml import html
page = html.parse('http://directory.ccnecommunity.org/reports/rptAccreditedPrograms_New.asp?sort=institution')
# print(page.content)
unis = page.xpath('//tr/td[#valign="top" and #style="width: 50%;padding-right:15px"]/h3/text()')
print(unis.__len__())
with open('workfile.txt', 'w') as f:
for uni in unis:
f.write(uni + '\n')
The website right here (http://directory.ccnecommunity.org/reports/rptAccreditedPrograms_New.asp?sort=institution#Z) is full of universities.
The problem is that it parses till the letter 'H' (244 unis).
I can't understand why, as I see it parses all the HTML till the end.
I also documented my self that 244 is not a limit of a list or anything in python3.
That HTML page simply isn't HTML, it's totally broken. But the following will do what you want. It uses the BeautifulSoup parser.
from lxml.html.soupparser import parse
import urllib
url = 'http://directory.ccnecommunity.org/reports/rptAccreditedPrograms_New.asp?sort=institution'
page = parse(urllib.request.urlopen(url))
unis = page.xpath('//tr/td[#valign="top" and #style="width: 50%;padding-right:15px"]/h3/text()')
See http://lxml.de/lxmlhtml.html#really-broken-pages for more info.
For web-scraping i recommend you to use BeautifulSoup 4
With bs4 this is easily done:
from bs4 import BeautifulSoup
import urllib.request
universities = []
result = urllib.request.urlopen('http://directory.ccnecommunity.org/reports/rptAccreditedPrograms_New.asp?sort=institution#Z')
soup = BeautifulSoup(result.read(),'html.parser')
table = soup.find_all(lambda tag: tag.name=='table')
for t in table:
rows = t.find_all(lambda tag: tag.name=='tr')
for r in rows:
# there are also the A-Z headers -> check length
# there are also empty headers -> check isspace()
headers = r.find_all(lambda tag: tag.name=='h3' and tag.text.isspace()==False and len(tag.text.strip()) > 2)
for h in headers:
universities.append(h.text)
I am trying to make a python script that reads crunchyroll's page and gives me the ssid of the subtitle.
For example :- http://www.crunchyroll.com/i-cant-understand-what-my-husband-is-saying/episode-1-wriggling-memories-678035
Go to the source code and look for ssid,I want to extract the numbers after ssid of this element
English (US)
I want to extract "154757", but I can't seem to get my script working
This is my current script:
import feedparser
import re
import urllib2
from urllib2 import urlopen
from bs4 import BeautifulSoup
feed = feedparser.parse('http://www.crunchyroll.com/rss/anime')
url1 = feed['entries'][0]['link']
soup = BeautifulSoup(urlopen(url1), 'html.parser')
How can I modify my code to search and extract that particular number?
This should get you started with being able to extract the ssid for each entry. Note that some of those link don't have any ssid so you'll have to account for that with some error catching. No need for re or the urllib2 modules here.
import feedparser
import requests
from bs4 import BeautifulSoup
d = feedparser.parse('http://www.crunchyroll.com/rss/anime')
for url in d.entries:
#print url.link
r = requests.get(url.link)
soup = BeautifulSoup(r.text)
#print soup
subtitles = soup.find_all('span',{'class':'showmedia-subtitle-text'})
for ssid in subtitles:
x = ssid.findAll('a')
for a in x:
print a['href']
Output:
--snip--
/i-cant-understand-what-my-husband-is-saying/episode-12-baby-skip-beat-678057?ssid=166035
/i-cant-understand-what-my-husband-is-saying/episode-12-baby-skip-beat-678057?ssid=165817
/i-cant-understand-what-my-husband-is-saying/episode-12-baby-skip-beat-678057?ssid=165819
/i-cant-understand-what-my-husband-is-saying/episode-12-baby-skip-beat-678057?ssid=166783
/i-cant-understand-what-my-husband-is-saying/episode-12-baby-skip-beat-678057?ssid=165839
/i-cant-understand-what-my-husband-is-saying/episode-12-baby-skip-beat-678057?ssid=165989
/i-cant-understand-what-my-husband-is-saying/episode-12-baby-skip-beat-678057?ssid=166051
/urawa-no-usagi-chan/episode-11-if-i-retort-i-lose-678873?ssid=166011
/urawa-no-usagi-chan/episode-11-if-i-retort-i-lose-678873?ssid=165995
/urawa-no-usagi-chan/episode-11-if-i-retort-i-lose-678873?ssid=165997
/urawa-no-usagi-chan/episode-11-if-i-retort-i-lose-678873?ssid=166033
/urawa-no-usagi-chan/episode-11-if-i-retort-i-lose-678873?ssid=165825
/urawa-no-usagi-chan/episode-11-if-i-retort-i-lose-678873?ssid=166013
/urawa-no-usagi-chan/episode-11-if-i-retort-i-lose-678873?ssid=166009
/urawa-no-usagi-chan/episode-11-if-i-retort-i-lose-678873?ssid=166003
/etotama/episode-11-catrat-shuffle-678659?ssid=166007
/etotama/episode-11-catrat-shuffle-678659?ssid=165969
/etotama/episode-11-catrat-shuffle-678659?ssid=166489
/etotama/episode-11-catrat-shuffle-678659?ssid=166023
/etotama/episode-11-catrat-shuffle-678659?ssid=166015
/etotama/episode-11-catrat-shuffle-678659?ssid=166049
/etotama/episode-11-catrat-shuffle-678659?ssid=165993
/etotama/episode-11-catrat-shuffle-678659?ssid=165981
--snip--
There are more but I left them out for brevity. From these results you should be able to easily parse out the ssid with some slicing since it looks like the ssid are all 6 digits long. Doing something like:
print a['href'][-6:]
would do the trick and get you just the ssid.
I'm trying to read data from a website that contains only text. I'd like to read only the data that follows "&values". I've been able to open the entire website, but I don't know how to get rid of the extraneous data and I don't know any HTML. Any help would be much appreciated.
The contents of that url look like url parameters. You could use urllib.parse_qs to parse them into a dict:
import urllib2
import urlparse
url = 'http://www.tip.it/runescape/gec/price_graph.php?avg=1&start=1327715574&mainitem=10350&item=10350'
response = urllib2.urlopen(url)
content = response.read()
params = urlparse.parse_qs(content)
print(params['values'])
You may want to look into the re module (although if you do eventually move to HTML, regex is not the best solution). Here is a basic example that grabs the text after &values and returns the following number/comma/space combinations:
>>> import re
>>> import urllib2
>>> url = 'http://www.tip.it/runescape/gec/price_graph.php?avg=1&start=1327715574&mainitem=10350&item=10350'
>>> contents = urllib2.urlopen(url).read()
>>> values = re.findall(r'&values=([\d,\s]*)', contents)
>>> values[0].split(',')
['33900000', '33900000', '33900000', #continues....]
I'm doing a project for my school in which I would like to compare scam mails. I found this website: http://www.419scam.org/emails/
Now what I would like to do is to save every scam in apart documents then later on I can analyse them.
Here is my code so far:
import BeautifulSoup, urllib2
address='http://www.419scam.org/emails/'
html = urllib2.urlopen(address).read()
f = open('test.txt', 'wb')
f.write(html)
f.close()
This saves me the whole html file in a text format, now I would like to strip the file and save the content of the html links to the scams:
01
02
03
etc.
If i get that, I would still need to go a step further and open save another href. Any idea how do I do it in one python code?
Thank you!
You picked the right tool in BeautifulSoup. Technically you could do it all do it in one script, but you might want to segment it, because it looks like you'll be dealing with tens of thousands of e-mails, all of which are seperate requests - and that will take a while.
This page is gonna help you a lot, but here's just a little code snippet to get you started. This gets all of the html tags that are index pages for the e-mails, extracts their href links and appends a bit to the front of the url so they can be accessed directly.
from bs4 import BeautifulSoup
import re
import urllib2
soup = BeautifulSoup(urllib2.urlopen("http://www.419scam.org/emails/"))
tags = soup.find_all(href=re.compile("20......../index\.htm")
links = []
for t in tags:
links.append("http://www.419scam.org/emails/" + t['href'])
're' is a Python's regular expressions module. In the fifth line, I told BeautifulSoup to find all the tags in the soup whose href attribute match that regular expression. I chose this regular expression to get only the e-mail index pages rather than all of the href links on that page. I noticed that the index page links had that pattern for all of their URLs.
Having all the proper 'a' tags, I then looped through them, extracting the string from the href attribute by doing t['href'] and appending the rest of the URL to the front of the string, to get raw string URLs.
Reading through that documentation, you should get an idea of how to expand these techniques to grab the individual e-mails.
You might also find value in requests and lxml.html. Requests is another way to make http requests and lxml is an alternative for parsing xml and html content.
There are many ways to search the html document but you might want to start with cssselect.
import requests
from lxml.html import fromstring
url = 'http://www.419scam.org/emails/'
doc = fromstring(requests.get(url).content)
atags = doc.cssselect('a')
# using .get('href', '') syntax because not all a tags will have an href
hrefs = (a.attrib.get('href', '') for a in atags)
Or as suggested in the comments using .iterlinks(). Note that you will still need to filter if you only want 'a' tags. Either way the .make_links_absolute() call is probably going to be helpful. It is your homework though, so play around with it.
doc.make_links_absolute(base_url=url)
hrefs = (l[2] for l in doc.iterlinks() if l[0].tag == 'a')
Next up for you... how to loop through and open all of the individual spam links.
To get all links on the page you could use BeautifulSoup. Take a look at this page, it can help. It actually tells how to do exactly what you need.
To save all pages, you could do the same as what you do in your current code, but within a loop that would iterate over all links you'll have extracted and stored, say, in a list.
Heres a solution using lxml + XPath and urllib2 :
#!/usr/bin/env python2 -u
# -*- coding: utf8 -*-
import cookielib, urllib2
from lxml import etree
cj = cookielib.CookieJar()
opener = urllib2.build_opener(urllib2.HTTPCookieProcessor(cj))
page = opener.open("http://www.419scam.org/emails/")
page.addheaders = [('User-agent', 'Mozilla/5.0')]
reddit = etree.HTML(page.read())
# XPath expression : we get all links under body/p[2] containing *.htm
for node in reddit.xpath('/html/body/p[2]/a[contains(#href,".htm")]'):
for i in node.items():
url = 'http://www.419scam.org/emails/' + i[1]
page = opener.open(url)
page.addheaders = [('User-agent', 'Mozilla/5.0')]
lst = url.split('/')
try:
if lst[6]: # else it's a "month" link
filename = '/tmp/' + url.split('/')[4] + '-' + url.split('/')[5]
f = open(filename, 'w')
f.write(page.read())
f.close()
except:
pass
# vim:ts=4:sw=4
You could use HTML parser and specify the type of object you are searching for.
from HTMLParser import HTMLParser
import urllib2
class MyHTMLParser(HTMLParser):
def handle_starttag(self, tag, attrs):
if tag == 'a':
for attr in attrs:
if attr[0] == 'href':
print attr[1]
address='http://www.419scam.org/emails/'
html = urllib2.urlopen(address).read()
f = open('test.txt', 'wb')
f.write(html)
f.close()
parser = MyHTMLParser()
parser.feed(html)