Develop Reference

Python - print 2nd argument

I am new to Python and I've written this test-code for practicing purposes, in order to find and print email addresses from various web pages:
def FindEmails(*urls):
for i in urls:
totalemails = []
req = urllib2.Request(i)
aResp = urllib2.urlopen(req)
webpage = aResp.read()
patt1 = '(\w+[-\w]\w+#\w+[.]\w+[.\w+]\w+)'
patt2 = '(\w+[\w]\w+#\w+[.]\w+)'
regexlist = [patt1,patt2]
for regex in regexlist:
match = re.search(regex,webpage)
if match:
totalemails.append(match.group())
break
#return totalemails
print "Mails from webpages are: %s " % totalemails
if __name__== "__main__":
FindEmails('https://www.urltest1.com', 'https://www.urltest2.com')
When I run it, it prints only one argument.
My goal is to print the emails acquired from webpages and store them in a list, separated by commas.
Thanks in advance.

The problem here is the line: totalemails = []. Here, you are re-instantiating the the variables totalemails to have zero entries. So, in each iteration, it only has one entry inside it. After the last iteration, you'll end up with just the last entry in the list. To get a list of all emails, you need to put the variable outside of the for loop.
Example:
def FindEmails(*urls):
totalemails = []
for i in urls:
req = urllib2.Request(i)
....

Issue with web crawler: IndexError: string index out of range

I am making a web crawler. I'm not using scrapy or anything, I'm trying to have my script do most things. I have tried doing a search for the issue however I can't seem to find anything that helps with the error. I've tried switching around some of the variable to try and narrow down the problem. I am getting an error on line 24 saying IndexError: string index out of range. The functions run on the first url, (the original url) then the second and fail on the third in the original array. I'm lost, any help would be appreciated greatly! Note, I'm only printing all of them for testing, I'll eventually have them printed to a text file.
import requests
from bs4 import BeautifulSoup
# creating requests from user input
url = raw_input("Please enter a domain to crawl, without the 'http://www' part : ")
def makeRequest(url):
r = requests.get('http://' + url)
# Adding in BS4 for finding a tags in HTML
soup = BeautifulSoup(r.content, 'html.parser')
# Writes a as the link found in the href
output = soup.find_all('a')
return output
def makeFilter(link):
# Creating array for our links
found_link = []
for a in link:
a = a.get('href')
a_string = str(a)
# if statement to filter our links
if a_string[0] == '/': # this is the line with the error
# Realtive Links
found_link.append(a_string)
if 'http://' + url in a_string:
# Links from the same site
found_link.append(a_string)
if 'https://' + url in a_string:
# Links from the same site with SSL
found_link.append(a_string)
if 'http://www.' + url in a_string:
# Links from the same site
found_link.append(a_string)
if 'https://www.' + url in a_string:
# Links from the same site with SSL
found_link.append(a_string)
#else:
# found_link.write(a_string + '\n') # testing only
output = found_link
return output
# Function for removing duplicates
def remove_duplicates(values):
output = []
seen = set()
for value in values:
if value not in seen:
output.append(value)
seen.add(value)
return output
# Run the function with our list in this order -> Makes the request -> Filters the links -> Removes duplicates
def createURLList(values):
requests = makeRequest(values)
new_list = makeFilter(requests)
filtered_list = remove_duplicates(new_list)
return filtered_list
result = createURLList(url)
# print result
# for verifying and crawling resulting pages
for b in result:
sub_directories = createURLList(url + b)
crawler = []
crawler.append(sub_directories)
print crawler

After a_string = str(a) try adding:
if not a_string:
continue

Web crawler does not open all links in a page

I'am trying to build a web crawler using beautifulsoup and urllib. The crawler is working, but it does not open all the pages in a site. It opens the first link and goes to that link, opens the first link of that page and so on.
Here's my code:
from bs4 import BeautifulSoup
from urllib.request import urlopen
from urllib.parse import urljoin
import json, sys
sys.setrecursionlimit(10000)
url = input('enter url ')
d = {}
d_2 = {}
l = []
url_base = url
count = 0
def f(url):
global count
global url_base
if count <= 100:
print("count: " + str(count))
print('now looking into: '+url+'\n')
count += 1
l.append(url)
html = urlopen(url).read()
soup = BeautifulSoup(html, "html.parser")
d[count] = soup
tags = soup('a')
for tag in tags:
meow = tag.get('href',None)
if (urljoin(url, meow) in l):
print("Skipping this one: " + urljoin(url,meow))
elif "mailto" in urljoin(url,meow):
print("Skipping this one with a mailer")
elif meow == None:
print("skipping 'None'")
elif meow.startswith('http') == False:
f(urljoin(url, meow))
else:
f(meow)
else:
return
f(url)
print('\n\n\n\n\n')
print('Scrapping Completed')
print('\n\n\n\n\n')

The reason you're seeing this behavior is due to when the code recursively calls your function. As soon as the code finds a valid link, the function f gets called again preventing the rest of the for loop from running until it returns.
What you're doing is a depth first search, but the internet is very deep. You want to do a breadth first search instead.
Probably the easiest way to modify your code to do that is to have a global list of links to follow. Have the for loop append all the scraped links to the end of this list and then outside of the for loop, remove the first element of the list and follow that link.
You may have to change your logic slightly for your max count.

If count reaches 100, no further links will be opened. Therefore I think you should decrease count by one after leaving the for loop. If you do this, count would be something like the current link depth (and 100 would be the maximum link depth).
If the variable count should refer to the number of opened links, then you might want to control the link depth in another way.

Verify if link starts with http when using lxml and xpath in Python

I am trying to print all the links from multiple pages using the following:
my_page = '//div[#class="product_info"]//table//tr[7]//td[2]//a/#href'
Now, this works for most of the links, but in some cases I have something like:
To follow which isn't a link.
How can I omit these links ? What condition should I use when using:
# some more code
EMPTY = ''
my_page = '//div[#class="product_info"]//table//tr[7]//td[2]//a/#href'
for part in dom1.xpath(my_page):
FINAL_URL = urlparse.urljoin(url, part)
if part == EMPTY:
continue
print part

To filter those links that start with https:// or http://, simply add a condition in your loop:
# some more code
EMPTY = ''
other_links = set()
processed_links = set()
my_page = '//div[#class="product_info"]//table//tr[7]//td[2]//a/#href'
for part in dom1.xpath(my_page):
if part[:4] == 'http':
if part not in processed_links:
processed_links.add(part)
FINAL_URL = urlparse.urljoin(url, part)
else:
other_links.add(part)
I've also added some code so that:
You collect all the other links that are not processed.
If the same (valid) link appears in the page more than once, you only process it once.

Recursive function gives no output

I'm scraping all the URL of my domain with recursive function.
But it outputs nothing, without any error.
#usr/bin/python
from bs4 import BeautifulSoup
import requests
import tldextract
def scrape(url):
for links in url:
main_domain = tldextract.extract(links)
r = requests.get(links)
data = r.text
soup = BeautifulSoup(data)
for href in soup.find_all('a'):
href = href.get('href')
if not href:
continue
link_domain = tldextract.extract(href)
if link_domain.domain == main_domain.domain :
problem.append(href)
elif not href == '#' and link_domain.tld == '':
new = 'http://www.'+ main_domain.domain + '.' + main_domain.tld + '/' + href
problem.append(new)
return len(problem)
return scrape(problem)
problem = ["http://xyzdomain.com"]
print(scrape(problem))
When I create a new list, it works, but I don't want to make a list every time for every loop.

You need to structure your code so that it meets the pattern for recursion as your current code doesn't - you also should not call variables the same name as libraries, e.g. href = href.get() because this will usually stop the library working as it becomes the variable, your code as it currently is will only ever return the len() as this return is unconditionally reached before: return scrap(problem).:
def Recursive(Factorable_problem)
if Factorable_problem is Simplest_Case:
return AnswerToSimplestCase
else:
return Rule_For_Generating_From_Simpler_Case(Recursive(Simpler_Case))
for example:
def Factorial(n):
""" Recursively Generate Factorials """
if n < 2:
return 1
else:
return n * Factorial(n-1)

Hello I've made a none recursive version of this that appears to get all the links on the same domain.
The code below I've tested using the problem included in the code. When I'd solved the problems with the recursive version the next problem was hitting the recursion depth limit so I rewrote it so it ran in an iterative fashion, the code and result below:
from bs4 import BeautifulSoup
import requests
import tldextract
def print_domain_info(d):
print "Main Domain:{0} \nSub Domain:{1} \nSuffix:{2}".format(d.domain,d.subdomain,d.suffix)
SEARCHED_URLS = []
problem = [ "http://Noelkd.neocities.org/", "http://youpi.neocities.org/"]
while problem:
# Get a link from the stack of links
link = problem.pop()
# Check we haven't been to this address before
if link in SEARCHED_URLS:
continue
# We don't want to come back here again after this point
SEARCHED_URLS.append(link)
# Try and get the website
try:
req = requests.get(link)
except:
# If its not working i don't care for it
print "borked website found: {0}".format(link)
continue
# Now we get to this point worth printing something
print "Trying to parse:{0}".format(link)
print "Status Code:{0} Thats: {1}".format(req.status_code, "A-OK" if req.status_code == 200 else "SOMTHINGS UP" )
# Get the domain info
dInfo = tldextract.extract(link)
print_domain_info(dInfo)
# I like utf-8
data = req.text.encode("utf-8")
print "Lenght Of Data Retrived:{0}".format(len(data)) # More info
soup = BeautifulSoup(data) # This was here before so i left it.
print "Found {0} link{1}".format(len(soup.find_all('a')),"s" if len(soup.find_all('a')) > 1 else "")
FOUND_THIS_ITERATION = [] # Getting the same links over and over was boring
found_links = [x for x in soup.find_all('a') if x.get('href') not in SEARCHED_URLS] # Find me all the links i don't got
for href in found_links:
href = href.get('href') # You wrote this seems to work well
if not href:
continue
link_domain = tldextract.extract(href)
if link_domain.domain == dInfo.domain: # JUST FINDING STUFF ON SAME DOMAIN RIGHT?!
if href not in FOUND_THIS_ITERATION: # I'ma check you out next time
print "Check out this link: {0}".format(href)
print_domain_info(link_domain)
FOUND_THIS_ITERATION.append(href)
problem.append(href)
else: # I got you already
print "DUPE LINK!"
else:
print "Not on same domain moving on"
# Count down
print "We have {0} more sites to search".format(len(problem))
if problem:
continue
else:
print "Its been fun"
print "Lets see the URLS we've visited:"
for url in SEARCHED_URLS:
print url
Which prints, after a lot of other logging loads of neocities websites!
What's happening is the script is popping a value of the list of websites yet to visit, it then gets all the links on the page which are on the same domain. If those links are to pages we haven't visited we add the link to the list of links to be visited. After we do that we pop the next page and do the same thing again until there are no pages left to visit.
Think this is what your looking for, get back to us in the comments if this doesn't work in the way that you want or if anyone can improve please leave a comment.