Reputation: 65
My code is to search a Link passed in the command prompt, get the HTML code for the webpage at the Link, search the HTML code for links on the webpage, and then repeat these steps for the links found. I hope that is clear.
It should print out any links that cause errors.
Some more needed info:
The max visits it can do is 100. If a website has an error, a None value is returned.
Python3 is what I am using
eg:
s = readwebpage(url)... # This line of code gets the HTML code for the link(url) passed in its argument.... if the link has an error, s = None.
The HTML code for that website has links that end in p2.html
, p3.html
, p4.html
, and p5.html
on its webpage. My code reads all of these, but it does not visit these links individually to search for more links. If it did this, it should search through these links and find a link that ends in p10.html, and then it should report that the link ending with p10.html has errors. Obviously it doesn't do that at the moment, and it's giving me a hard time.
My code..
url = args.url[0]
url_list = [url]
checkedURLs = []
AmountVisited = 0
while (url_list and AmountVisited<maxhits):
url = url_list.pop()
s = readwebpage(url)
print("testing url: http",url) #Print the url being tested, this code is here only for testing..
AmountVisited = AmountVisited + 1
if s == None:
print("* bad reference to http", url)
else:
urls_list = re.findall(r'href="http([\s:]?[^\'" >]+)', s) #Creates a list of all links in HTML code starting with...
while urls_list: #... http or https
insert = urls_list.pop()
while(insert in checkedURLs and urls_list):
insert = urls_list.pop()
url_list.append(insert)
checkedURLs = insert
Please help :)
Upvotes: 0
Views: 767
Reputation: 3335
http
outside your capture group, and [\s:]
matches "some sort of whitespace (ie \s
) or :"I'd change the regex to: urls_list = re.findall(r'href="(.*)"',s)
. Also known as "match anything in quotes, after href=". If you absolutely need to ensure the http[s]://, use r'href="(https?://.*)"'
(s?
=> one or zero s
)
EDIT: And with actually working regex, using a non-greedly glom: href=(?P<q>[\'"])(https?://.*?)(?P=q)'
(Also, uh, while it's not technically necessary in your case because re
caches, I think it's good practice to get into the habit of using re.compile
.)
Upvotes: 0
Reputation: 3898
Here is the code you wanted. However, please, stop using regexes for parsing HTML. BeautifulSoup is the way to go for that.
import re
from urllib import urlopen
def readwebpage(url):
print "testing ",current
return urlopen(url).read()
url = 'http://xrisk.esy.es' #put starting url here
yet_to_visit= [url]
visited_urls = []
AmountVisited = 0
maxhits = 10
while (yet_to_visit and AmountVisited<maxhits):
print yet_to_visit
current = yet_to_visit.pop()
AmountVisited = AmountVisited + 1
html = readwebpage(current)
if html == None:
print "* bad reference to http", current
else:
r = re.compile('(?<=href=").*?(?=")')
links = re.findall(r,html) #Creates a list of all links in HTML code starting with...
for u in links:
if u in visited_urls:
continue
elif u.find('http')!=-1:
yet_to_visit.append(u)
print links
visited_urls.append(current)
Upvotes: 1
Reputation: 4399
Not Python but since you mentioned you aren't tied strictly to regex
, I think you might find some use in using wget
for this.
wget --spider -o C:\wget.log -e robots=off -w 1 -r -l 10 http://www.stackoverflow.com
Broken down:
--spider
: When invoked with this option, Wget will behave as a Web spider, which means that it will not download the pages, just check that they are there.
-o C:\wget.log
: Log all messages to C:\wget.log.
-e robots=off
: Ignore robots.txt
-w 1
: set a wait time of 1 second
-r
: set recursive search on
-l 10
: sets the recursive depth to 10, meaning wget will only go as deep as 10 levels in, this may need to change depending on your max requests
http://www.stackoverflow.com
: the URL you want to start with
Once complete, you can review the wget.log
entries to determine which links had errors by searching for something like HTTP status codes 404
, etc.
Upvotes: 0