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meaning of " for a in html('.l')" in python

I have been looking in the source code of howdoi.
https://github.com/gleitz/howdoi
In here the extract_links_from_bing and extract_links_from_google had these kind of syntax.
I tried to search online everything related to xml, element trees but the contructor like syntax is not found anywhere.
Here is the function
def _extract_links_from_bing(html):
html.remove_namespaces()
return [a.attrib['href'] for a in html('.b_algo')('h2')('a')]
def _extract_links_from_google(html):
return [a.attrib['href'] for a in html('.l')] or \
[a.attrib['href'] for a in html('.r')('a')]
My question is how does html('.b_algo')('h2')('a') iterate. Any links related to similar syntax will be appreciated.
Thanks for reading.

That project is using PyQuery, not xml etree.
Note that html is coming from _get_links():
def _get_links(query):
search_engine = os.getenv('HOWDOI_SEARCH_ENGINE', 'google')
search_url = _get_search_url(search_engine)
result = _get_result(search_url.format(URL, url_quote(query)))
html = pq(result)
return _extract_links(html, search_engine)
and pq comes from here:
from pyquery import PyQuery as pq
A PyQuery object can be used like $ from jquery. This is the function call syntax you're referring to.
From their Quickstart:
>>> d("#hello")
[<p#hello.hello>]
>>> p = d("#hello")
>>> print(p.html())
Hello world !
>>> p.html("you know <a href='http://python.org/'>Python</a> rocks")
[<p#hello.hello>]
>>> print(p.html())
you know Python rocks
>>> print(p.text())
you know Python rocks

Handling web links in Python

I have some links stored in a file which looks like this:
http://r14---sn-p5qlsnss.googlevideo.com/videoplayback?itag=22&id=o-AOtM1kWozUiJKP2ENWH989ZIfJaZNPVvXTrBkXx40lG5&key=yt6&ip=159.253.144.86&lmt=1480060612064057&dur=1047.870&mv=m&source=youtube&ms=au&ei=DtN8WLfwFsKb1gKXho6YDw&expire=1484597102&mn=sn-p5qlsnss&mm=31&ipbits=0&nh=IgpwcjAzLmlhZDA3KgkxMjcuMC4wLjE&initcwndbps=4717500&mt=1484575249&pl=24&signature=1ECAB2B56C30CBF760721A1A26A7E80963DB36B8.6336B2C9C41DB53C8FA1D2A037793275F57C4825&ratebypass=yes&mime=video%2Fmp4&upn=tUcEt34Qe6c&sparams=dur%2Cei%2Cid%2Cinitcwndbps%2Cip%2Cipbits%2Citag%2Clmt%2Cmime%2Cmm%2Cmn%2Cms%2Cmv%2Cnh%2Cpl%2Cratebypass%2Csource%2Cupn%2Cexpire&title=600ft+UFO+Crash+Site+Discovered+On+Mars%21+11%2F23%2F16
At the end of the link we have the video's title. I want to read this link from a file and get the video's title in a proper format (with those '+' and '%' signs properly resolved). How do I do that?
I cannot use raw cgi as suggested here since the link is read from a file and not submitted by a form. Any idea?

There's super convenient urllib.parse.parse_qs for python 3, but if you're using python 2, you might have to dig out the title string first.
import urllib
url = 'http://r14---sn-p5qlsnss.googlevideo.com/videoplayback?itag=22&id=o-AOtM1kWozUiJKP2ENWH989ZIfJaZNPVvXTrBkXx40lG5&key=yt6&ip=159.253.144.86&lmt=1480060612064057&dur=1047.870&mv=m&source=youtube&ms=au&ei=DtN8WLfwFsKb1gKXho6YDw&expire=1484597102&mn=sn-p5qlsnss&mm=31&ipbits=0&nh=IgpwcjAzLmlhZDA3KgkxMjcuMC4wLjE&initcwndbps=4717500&mt=1484575249&pl=24&signature=1ECAB2B56C30CBF760721A1A26A7E80963DB36B8.6336B2C9C41DB53C8FA1D2A037793275F57C4825&ratebypass=yes&mime=video%2Fmp4&upn=tUcEt34Qe6c&sparams=dur%2Cei%2Cid%2Cinitcwndbps%2Cip%2Cipbits%2Citag%2Clmt%2Cmime%2Cmm%2Cmn%2Cms%2Cmv%2Cnh%2Cpl%2Cratebypass%2Csource%2Cupn%2Cexpire&title=600ft+UFO+Crash+Site+Discovered+On+Mars%21+11%2F23%2F16'
title = url[url.rfind('&title=') + 7:]
print urllib.unquote_plus(title)
Note: thanks to bereal for pointing out parse_qs is also available in python 2, so just:
import urlparse
print urlparse.parse_qs(url)['title'][0]
'600ft UFO Crash Site Discovered On Mars! 11/23/16'

You could use urllib.parse.parse_qs and give it the string:
In [17]: urllib.parse.parse_qs(s)
Out[17]:
{'dur': ['1047.870'],
'ei': ['DtN8WLfwFsKb1gKXho6YDw'],
'expire': ['1484597102'],
'http://r14---sn-p5qlsnss.googlevideo.com/videoplayback?itag': ['22'],
[.. and so on ..]
'source': ['youtube'],
'sparams': ['dur,ei,id,initcwndbps,ip,ipbits,itag,lmt,mime,mm,mn,ms,mv,nh,pl,ratebypass,source,upn,expire'],
'title': ['600ft UFO Crash Site Discovered On Mars! 11/23/16'],
'upn': ['tUcEt34Qe6c']}
In [18]: urllib.parse.parse_qs(s)["title"][0]
Out[18]: '600ft UFO Crash Site Discovered On Mars! 11/23/16'

Purl can fit your needs:
import purl
u = purl.URL('http://r14---sn-p5qlsnss.googlevideo.com/videoplayback?itag=22&id=o-AOtM1kWozUiJKP2ENWH989ZIfJaZNPVvXTrBkXx40lG5&key=yt6&ip=159.253.144.86&lmt=1480060612064057&dur=1047.870&mv=m&source=youtube&ms=au&ei=DtN8WLfwFsKb1gKXho6YDw&expire=1484597102&mn=sn-p5qlsnss&mm=31&ipbits=0&nh=IgpwcjAzLmlhZDA3KgkxMjcuMC4wLjE&initcwndbps=4717500&mt=1484575249&pl=24&signature=1ECAB2B56C30CBF760721A1A26A7E80963DB36B8.6336B2C9C41DB53C8FA1D2A037793275F57C4825&ratebypass=yes&mime=video%2Fmp4&upn=tUcEt34Qe6c&sparams=dur%2Cei%2Cid%2Cinitcwndbps%2Cip%2Cipbits%2Citag%2Clmt%2Cmime%2Cmm%2Cmn%2Cms%2Cmv%2Cnh%2Cpl%2Cratebypass%2Csource%2Cupn%2Cexpire&title=600ft+UFO+Crash+Site+Discovered+On+Mars%21+11%2F23%2F16')
print(u.query_param('title'))

use urlparse.parse_qs:
try:
from urlparse import urlparse # for python2
except:
from urllib import parse as urlparse # for python3
rv = urlparse.parse_qs(link)
title = rv['title'][0]

import urllib
a = "http://r14---sn-p5qlsnss.googlevideo.com/videoplayback?itag=22&id=o-AOtM1kWozUiJKP2ENWH989ZIfJaZNPVvXTrBkXx40lG5&key=yt6&ip=159.253.144.86&lmt=1480060612064057&dur=1047.870&mv=m&source=youtube&ms=au&ei=DtN8WLfwFsKb1gKXho6YDw&expire=1484597102&mn=sn-p5qlsnss&mm=31&ipbits=0&nh=IgpwcjAzLmlhZDA3KgkxMjcuMC4wLjE&initcwndbps=4717500&mt=1484575249&pl=24&signature=1ECAB2B56C30CBF760721A1A26A7E80963DB36B8.6336B2C9C41DB53C8FA1D2A037793275F57C4825&ratebypass=yes&mime=video%2Fmp4&upn=tUcEt34Qe6c&sparams=dur%2Cei%2Cid%2Cinitcwndbps%2Cip%2Cipbits%2Citag%2Clmt%2Cmime%2Cmm%2Cmn%2Cms%2Cmv%2Cnh%2Cpl%2Cratebypass%2Csource%2Cupn%2Cexpire&title=600ft+UFO+Crash+Site+Discovered+On+Mars%21+11%2F23%2F16"
b = a.split('=')[-1]
print urllib.unquote_plus(b)

Python- regular expression to print word within link

I am using Jupyter Notebook to get docid=PE209374738 as my output using reg ex. It is currently stored in a dictionary in this format:
{'Url': 'https://backtoschool.com/document.php?docid=PE209374738&datasource=PHE&vid=3326&referrer=api'}.
This is my code:
results= xmldoc.getElementsByTagName("result")
dict= {}
for a in results:
url= 'Url'
dict[url] = a.getElementsByTagName("url")[0].childNodes[0].nodeValue
docid= re.search(r'\?(.*?)&')
Does anyone have any suggestions on how to print that id?

The standard library already has methods for parsing URLs properly, no need for regex.
In Python 3:
from urllib.parse import urlparse, parse_qs
url = 'https://backtoschool.com/document.php?docid=PE209374738&datasource=PHE&vid=3326&referrer=api'
print(parse_qs(urlparse(url).query)['docid'][0]) # PE209374738
In Python 2 the first line is:
from urlparse import urlparse, parse_qs

#alex-hall is correct, you probably should better parse this using a proper URL parser.
That said, your original question was about doing it with using regexps, so here is the solution (which you nearly nailed already):
s = 'https://backtoschool.com/document.php?docid=PE209374738&datasource=PHE&vid=3326&referrer=api'
m = re.search(r'\?docid=(.*?)&', s)
print m.groups()[0]
This will print the desired PE209374738.

How to validate LinkedIn public profile url regular expression in python

I jus want to validate the linkedin public profile url. I tried the concept like below
a = "https://in.linkedin.com/afadasdf"
p = re.compile('(http(s?)://|[a-zA-Z0-9\-]+\.|[linkedin])[linkedin/~\-]+\.[a-zA-Z0-9/~\-_,&=\?\.;]+[^\.,\s<]')
p.match(a)
The above concept is working fine. But when i give the url https://www.linkedin.com means that it's not working. Can anyone help me to validate both concepts.

It is the oring between the http(s) and www. which has given you the above problem. You could change them to * (i.e. 0 or more).
import re
a = "https://www.linkedin.com/afadasdf"
p = re.compile('((http(s?)://)*([a-zA-Z0-9\-])*\.|[linkedin])[linkedin/~\-]+\.[a-zA-Z0-9/~\-_,&=\?\.;]+[^\.,\s<]')
print p.match(a)
Although you might want to restrict it to www rather than any numbers or letters? So maybe:
p = re.compile('((http(s?)://)*([www])*\.|[linkedin])[linkedin/~\-]+\.[a-zA-Z0-9/~\-_,&=\?\.;]+[^\.,\s<]')

This pattern may help.
^((http|https):\/\/)?+(www.linkedin.com\/)+[a-z]+(\/)+[a-zA-Z0-9-]{5,30}+$
I have tested it and it works fine for me.

Instead of matching the url with a regex you could use the urllib module:
In [1]: import urllib
In [2]: u = "https://in.linkedin.com/afadasdf"
In [3]: urllib.parse.urlparse(u)
Out[3]: ParseResult(scheme='https', netloc='in.linkedin.com', path='/afadasdf', params='', query='', fragment='')
Now you can check for the netloc and path property.

urllib2 file name

If I open a file using urllib2, like so:
remotefile = urllib2.urlopen('http://example.com/somefile.zip')
Is there an easy way to get the file name other then parsing the original URL?
EDIT: changed openfile to urlopen... not sure how that happened.
EDIT2: I ended up using:
filename = url.split('/')[-1].split('#')[0].split('?')[0]
Unless I'm mistaken, this should strip out all potential queries as well.

Did you mean urllib2.urlopen?
You could potentially lift the intended filename if the server was sending a Content-Disposition header by checking remotefile.info()['Content-Disposition'], but as it is I think you'll just have to parse the url.
You could use urlparse.urlsplit, but if you have any URLs like at the second example, you'll end up having to pull the file name out yourself anyway:
>>> urlparse.urlsplit('http://example.com/somefile.zip')
('http', 'example.com', '/somefile.zip', '', '')
>>> urlparse.urlsplit('http://example.com/somedir/somefile.zip')
('http', 'example.com', '/somedir/somefile.zip', '', '')
Might as well just do this:
>>> 'http://example.com/somefile.zip'.split('/')[-1]
'somefile.zip'
>>> 'http://example.com/somedir/somefile.zip'.split('/')[-1]
'somefile.zip'

If you only want the file name itself, assuming that there's no query variables at the end like http://example.com/somedir/somefile.zip?foo=bar then you can use os.path.basename for this:
[user#host]$ python
Python 2.5.1 (r251:54869, Apr 18 2007, 22:08:04)
Type "help", "copyright", "credits" or "license" for more information.
>>> import os
>>> os.path.basename("http://example.com/somefile.zip")
'somefile.zip'
>>> os.path.basename("http://example.com/somedir/somefile.zip")
'somefile.zip'
>>> os.path.basename("http://example.com/somedir/somefile.zip?foo=bar")
'somefile.zip?foo=bar'
Some other posters mentioned using urlparse, which will work, but you'd still need to strip the leading directory from the file name. If you use os.path.basename() then you don't have to worry about that, since it returns only the final part of the URL or file path.

I think that "the file name" isn't a very well defined concept when it comes to http transfers. The server might (but is not required to) provide one as "content-disposition" header, you can try to get that with remotefile.headers['Content-Disposition']. If this fails, you probably have to parse the URI yourself.

Just saw this I normally do..
filename = url.split("?")[0].split("/")[-1]

Using urlsplit is the safest option:
url = 'http://example.com/somefile.zip'
urlparse.urlsplit(url).path.split('/')[-1]

Do you mean urllib2.urlopen? There is no function called openfile in the urllib2 module.
Anyway, use the urllib2.urlparse functions:
>>> from urllib2 import urlparse
>>> print urlparse.urlsplit('http://example.com/somefile.zip')
('http', 'example.com', '/somefile.zip', '', '')
Voila.

You could also combine both of the two best-rated answers :
Using urllib2.urlparse.urlsplit() to get the path part of the URL, and then os.path.basename for the actual file name.
Full code would be :
>>> remotefile=urllib2.urlopen(url)
>>> try:
>>> filename=remotefile.info()['Content-Disposition']
>>> except KeyError:
>>> filename=os.path.basename(urllib2.urlparse.urlsplit(url).path)

The os.path.basename function works not only for file paths, but also for urls, so you don't have to manually parse the URL yourself. Also, it's important to note that you should use result.url instead of the original url in order to follow redirect responses:
import os
import urllib2
result = urllib2.urlopen(url)
real_url = urllib2.urlparse.urlparse(result.url)
filename = os.path.basename(real_url.path)

I guess it depends what you mean by parsing. There is no way to get the filename without parsing the URL, i.e. the remote server doesn't give you a filename. However, you don't have to do much yourself, there's the urlparse module:
In [9]: urlparse.urlparse('http://example.com/somefile.zip')
Out[9]: ('http', 'example.com', '/somefile.zip', '', '', '')

not that I know of.
but you can parse it easy enough like this:
url = 'http://example.com/somefile.zip'
print url.split('/')[-1]

using requests, but you can do it easy with urllib(2)
import requests
from urllib import unquote
from urlparse import urlparse
sample = requests.get(url)
if sample.status_code == 200:
#has_key not work here, and this help avoid problem with names
if filename == False:
if 'content-disposition' in sample.headers.keys():
filename = sample.headers['content-disposition'].split('filename=')[-1].replace('"','').replace(';','')
else:
filename = urlparse(sample.url).query.split('/')[-1].split('=')[-1].split('&')[-1]
if not filename:
if url.split('/')[-1] != '':
filename = sample.url.split('/')[-1].split('=')[-1].split('&')[-1]
filename = unquote(filename)

You probably can use simple regular expression here. Something like:
In [26]: import re
In [27]: pat = re.compile('.+[\/\?#=]([\w-]+\.[\w-]+(?:\.[\w-]+)?$)')
In [28]: test_set
['http://www.google.com/a341.tar.gz',
'http://www.google.com/a341.gz',
'http://www.google.com/asdasd/aadssd.gz',
'http://www.google.com/asdasd?aadssd.gz',
'http://www.google.com/asdasd#blah.gz',
'http://www.google.com/asdasd?filename=xxxbl.gz']
In [30]: for url in test_set:
....: match = pat.match(url)
....: if match and match.groups():
....: print(match.groups()[0])
....:
a341.tar.gz
a341.gz
aadssd.gz
aadssd.gz
blah.gz
xxxbl.gz

Using PurePosixPath which is not operating system—dependent and handles urls gracefully is the pythonic solution:
>>> from pathlib import PurePosixPath
>>> path = PurePosixPath('http://example.com/somefile.zip')
>>> path.name
'somefile.zip'
>>> path = PurePosixPath('http://example.com/nested/somefile.zip')
>>> path.name
'somefile.zip'
Notice how there is no network traffic here or anything (i.e. those urls don't go anywhere) - just using standard parsing rules.

import os,urllib2
resp = urllib2.urlopen('http://www.example.com/index.html')
my_url = resp.geturl()
os.path.split(my_url)[1]
# 'index.html'
This is not openfile, but maybe still helps :)