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I want to get a html code value from many urls for the same domain and as example
the html code is the name
and the domain is facebook
and the urls is just like
https://www.facebook.com/mohamed.nazem2
so if you opened that url you will see the name is Mohamed Nazem
at shown by the code :
Mohamed Nazem (ناظِم)
as so that facebook url
https://www.facebook.com/zuck
Mark Zuckerberg
so the value at the first url was >Mohamed Nazem<
and the second url it's Mark Zuckerberg
hopefully you got what i thinking in..
To fetch the HTML page for each url you will need to use something like the requests library. To install it, use pip install requests and then in your code use it like so:
import requests
response = requests.get('https://facebook.com/zuck')
print(response.data)
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I'm using the below code to extract full HTML:
cont = await page1.content()
The website I intend to extract from is:
https://www.mohmal.com/en
which is a website to make temporary email accounts. The exact thing I want to do is reading the content of received emails, but by using the above code, I could not extract inner frame HTML where received emails contents placed within it. How can I do so?
Did you try using urllib?
You can use the urllib module to read html websites.
from urllib.request import urlopen
f = urlopen("https://www.google.com")
print(f.read())
f.close()
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Given a FB url like:
https://www.facebook.com/photo.php?fbid[LONGUSERID]&set=a.313002535549859&type=3&theater
How could I extract the real photo URL using PHP or Python?
Normally the actual URL looks like this ( as seen in Chrome Network tab)
https://scontent.fbru1-1.fna.fbcdn.net/v/t31.0-1/cp0/p32x32/11942095_139657766378816_623531952343456734_o.jpg?_nc_cat=106&_nc_sid=0081f9&_nc_ohc=VpijQtyWbUQAX-fsPMj&_nc_ht=scontent.fbru1-1.fna&oh=eb4435eed183716c807b405d0d57c3a4&oe=5F674BAB
But is there a way to automate this extraction this with script? Any example would be appreciated.
The simplest example.
I just got an HTML page, divided the text by double quotes into lines. Then I checked to see if the JPG extension was on the line.
import requests
from html import unescape
from urllib.parse import unquote
url = "https://www.facebook.com/photo.php?fbid=445552432123146"
response = requests.get(url)
if response:
lines = response.text.split('\"')
for line in lines:
if ".jpg" in line:
print(unquote(unescape(line)))
else:
print("fail!")
With the help of Selenium you can already search for elements in HTML code correctly.
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I get a link with requests.get and when I check history it's empty although the link redirects to another address when I open it with my browser. What is the problem?
import requests
r=requests.get('http://dir.iran.ir/home?p_p_id=webdirectorydisplay_WAR_webdirectoryportlet&p_p_lifecycle=0&p_p_state=exclusive&p_p_mode=view&_webdirectorydisplay_WAR_webdirectoryportlet_itemEntryId=14439&_webdirectorydisplay_WAR_webdirectoryportlet_cmd=redirectToLink')
result=r.history
but result equal with empty list
and final link is http://www.dps.ir/
You should check the result of that URL first.
>>> r.content
'<script type="text/javascript">window.location.href="http://www.dps.ir";</script> '
Requests library doesn't provide ability to execute Javascript, so that explains why there is no history.
PS: Btw you could give phantomjs a shot.
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For example, I tried getting Python to read the following filtered page
http://www.hearthpwn.com/cards?filter-attack-val=1&filter-attack-op=1&display=1
but Python only gets the unfiltered page http://www.hearthpwn.com/cards instead.
The standard library urllib2 normally follows redirects. If retrieving this URL used to work without being redirected, then the site has changed.
Although you can prevent following the redirect within urllib2 (by providing an alternative HTTP handler), I recommend using requests, where you can do:
import requests
r = requests.get('http://www.hearthpwn.com/cards?filter-attack-val=1'
'&filter-attack-op=1&display=1', allow_redirects=False)
print(r)
giving you:
<Response [302]>
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So let's say I have this URL: https://www.python.org/
and I want to download the page's source into a .txt file named python_source.txt
how would I do that?
Use urllib2, Here's how it's done:
response = urllib2.urlopen(url)
content = response.read()
Now you can save the content in any text file.
The python package urllib does just this. The documentation gives a very clear example on what you want to do.
import urllib.request
local_filename, headers = urllib.request.urlretrieve('http://python.org/')
html = open(local_filename)