I'm trying to get into CTF's and I found a cool website ment to practice some web based CTF skills called ctf.slothparadise.com. I've managed to get 4 of the Flags but two of them are giving me the finger and sadly I've had to dust off the good Ol' Python skills.
import urllib.error
import urllib.request
import urllib.parse
import urllib
import sys
while True:
about_page = urllib.request.urlopen("http://ctf.slothparadise.com/about.php").read()
if "KEY" in about_page:
print(about_page)
sys.exit(0)
ctf.slothpython.com/about.php is the page I'm programming for and it spits out the key in the source code every 1000 visitors. Instead of being a moron and refreshing it till 1000 I wrote that code in hopes it would keep opening the page until the phrase "KEY" appeared in the pages source code.
I'm getting this: (TypeError: 'str' does not support the buffer interface)
From what I know about TypeErrors I'm guessing that I may have "KEY" in the wrong format perhaps? I'm not really sure, I also may not even be using the right modules but the old urllib2 module I would typically use for this got split up into different modules so I'm learning as I go with these new modules.
Any help is appreciated in fixing this issue, also if my interpretaion of TypeErrors is wrong feel free to correct me.
The object returned by urlopen().read() acts like a context manager.
You are not using it correctly.
Try something like that:
import urllib.request
while True:
with urllib.request.urlopen('http://ctf.slothparadise.com/about.php') as response:
html = response.read()
if b"KEY" in html:
print(html)
sys.exit(0)
urllib.request.urlopen returns an http.client.HTTPResponse object and that object's read returns an encoded bytes object. How to decode may be in the returned http header, or in your case, embedded in an html meta tag. You likely don't want to parse the html for this particular test, so just look for the bytes object b'KEY'.
I don't know what you want to do with the data next, but if you want it to print nicely or scan the html, then you will have to do some parsing.
import urllib.error
import urllib.request
import urllib.parse
import urllib
import sys
while True:
about_page = urllib.request.urlopen("http://ctf.slothparadise.com/about.php").read()
if b"KEY" in about_page:
print(about_page)
sys.exit(0)
Make about_page a string with
about_page=str(urllib.request.urlopen("http://ctf.slothparadise.com/about.php").read())
This should make your code work. Hope this helps!!
Related
I have this current Python code...
import requests
import json
response = json.loads(requests.get("https://meta.multimc.org/v1/net.minecraftforge/index.json").text)
But I can't find out how to get a specific bit instead of the whole JSON page.
I'd like to get versions.version on that URL but I can't find out what piece of code to add to make it do that.
You iterate over the versions array in response['versions'] and add each version to a list.
import request
import json
response = json.loads(requests.get("https://meta.multimc.org/v1/net.minecraftforge/index.json").text)
versions=[]
for i in range(len(response['versions'])):
versions.append(response['versions'][i]['version'])
I would like to get the plain text (e.g. no html tags and entities) from a given URL.
What library should I use to do that as quickly as possible?
I've tried (maybe there is something faster or better than this):
import re
import mechanize
br = mechanize.Browser()
br.open("myurl.com")
vh = br.viewing_html
//<bound method Browser.viewing_html of <mechanize._mechanize.Browser instance at 0x01E015A8>>
Thanks
you can use HTML2Text if the site isnt working for you you can go to HTML2Text github Repo and get it for Python
or maybe try this:
import urllib
from bs4 import*
html = urllib.urlopen('myurl.com').read()
soup = BeautifulSoup(html)
text = soup.get_text()
print text
i dont know if it gets rid of all the js and stuff but it gets rid of the HTML
do some Google searches there are multiple other questions similar to this one
also maybe take a look at Read2Text
In Python 3, you can fetch the HTML as bytes and then convert to a string representation:
from urllib import request
text = request.urlopen('myurl.com').read().decode('utf8')
I am trying to learn python and also create a web utility. One task I am trying to accomplish is creating a single html file which can be run locally but link to everything it needs to look like the original web page. (if you are going to ask why i want this, its because it may act of a part of a utility i am creating, or if not, just for education) So i have two questions, a theoretical one and a practical one:
1) Is this, for visual (as opposed to functional) purposes, possible? Can a html page work offline while linking to everything it needs online? or if their something fundamental about having the html file itself execute on the web server which does not allow this to be possible? How far can I go with it?
2) I have started a python script which de-relativises (made that one up) linked elements on a html page, but I am a noob so most likely I missed some elements or attributes which would also link to outside resources. I have noticed after trying a few pages that the one in the code below does not work properly, their appears to be a .js file which is not linking correctly. (the first of many problems to come) Assuming the answer to my first question was at least a partial yes, can anyone help me fix the code for this website?
Thank you.
Update, I missed the script tag on this, but even after I added it it still does not work correctly.
import lxml
import sys
from lxml import etree
from StringIO import StringIO
from lxml.html import fromstring, tostring
import urllib2
from urlparse import urljoin
site = "www.script-tutorials.com/advance-php-login-system-tutorial/"
output_filename = "output.html"
def download(site):
response = urllib2.urlopen("http://"+site)
html_input = response.read()
return html_input
def derealitivise(site, html_input):
active_html = lxml.html.fromstring(html_input)
for element in tags_to_derealitivise:
for tag in active_html.xpath(str(element+"[#"+"src"+"]")):
tag.attrib["src"] = urljoin("http://"+site, tag.attrib.get("src"))
for tag in active_html.xpath(str(element+"[#"+"href"+"]")):
tag.attrib["href"] = urljoin("http://"+site, tag.attrib.get("href"))
return lxml.html.tostring(active_html)
active_html = ""
tags_to_derealitivise = ("//img", "//a", "//link", "//embed", "//audio", "//video", "//script")
print "downloading..."
active_html = download(site)
active_html = derealitivise(site, active_html)
print "writing file..."
output_file = open (output_filename, "w")
output_file.write(active_html)
output_file.close()
Furthermore, I could make the code more through by checking all of the elements...
It would look kind of like this, but I do not know the exact way to iterate through all of the elements. This is a seperate problem, and I will most likely figure it out by the time anyone responds...:
def derealitivise(site, html_input):
active_html = lxml.html.fromstring(html_input)
for element in active_html.xpath:
for tag in active_html.xpath(str(element+"[#"+"src"+"]")):
tag.attrib["src"] = urljoin("http://"+site, tag.attrib.get("src"))
for tag in active_html.xpath(str(element+"[#"+"href"+"]")):
tag.attrib["href"] = urljoin("http://"+site, tag.attrib.get("href"))
return lxml.html.tostring(active_html)
update
Thanks to Burhan Khalid's solution, which seemed too simple to be viable at first glance, I got it working. The code is so simple most of you will most likely not require it, but I will post it anyway incase it helps:
import lxml
import sys
from lxml import etree
from StringIO import StringIO
from lxml.html import fromstring, tostring
import urllib2
from urlparse import urljoin
site = "www.script-tutorials.com/advance-php-login-system-tutorial/"
output_filename = "output.html"
def download(site):
response = urllib2.urlopen(site)
html_input = response.read()
return html_input
def derealitivise(site, html_input):
active_html = html_input.replace('<head>', '<head> <base href='+site+'>')
return active_html
active_html = ""
print "downloading..."
active_html = download(site)
active_html = derealitivise(site, active_html)
print "writing file..."
output_file = open (output_filename, "w")
output_file.write(active_html)
output_file.close()
Despite all of this, and its great simplicity, the .js object running on the website I have listed in the script still will not load correctly. Does anyone know if this is possible to fix?
while i am trying to make only the html file offline, while using the
linked resources over the web.
This is a two step process:
Copy the HTML file and save it to your local directory.
Add a BASE tag in the HEAD section, and point the href attribute of it to the absolute URL.
Since you want to learn how to do it yourself, I will leave it at that.
#Burhan has an easy answer using <base href="..."> tag in the <head>, and it works as you have found out. I ran the script you posted, and the page downloaded fine. As you noticed, some of the JavaScript now fails. This can be for multiple reasons.
If you are opening the HTML file as a local file:/// URL, the page may not work. Many browsers heavily sandbox local HTML files, not allowing them to perform network requests or examine local files.
The page may perform XmlHTTPRequests or other network operations to the remote site, which will be denied for cross domain scripting reasons. Looking in the JS console, I see the following errors for the script you posted:
XMLHttpRequest cannot load http://www.script-tutorials.com/menus.php?give=menu. Origin http://localhost:8000 is not allowed by Access-Control-Allow-Origin.
Unfortunately, if you do not have control of www.script-tutorials.com, there is no easy way around this.
I want to download some files from this site: http://www.emuparadise.me/soundtracks/highquality/index.php
But I only want to get certain ones.
Is there a way to write a python script to do this? I have intermediate knowledge of python
I'm just looking for a bit of guidance, please point me towards a wiki or library to accomplish this
thanks,
Shrub
Here's a link to my code
I looked at the page. The links seem to redirect to another page, where the file is hosted, clicking which downloads the file.
I would use mechanize to follow the required links to the right page, and then use BeautifulSoup or lxml to parse the resultant page to get the filename.
Then it's a simple matter of opening the file using urlopen and writing its contents out into a local file like so:
f = open(localFilePath, 'w')
f.write(urlopen(remoteFilePath).read())
f.close()
Hope that helps
Make a url request for the page. Once you have the source, filter out and get urls.
The files you want to download are urls that contain a specific extension. It is with this that you can do a regular expression search for all urls that match your criteria.
After filtration, then do a url request for each matched url's data and write it to memory.
Sample code:
#!/usr/bin/python
import re
import sys
import urllib
#Your sample url
sampleUrl = "http://stackoverflow.com"
urlAddInfo = urllib.urlopen(sampleUrl)
data = urlAddInfo.read()
#Sample extensions we'll be looking for: pngs and pdfs
TARGET_EXTENSIONS = "(png|pdf)"
targetCompile = re.compile(TARGET_EXTENSIONS, re.UNICODE|re.MULTILINE)
#Let's get all the urls: match criteria{no spaces or " in a url}
urls = re.findall('(https?://[^\s"]+)', data, re.UNICODE|re.MULTILINE)
#We want these folks
extensionMatches = filter(lambda url: url and targetCompile.search(url), urls)
#The rest of the unmatched urls for which the scrapping can also be repeated.
nonExtMatches = filter(lambda url: url and not targetCompile.search(url), urls)
def fileDl(targetUrl):
#Function to handle downloading of files.
#Arg: url => a String
#Output: Boolean to signify if file has been written to memory
#Validation of the url assumed, for the sake of keeping the illustration short
urlAddInfo = urllib.urlopen(targetUrl)
data = urlAddInfo.read()
fileNameSearch = re.search("([^\/\s]+)$", targetUrl) #Text right before the last slash '/'
if not fileNameSearch:
sys.stderr.write("Could not extract a filename from url '%s'\n"%(targetUrl))
return False
fileName = fileNameSearch.groups(1)[0]
with open(fileName, "wb") as f:
f.write(data)
sys.stderr.write("Wrote %s to memory\n"%(fileName))
return True
#Let's now download the matched files
dlResults = map(lambda fUrl: fileDl(fUrl), extensionMatches)
successfulDls = filter(lambda s: s, dlResults)
sys.stderr.write("Downloaded %d files from %s\n"%(len(successfulDls), sampleUrl))
#You can organize the above code into a function to repeat the process for each of the
#other urls and in that way you can make a crawler.
The above code is written mainly for Python2.X. However, I wrote a crawler that works on any version starting from 2.X
Why yes! 5 years later and, not only is this possible, but you've now got a lot of ways to do it.
I'm going to avoid code-examples here, because mainly want to help break your problem into segments and give you some options for exploration:
Segment 1: GET!
If you must stick to the stdlib, for either python2 or python3, urllib[n]* is what you're going to want to use to pull-down something from the internet.
So again, if you don't want dependencies on other packages:
urllib or urllib2 or maybe another urllib[n] I'm forgetting about.
If you don't have to restrict your imports to the Standard Library:
you're in luck!!!!! You've got:
requests with docs here. requests is the golden standard for gettin' stuff off the web with python. I suggest you use it.
uplink with docs here. It's relatively new & for more programmatic client interfaces.
aiohttp via asyncio with docs here. asyncio got included in python >= 3.5 only, and it's also extra confusing. That said, it if you're willing to put in the time it can be ridiculously efficient for exactly this use-case.
...I'd also be remiss not to mention one of my favorite tools for crawling:
fake_useragent repo here. Docs like seriously not necessary.
Segment 2: Parse!
So again, if you must stick to the stdlib and not install anything with pip, you get to use the extra-extra fun and secure (<==extreme-sarcasm) xml builtin module. Specifically, you get to use the:
xml.etree.ElementTree() with docs here.
It's worth noting that the ElementTree object is what the pip-downloadable lxml package is based on, and made make easier to use. If you want to recreate the wheel and write a bunch of your own complicated logic, using the default xml module is your option.
If you don't have to restrict your imports to the Standard Library:
lxml with docs here. As i said before, lxml is a wrapper around xml.etree that makes it human-usable & implements all those parsing tools you'd need to make yourself. However, as you can see by visiting the docs, it's not easy to use by itself. This brings us to...
BeautifulSoup aka bs4 with docs here. BeautifulSoup makes everything easier. It's my recommendation for this.
Segment 3: GET GET GET!
This section is nearly exactly the same as "Segment 1," except you have a bunch of links not one.
The only thing that changes between this section and "Segment 1" is my recommendation for what to use: aiohttp here will download way faster when dealing with several URLs because it's allows you to download them in parallel.**
* - (where n was decided-on from python-version to ptyhon-version in a somewhat frustratingly arbitrary manner. Look up which urllib[n] has .urlopen() as a top-level function. You can read more about this naming-convention clusterf**k here, here, and here.)
** - (This isn't totally true. It's more sort-of functionally-true at human timescales.)
I would use a combination of wget for downloading - http://www.thegeekstuff.com/2009/09/the-ultimate-wget-download-guide-with-15-awesome-examples/#more-1885 and BeautifulSoup http://www.crummy.com/software/BeautifulSoup/bs4/doc/ for parsing the downloaded file
i am really new to python. this is actually my first script with it and most of it is a copied example. i have an xml file that i need to parse out an attribute for. i got that part figured out but my issue is that that attribute does not always exist in the xml file. here is my code:
#!/usr/bin/python
#import library to do http requests:
import urllib2
import os
#import easy to use xml parser called minidom:
from xml.dom.minidom import parseString
#download the history:
history = urllib2.urlopen('http://192.168.1.1/example.xml')
#convert to string:
historydata = history.read()
history.close()
#parse the xml you downloaded
dom = parseString(historydata)
xmlTagHistory = dom.getElementsByTagName('loaded')[0].toxml()
xmlDataHistory=xmlTagHistory.replace('<loaded>','').replace('</loaded>','')
print xmlDataHistory
when the attribute doesnt exist i get a return of "IndexError: list index out of range". what i am attempting to do with this code is to get it to run a command if the attribute doesnt exist, or it is false. the other issue i will probably have is that there will be times when that attribute appears more than once so i would also need it to account for that scenario by NOT running the command if there is even one instance of "loaded" being true. as i said, i am really new at this so i could use all the help i can get. much appreciated.
Since dom.getElementsByTagName('loaded') returns a list, you can just check the list size with the len(list) function. Only if the list length is above 0, is it valid to do the [0] dereferencing.
An alternative is to wrap the code in try/exception pair and catch the parse exception.
http://docs.python.org/tutorial/errors.html
using the try and except you should be able to handle everything you need.