Run a .py file

to run a Python program, 2 solutions:

- In a command window, type: python programme.py
- Rename the .py to .pyw and double-click on it (but in this case, you won't see the console)

Here’s what it looks like:

http://img72.imageshack.us/img72/8287/pythonmn9.jpg

Maybe an error in the code..

#!/usr/bin/env python # devloop 2006 import sys, socket, urllib2 if len(sys.argv) < 2: print "Usage: python statsliar.py url <referer_url>" sys.exit(1) url = sys.argv[1] if not url.startswith("http://"): print "Not a valid url" sys.exit(1) if len(sys.argv) == 3: if sys.argv[2].startswith("http://"): referer = sys.argv[2] else: print "Not a valid referer" sys.exit(1) else: referer = "" head = {'User-agent': 'Mozilla/4.0 (compatible; MSIE 5.5; Windows NT)', 'Referer': 'http://proxy-list.org/downloadproxylist.php?sp=-1&pp=any&pt=any&pc=any&ps=any'} socket.setdefaulttimeout(10) try: req = urllib2.Request("http://proxy-list.org/downloadproxylist.php?sp=-1&pp=any&pt=any&pc=any&ps=any", headers=head) proxylist = urllib2.urlopen(req).readlines() except IOError: print "Error reading proxy list" sys.exit(1) print "Fetching url " + url head = {'User-agent': 'Mozilla/4.0 (compatible; MSIE 5.5; Windows NT)'} if referer != "": head["Referer"] = referer print "Using referer " + referer for prox in proxylist: prox = prox.strip() try: req = urllib2.Request(url, headers=head) req.set_proxy(prox, 'http') urllib2.urlopen(req) except IOError: continue print prox</referer_url>

Here I get started and he tells me "Usage statsliar.py python url <referer_url>

Here he indicates how to use the software.
It means that you need to pass a URL in the command line.

Example: python statsliar.py https://www.toto.com/

If I understand correctly, this script is used to manipulate the stats of a site using proxies.

Hmmpf... :-/
What I think about it, huh...

I have Python 2.2.3

That must be why.

Uninstall your version of Python and install 2.5.1:
https://www.python.org/ftp/python/2.5.1/python-2.5.1.msi

It’s that the program displays things in the console.
So it needs to be run in a console (and not as .pyw).

"Python : Impossible d'ouvrir le fichier "statliars.py""

Did you navigate to the directory of your .py program before typing the command?

No, how should I do it?

Use the CD command to change the directory and place yourself in the directory where your script is located.

Here I start and he tells me "Usage statsliar.py python url <referer_url>"</referer_url>

Yes, but why do they give me an error after?

No idea.

At my place, the line socket.setdefaulttimeout(10)
does not cause any error.

What version of Python do you have?
Type: python --version

Here's how to add Python to the PATH.

Right-click on My Computer > Properties
Advanced tab
"Environment Variables" button

In the lower part (System variables)
select PATH and click on the "Edit" button.

In "Variable value", go to the end of the line and add:

;c:\python25

(Don't forget the semicolon just before c:\python25)

Confirm everything by clicking OK.

You may need to restart for the changes to take effect.

Hello, CMD does not recognize the Python command, what should I do???

Yes, but it doesn't execute (nothing launches in .pyw).

And how?