Usage

add an url to this address to get a proxyfied response
example: http://proxpix.tshirtman.fr/http://s1.directupload.net/images/110909/wfm2q82v.png

About

Developped using python and the flask microframework.
my source code is:
#!/usr/bin/env python
from flask import Flask
from flask import make_response
from flask import request
from flask import escape

app = Flask(__name__)

import urllib2
import mimetypes

@app.route('/')
def main():
    code = escape(open(__file__).read())
    return '''
    <!DOCTYPE html>
    <html><head><title></title></head><body>
    <h1>Usage</h1>
    <p>
    add an url to this address to get a proxyfied response<br/>
    example: '''+request.base_url+'''http://s1.directupload.net/images/110909/wfm2q82v.png

    </p>
    <h1>About</h1>
    Developped using <a href='http://python.org'>python</a> and the <a href='http://flask.pocoo.org'>flask</a> microframework.<br />

    my <a href='/code'>source code</a> is:
    <pre>''' + str(code) + '''
    </pre>

    (Yeah, most of the code is there so it's able to show its own code :])<br />
    released in WTFPL
    </body>
    </html>
    '''

@app.route('/<path:url>')
def proxy(url):
    if not url.startswith('http://'):
        url = 'http://'+url

    r = make_response(urllib2.urlopen(url).read())
    r.mimetype = mimetypes.guess_type(url)[0]
    return r

@app.route('/code')
def code():
    r = make_response(open(__file__).read())
    r.mimetype = 'text/plain'
    return r

if __name__ == '__main__':
    app.run(host='0.0.0.0', port=5001, debug=False)


    
(Yeah, most of the code is there so it's able to show its own code :])
released in WTFPL