pyramid

Possible Duplicate: Python, safe, sandbox I'm building a corporate web system in Python which allows scripts to be uploaded and run serverside. Given I'm already developing in Python and its such a nice simple language, it seems like a good language to write the scripts in. However, there is a security hazard there, I want to block all function calls except a limited subset. Is there a mechanism I can use to do this, or some other technique? Do I need to use something else, Lua perhaps? I'm developing in Pyramid/Pylons. This is a terrible idea, but just to let you know about the option: You

I have a table where I would like to fetch all the column names however after browsing the interwebs I could not find a way that works. This is what my table looks like: class myTable(Base): __tablename__ = 'myTable' col1 = Column(Integer, primary_key=True) col2 = Column(Unicode(10)) col3 = Column(Integer) col4 = Column(Numeric(10, 6)) col5 = Column(Numeric(6,3)) col6 = Column(Numeric(6,3)) child = relationship('tChild', backref=backref('children')) I would like to be able to print all the column names from a for loop. ex: "col1", "col2", "col3".... etc This is pretty easy with regular sql but

Currently, I'm just serving files like this: # view callable def export(request): response = Response(content_type='application/csv') # use datetime in filename to avoid collisions f = open('/temp/XML_Export_%s.xml' % datetime.now(), 'r') # this is where I usually put stuff in the file response.app_iter = f response.headers['Content-Disposition'] = ("attachment; filename=Export.xml") return response The problem with this is that I can't close or, even better, delete the file after the response has been returned. The file gets orphaned. I can think of some hacky ways around this, but I'm hoping

I have a custom 404 view defined in my Pyramid app: @view_config(context=HTTPNotFound, renderer='404.pt') def not_found(self, request): return {} It works fine, except that the HTTP status code sent with the content is 200 OK, which is not OK by any means. I'm having the same problem with 403 Forbidden. How can I get Pyramid to send the correct status code? The exception view is a separate view that provides a spot for you to do whatever you want. Just like any view that uses a renderer, you can affect the response object via request.response to modify its behavior. The renderer then fills in

I'm having problems with trying to catch an error. I'm using Pyramid/SQLAlchemy and made a sign up form with email as the primary key. The problem is when a duplicate email is entered it raises a IntegrityError, so I'm trying to catch that error and provide a message but no matter what I do I can't catch it, the error keeps appearing. try: new_user = Users(email, firstname, lastname, password) DBSession.add(new_user) return HTTPFound(location = request.route_url('new')) except IntegrityError: message1 = "Yikes! Your email already exists in our system. Did you forget your password?" I get the

I've got Apache2 running with mod_wsgi installed. I've confirmed that mod_wsgi actually works by following this . The problem comes when I try to get Pyramid running. I get an Internal Server Error and my Apache error log contains the exception: AssertionError: The EvalException middleware is not usable in a multi-process environment Here's my VHost: <VirtualHost *:80> ServerName pyramidtest.dev DocumentRoot /srv/pyramidtest.dev/www/ AssignUserID pyramidtest nogroup WSGIScriptAlias / /srv/pyramidtest.dev/pyramid/load.wsgi </VirtualHost> Here's my load.wsgi : import site site.addsitedir('/opt