This post from Graham Dumpleton (wrote mod_wsgi) goes into detail of
why that happens.
Extremely useful reading...
http://blog.dscpl.com.au/2010/03/improved-wsgi-script-for-use-with.html
Thanks,
Richard Shebora
On Tue, Jun 1, 2010 at 10:15 AM, Thomas Lionel Smets wrote:
>
> If I put at the end
If I put at the end of my settings.py file, the following code :
if (DEBUG):
print "Project name : " + PRJ_NAME
print "Root directory : " + ROOT_DIR
The lines are printed twice :
Project name : XXX
Root directory : /Users/tsmets/Documents/python/XXX/site/
Project name : XXX
Root directory
results.append({'URL': '/feed/%s/show/%s' % (item.rss, item.rssfeed),})
On Thu, Aug 21, 2008 at 23:29, Bobby Roberts <[EMAIL PROTECTED]> wrote:
>
> hi.
>
> I'm trying to figure out how to dynamically replace things in a string
> as follows:
>
>
>results.append({'URL': '/feed/%s/show/%s' (%ite
hi.
I'm trying to figure out how to dynamically replace things in a string
as follows:
results.append({'URL': '/feed/%s/show/%s' (%item.rss,
%item.rssfeed),})
I think i'm close but it's not the right syntax. can anyone out there
help?
--~--~-~--~~~---~--~---
> What I'm doing is to load a model from a template.
> I have in a template something like this:
> {% block menu %}
> {% block rss %}
>
> I want to load "menu" and "rss", but I think that is too much work for
> now, I have no idea how to implement it and I probably won't do it.
> Thank you very m
Thank you!
What I'm doing is to load a model from a template.
I have in a template something like this:
{% block menu %}
{% block rss %}
I want to load "menu" and "rss", but I think that is too much work for
now, I have no idea how to implement it and I probably won't do it.
Thank you very much f
On Tue, 2007-04-10 at 18:03 -0500, Jeremy Dunck wrote:
> On 4/10/07, Todd O'Bryan <[EMAIL PROTECTED]> wrote:
> ...
> > The key is that he wanted to use the string name of the class, not the
> > class itself. Assuming that Foo is available (i.e., is local to the code
> > you're running or has been
This should work:
class Foo:
def hello(self):
return "Hello World"
object = Foo()
print object.hello()
More information:
http://www.diveintopython.org/object_oriented_framework/index.html
Ciao,
- Matt
On Apr 11, 12:22 am, "Grupo Django" <[EMAIL PROTECTED]> wrote:
> I know this i
On 4/10/07, Todd O'Bryan <[EMAIL PROTECTED]> wrote:
...
> The key is that he wanted to use the string name of the class, not the
> class itself. Assuming that Foo is available (i.e., is local to the code
> you're running or has been imported), this should work:
>
> o = locals()['Foo']()
Oh. In t
On Tue, 2007-04-10 at 17:50 -0500, Jeremy Dunck wrote:
> On 4/10/07, Grupo Django <[EMAIL PROTECTED]> wrote:
> >
> > I know this is not the right place for asking about python, but it's a
> > simple question.
> > I need to load an object given in a string. Example:
> >
> > #I have a class called f
On 4/10/07, Grupo Django <[EMAIL PROTECTED]> wrote:
>
> I know this is not the right place for asking about python, but it's a
> simple question.
> I need to load an object given in a string. Example:
>
> #I have a class called foo
> class foo:
> def Hello():
> return "Hello World"
>
>
I know this is not the right place for asking about python, but it's a
simple question.
I need to load an object given in a string. Example:
#I have a class called foo
class foo:
def Hello():
return "Hello World"
object = 'foo'
print object.Hello()
Something like this.
Is it possib
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