At Wednesday 26/7/2006 18:57, KJZZ Webmaster wrote:
And you were right! This is not seen as as a valid dictionary (nor is it
seen as a string):
stuff = container.externalmethod(feedurl=feedurl)
test = same_type(stuff, {})
print test
return printed
returns "0"
However, interestingly enough, if
I found a way to test if the "ExternalMethod" really returned a dict here:
http://infrae.com/mailman_/pipermail/silva-dev/2004q2/001091.html
And you were right! This is not seen as as a valid dictionary (nor is it
seen as a string):
stuff = container.externalmethod(feedurl=feedurl)
test = same
Dieter,
Appreciate your help.
I found a way to test if the "ExternalMethod" really returned a dict here:
http://infrae.com/mailman_/pipermail/silva-dev/2004q2/001091.html
And you were right! This is not seen as as a valid dictionary (nor is it
seen as a string):
stuff = container.externalmetho
[EMAIL PROTECTED] wrote at 2006-7-24 11:13 -0700:
>I'm trying to move our sites from Zope 2.5.1 to Zope 2.8.1-final and I'm
>receiving an error with this script:
>
>http://www.zope.org/Members/johntynan/feedparser/
>
>In particular, I receive the following error, when using the attached page
>tem
I'm trying to move our sites from Zope 2.5.1 to Zope 2.8.1-final and I'm
receiving an error with this script:
http://www.zope.org/Members/johntynan/feedparser/
In particular, I receive the following error, when using the attached page
template "nprnews.pt":
"Unauthorized: You are not allowed
I'm trying to move our sites from Zope 2.5.1 to Zope 2.8.1-final and I'm
receiving an error with this script:
http://www.zope.org/Members/johntynan/feedparser/
In particular, I receive the following error, when using the attached page
template "nprnews.pt":
"Unauthorized: You are not allowed
