Re: [Zope] Dictionary Key Error from Parsed RSS
- Original Message - From: Tom Von Lahndorff [EMAIL PROTECTED] To: zope@zope.org Sent: Friday, June 15, 2007 10:18 AM Subject: [Zope] Dictionary Key Error from Parsed RSS Hi everyone, I've used Universal Feed Parser to parse an RSS feed. Then used these instructions below to get the parsed data into Zope via an External Method. Everything seems to work fine except that I keep getting an error whenever I try to access a key from the returned data. The resulting data is returned as a dictionary and if I render it as is it works fine, however in both ZPT and DTML, when trying to access a key, for instance feed or href from the dictionary I'm prompted to login, which I'm unable to do with my admin login. After hitting cancel I get a Key Error that says You are not able to access 'href' (or whatever) in this context. Is there any way to somehow let Zope know that this data is 'ok'? Thanks in advance. You should post the full Key Error message with traceback (from error_log in the root folder of the zmi). Jonathan ___ Zope maillist - Zope@zope.org http://mail.zope.org/mailman/listinfo/zope ** No cross posts or HTML encoding! ** (Related lists - http://mail.zope.org/mailman/listinfo/zope-announce http://mail.zope.org/mailman/listinfo/zope-dev )
Re: [Zope] Dictionary Key Error from Parsed RSS
--On 15. Juni 2007 10:18:20 -0400 Tom Von Lahndorff [EMAIL PROTECTED] wrote: I'm prompted to login, - use VerboseSecurity which I'm unable to do with my admin login. After hitting cancel I get a Key Error that says You are not able to access 'href' (or whatever) in this context. You must provide the full traceback. -aj pgp3DhwdmKNke.pgp Description: PGP signature ___ Zope maillist - Zope@zope.org http://mail.zope.org/mailman/listinfo/zope ** No cross posts or HTML encoding! ** (Related lists - http://mail.zope.org/mailman/listinfo/zope-announce http://mail.zope.org/mailman/listinfo/zope-dev )
[Zope] Dictionary Key Error from Parsed RSS
Hi everyone, I've used Universal Feed Parser to parse an RSS feed. Then used these instructions below to get the parsed data into Zope via an External Method. Everything seems to work fine except that I keep getting an error whenever I try to access a key from the returned data. The resulting data is returned as a dictionary and if I render it as is it works fine, however in both ZPT and DTML, when trying to access a key, for instance feed or href from the dictionary I'm prompted to login, which I'm unable to do with my admin login. After hitting cancel I get a Key Error that says You are not able to access 'href' (or whatever) in this context. Is there any way to somehow let Zope know that this data is 'ok'? Thanks in advance. Pre-requisites == - Zope 2.5.x (developed and tested with Zope 2.5.1, but other minor versions should work as well) - feedparser.py 3.0 (http://diveintomark.org/projects/feed_parser/) developed and tested with 3.0 beta 22, the latest version Installation 1. Installing feedparser.py --- 1.1. Unzip the archive (preserving path names) into /Zopesite/bin/lib/ 1.2. open a command prompt window and go to /Zopesite/bin/lib/ feedparser/ 1.3. enter the command (sans quotes): '..\..\python setup.py install' 1.4. after installation, enter the command: '..\..\python feedparsertest.py' 1.5. wait for testing to complete. On my laptop, 2051 tests ran in 118.481s 2. Installing the external method - 2.1 place the newsfeed.py module in the /Zopesite/Extensions/ directory 2.2 Add an external method to your zope site with an id and Module of 'newsfeed', and a function of 'parsefeed' 3. Installing the feed template --- 3.1 Add a Page template to your Zope site with the contents of the nprnews.pt Page Template 3.2 examine the generic template rendering of NPR News, all fields from the feed should be available. ___ Zope maillist - Zope@zope.org http://mail.zope.org/mailman/listinfo/zope ** No cross posts or HTML encoding! ** (Related lists - http://mail.zope.org/mailman/listinfo/zope-announce http://mail.zope.org/mailman/listinfo/zope-dev )
Re: [Zope] Dictionary Key Error from Parsed RSS
dtml-let news=newsfeed(feedurl='http://www.modscape.com/blog/ rss.xml') dtml-var news.feed.href /dtml-let gives: Site Error An error was encountered while publishing this resource. Error Type: Unauthorized Error Value: You are not allowed to access 'feed' in this context Troubleshooting Suggestions * The URL may be incorrect. * The parameters passed to this resource may be incorrect. * A resource that this resource relies on may be encountering an error. For more detailed information about the error, please refer to the error log. If the error persists please contact the site maintainer. Thank you for your patience. Traceback (innermost last): * Module ZPublisher.Publish, line 115, in publish * Module ZPublisher.mapply, line 88, in mapply * Module ZPublisher.Publish, line 41, in call_object * Module OFS.DTMLMethod, line 144, in __call__ DTMLMethod at /feeds/nprnews.html URL: http://localhost:8080/feeds/nprnews.html/manage_main Physical Path:/feeds/nprnews.html * Module DocumentTemplate.DT_String, line 476, in __call__ * Module DocumentTemplate.DT_Let, line 76, in render * Module DocumentTemplate.DT_Util, line 196, in eval __traceback_info__: news * Module string, line 1, in expression Unauthorized: You are not allowed to access 'feed' in this context On Jun 15, 2007, at 10:21 AM, Jonathan wrote: - Original Message - From: Tom Von Lahndorff [EMAIL PROTECTED] To: zope@zope.org Sent: Friday, June 15, 2007 10:18 AM Subject: [Zope] Dictionary Key Error from Parsed RSS Hi everyone, I've used Universal Feed Parser to parse an RSS feed. Then used these instructions below to get the parsed data into Zope via an External Method. Everything seems to work fine except that I keep getting an error whenever I try to access a key from the returned data. The resulting data is returned as a dictionary and if I render it as is it works fine, however in both ZPT and DTML, when trying to access a key, for instance feed or href from the dictionary I'm prompted to login, which I'm unable to do with my admin login. After hitting cancel I get a Key Error that says You are not able to access 'href' (or whatever) in this context. Is there any way to somehow let Zope know that this data is 'ok'? Thanks in advance. You should post the full Key Error message with traceback (from error_log in the root folder of the zmi). Jonathan ___ Zope maillist - Zope@zope.org http://mail.zope.org/mailman/listinfo/zope ** No cross posts or HTML encoding! ** (Related lists - http://mail.zope.org/mailman/listinfo/zope-announce http://mail.zope.org/mailman/listinfo/zope-dev )
Re: [Zope] Dictionary Key Error from Parsed RSS
--On 15. Juni 2007 11:45:49 -0400 Tom Von Lahndorff [EMAIL PROTECTED] wrote: dtml-let news=newsfeed(feedurl='http://www.modscape.com/blog/rss.xml') dtml-var news.feed.href /dtml-let gives: Site Error An error was encountered while publishing this resource. Error Type: Unauthorized Error Value: You are not allowed to access 'feed' in this context Likely because the module has no security assertions. All your stuff runs under the conditions of Restricted Python. The standard Zope security applies. -aj pgpZxlGOV6FfQ.pgp Description: PGP signature ___ Zope maillist - Zope@zope.org http://mail.zope.org/mailman/listinfo/zope ** No cross posts or HTML encoding! ** (Related lists - http://mail.zope.org/mailman/listinfo/zope-announce http://mail.zope.org/mailman/listinfo/zope-dev )
Re: [Zope] Dictionary Key Error from Parsed RSS
shouldnt this prevent that?
def dictify(p):
Since the FeedParser returns a dictionary-like class instance,
rather than
a real dictionary, We must iterate through all nested instances
and convert
them to dicts to avoid security problems within Zope. I have no
idea *why*
feedparser doesn't just use real nested dicts in the first place.
newdict = {}
if type(p) in [types.DictionaryType, types.InstanceType]:
d = p # for clarity
for k,v in d.items():
newdict[k] = dictify(v)
elif type(p) == types.ListType:
l = p # for clarity
for ss,v in zip(xrange(len(l)), l):
l[ss] = dictify(v)
return l
else: return p
return newdict
On Jun 15, 2007, at 11:49 AM, Andreas Jung wrote:
--On 15. Juni 2007 11:45:49 -0400 Tom Von Lahndorff
[EMAIL PROTECTED] wrote:
dtml-let news=newsfeed(feedurl='http://www.modscape.com/blog/
rss.xml')
dtml-var news.feed.href
/dtml-let
gives:
Site Error
An error was encountered while publishing this resource.
Error Type: Unauthorized
Error Value: You are not allowed to access 'feed' in this context
Likely because the module has no security assertions. All your
stuff runs
under the conditions of Restricted Python. The standard Zope
security applies.
-aj
___
Zope maillist - Zope@zope.org
http://mail.zope.org/mailman/listinfo/zope
** No cross posts or HTML encoding! **
(Related lists -
http://mail.zope.org/mailman/listinfo/zope-announce
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Re: [Zope] Dictionary Key Error from Parsed RSS
No idea what this method should tell use?! Your code uses
news.feed.href...nothing's like that can be found here in the code.
We can not guess what this snippet is doing and where and why.
-aj
--On 15. Juni 2007 12:02:48 -0400 Tom Von Lahndorff [EMAIL PROTECTED]
wrote:
shouldnt this prevent that?
def dictify(p):
Since the FeedParser returns a dictionary-like class instance,
rather than
a real dictionary, We must iterate through all nested instances and
convert
them to dicts to avoid security problems within Zope. I have no idea
*why*
feedparser doesn't just use real nested dicts in the first place.
newdict = {}
if type(p) in [types.DictionaryType, types.InstanceType]:
d = p # for clarity
for k,v in d.items():
newdict[k] = dictify(v)
elif type(p) == types.ListType:
l = p # for clarity
for ss,v in zip(xrange(len(l)), l):
l[ss] = dictify(v)
return l
else: return p
return newdict
On Jun 15, 2007, at 11:49 AM, Andreas Jung wrote:
--On 15. Juni 2007 11:45:49 -0400 Tom Von Lahndorff
[EMAIL PROTECTED] wrote:
dtml-let news=newsfeed(feedurl='http://www.modscape.com/blog/
rss.xml')
dtml-var news.feed.href
/dtml-let
gives:
Site Error
An error was encountered while publishing this resource.
Error Type: Unauthorized
Error Value: You are not allowed to access 'feed' in this context
Likely because the module has no security assertions. All your
stuff runs
under the conditions of Restricted Python. The standard Zope
security applies.
-aj
--
ZOPYX Ltd. Co. KG - Charlottenstr. 37/1 - 72070 Tübingen - Germany
Web: www.zopyx.com - Email: [EMAIL PROTECTED] - Phone +49 - 7071 - 793376
Registergericht: Amtsgericht Stuttgart, Handelsregister A 381535
Geschäftsführer/Gesellschafter: ZOPYX Limited, Birmingham, UK
E-Publishing, Python, Zope Plone development, Consulting
pgpiQY7pxdE3Q.pgp
Description: PGP signature
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Re: [Zope] Dictionary Key Error from Parsed RSS
Here's a snippet of the rendered RSS feed:
{'feed': {'subtitle': u'', 'links': [{'href': u'http://
www.modscape.com/blog', 'type': 'text/html', 'rel': 'alternate'}],
'title': u'modscape.com Home', 'subtitle_detail': {'base': 'http://
www.modscape.com/blog/rss.xml', 'type': 'text/html', 'value': u'',
'language': None}, 'title_detail': {'base': 'http://www.modscape.com/
blog/rss.xml', 'type': 'text/plain', 'value': u'modscape.com Home',
'language': None}, 'link': u'http://www.modscape.com/blog'},
'status': 200, 'updated': (2007, 6, 15, 16, 48, 28, 4, 166, 0),
'version': 'rss20', 'encoding': 'iso-8859-15', 'bozo': 0, 'headers':
{'content-length': '18486', 'via': '1.0 px04.bc.fw.cbs.net:80
(squid)', 'x-cache': 'MISS from px04.bc.fw.cbs.net', 'x-powered-by':
'Zope (www.zope.org), Python (www.python.org)', 'x-cache-lookup':
'HIT from px04.bc.fw.cbs.net:80', 'expires': 'Fri, 15 Jun 2007
17:48:28 GMT', 'server': 'Apache/2.0.52 (Red Hat) DAV/2 SVN/1.2.3
mod_ssl/2.0.52 OpenSSL/0.9.7a PHP/5.1.2 mod_fastcgi/2.4.2', 'last-
modified': 'Fri, 15 Jun 2007 16:48:28 GMT', 'connection': 'close',
'cache-control': 'max-age=3600', 'date': 'Fri, 15 Jun 2007 16:48:28
GMT', 'content-type': 'text/xml; charset=iso-8859-15'}, 'etag': None,
'href': 'http://www.modscape.com/blog/rss.xml', 'namespaces': {},
'entries': [{'updated': u'Mon, Sep 04 2006, 06:23PM',
'updated_parsed': None, 'links': [{'href': u'http://www.modscape.com/
blog/index_html?showall=yes#id1157412223', 'type': 'text/html',
'rel': 'alternate'}], 'title': u'Ubuntu: Linux For Human Beings',
'summary_detail': {'base': 'http://www.modscape.com/blog/rss.xml',
'type': 'text/html', 'value': u'img src=http://www.modscape.com/
blog/1157412223/ubuntu.jpg width=150 alt=ubuntu.jpg
align=right hspace=10 /\n\nTom wrote:br /br /\n\na
href=http://www.ubuntu.com/;Ubuntu/a is an easy
On Jun 15, 2007, at 12:17 PM, Andreas Jung wrote:
No idea what this method should tell use?! Your code uses
news.feed.href...nothing's like that can be found here in the code.
We can not guess what this snippet is doing and where and why.
-aj
--On 15. Juni 2007 12:02:48 -0400 Tom Von Lahndorff
[EMAIL PROTECTED] wrote:
shouldnt this prevent that?
def dictify(p):
Since the FeedParser returns a dictionary-like class instance,
rather than
a real dictionary, We must iterate through all nested
instances and
convert
them to dicts to avoid security problems within Zope. I have
no idea
*why*
feedparser doesn't just use real nested dicts in the first
place.
newdict = {}
if type(p) in [types.DictionaryType, types.InstanceType]:
d = p # for clarity
for k,v in d.items():
newdict[k] = dictify(v)
elif type(p) == types.ListType:
l = p # for clarity
for ss,v in zip(xrange(len(l)), l):
l[ss] = dictify(v)
return l
else: return p
return newdict
On Jun 15, 2007, at 11:49 AM, Andreas Jung wrote:
--On 15. Juni 2007 11:45:49 -0400 Tom Von Lahndorff
[EMAIL PROTECTED] wrote:
dtml-let news=newsfeed(feedurl='http://www.modscape.com/blog/
rss.xml')
dtml-var news.feed.href
/dtml-let
gives:
Site Error
An error was encountered while publishing this resource.
Error Type: Unauthorized
Error Value: You are not allowed to access 'feed' in this context
Likely because the module has no security assertions. All your
stuff runs
under the conditions of Restricted Python. The standard Zope
security applies.
-aj
--
ZOPYX Ltd. Co. KG - Charlottenstr. 37/1 - 72070 Tübingen - Germany
Web: www.zopyx.com - Email: [EMAIL PROTECTED] - Phone +49 - 7071 - 793376
Registergericht: Amtsgericht Stuttgart, Handelsregister A 381535
Geschäftsführer/Gesellschafter: ZOPYX Limited, Birmingham, UK
--
--
E-Publishing, Python, Zope Plone development, Consulting
___
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Re: [Zope] Dictionary Key Error from Parsed RSS
Tom Von Lahndorff wrote at 2007-6-15 11:45 -0400: dtml-let news=newsfeed(feedurl='http://www.modscape.com/blog/ rss.xml') dtml-var news.feed.href /dtml-let gives: Site Error An error was encountered while publishing this resource. Error Type: Unauthorized Error Value: You are not allowed to access 'feed' in this context ... If the error persists please contact the site maintainer. Thank you for your patience. Traceback (innermost last): * Module ZPublisher.Publish, line 115, in publish * Module ZPublisher.mapply, line 88, in mapply * Module ZPublisher.Publish, line 41, in call_object * Module OFS.DTMLMethod, line 144, in __call__ DTMLMethod at /feeds/nprnews.html URL: http://localhost:8080/feeds/nprnews.html/manage_main Physical Path:/feeds/nprnews.html * Module DocumentTemplate.DT_String, line 476, in __call__ * Module DocumentTemplate.DT_Let, line 76, in render * Module DocumentTemplate.DT_Util, line 196, in eval __traceback_info__: news * Module string, line 1, in expression Unauthorized: You are not allowed to access 'feed' in this context It is likely that the object looks like a dictionary (it uses a string representation like that of a true 'dict') but it is not a proper dictionary (but rather some instance of a class). Zope's restricted code treats proper dictionaries as unprotected (like it does for lists and tuples) but requires special security declarations for instances to be accessible in untrusted code. A solution for you might be to convert the RSS data into a proper dictionary. How this works in detail depends on the instance at hand. If it implementes an items method, then dict(instance.items()) would do it. Note that substructures, too, may look like Python types but are in fact instances. In this case, you would need to convert such substructures as well (similar to the top level object). -- Dieter ___ Zope maillist - Zope@zope.org http://mail.zope.org/mailman/listinfo/zope ** No cross posts or HTML encoding! ** (Related lists - http://mail.zope.org/mailman/listinfo/zope-announce http://mail.zope.org/mailman/listinfo/zope-dev )
