[Zope] Unauthorized Within a ZClass

jimherz@yahoo.com jimherz@yahoo.com
Thu, 01 Nov 2001 16:41:43 -0000


Hello All,

I've created a ZClass that contains a DTML Document object within it. 
I'm able to create, manipulate, view, etc. the DTML Document object
within the ZClass management environment; however, once I create a
class instance and try to view the DTML Document component, I get an
unauthorized error.  I'm logged in as the manager, and I've also tried
manually allowing all the permissions for the ZClass instance, but
this still has not helped.  Any ideas?

I'm using Zope 2.4.1 on Redhat Linux 7.0

Here's the traceback data:

Unauthorized

  You are not authorized to access parm1. 

  Traceback (innermost last):
    File
/usr/local/zope/Zope-2.4.1/lib/python/ZPublisher/Publish.py, 		line
223, in publish_module
    File
/usr/local/zope/Zope-2.4.1/lib/python/ZPublisher/Publish.py, 		line
187, in publish
    File
/usr/local/zope/Zope-2.4.1/lib/python/ZPublisher/Publish.py, 		line
171, in publish
    File /usr/local/zope/Zope-2.4.1/lib/python/ZPublisher/mapply.py, 	
line 160, in mapply
    (Object: parmForm)
    File
/usr/local/zope/Zope-2.4.1/lib/python/ZPublisher/Publish.py, 		line
112, in call_object
    (Object: parmForm)
    File /usr/local/zope/Zope-2.4.1/lib/python/OFS/DTMLMethod.py,
line 		194, in __call__
    (Object: parmForm)
    File
/usr/local/zope/Zope-2.4.1/lib/python/DocumentTemplate/DT_String.py,
line 546, in __call__
    (Object: parmForm)
    File
/usr/local/zope/Zope-2.4.1/lib/python/AccessControl/DTML.py, 		line
101, in guarded_getattr
    (Object: parmForm)
    File
/usr/local/zope/Zope-2.4.1/lib/python/AccessControl/ZopeGuards.py,
line 120, in guarded_getattr
    (Object: LockableItem)
    File
/usr/local/zope/Zope-2.4.1/lib/python/AccessControl/ZopeGuards.py,
line 103, in aq_validate
    (Object: LockableItem)
    File
/usr/local/zope/Zope-2.4.1/lib/python/AccessControl/SecurityManager.py,
line 149, in validate
    File
/usr/local/zope/Zope-2.4.1/lib/python/AccessControl/ZopeSecurityPolicy.py,
line 229, in validate
  Unauthorized: (see above)