openerp-india team mailing list archive

[Andreas Brueckl <1015126@xxxxxxxxxxxxxxxxxx>]

Public bug reported:

I am using a server-action (ir.actions.server) and want to call it via a menu item.
The user also has all rights to access the model of the server action

Expected Result:
User sees menu item with the server action

Actual Result:
User does not see the item with the server action

Problem:
Method _filter_visible_menus of ir_ui_menu.py calls modelaccess.check() with a browse_record(model_id, ..) as parameter.
Then method "check()" in ir_model.py wants to obtain the name of the model with "model.name".
The correct way would be to use "model.model" instead of model.name because this is the correct model name.

I am using OpenERP 6.0.3, but the problem should also exist in the
current trunk.

** Affects: openobject-server
     Importance: Undecided
         Status: New

** Project changed: openobject-client => openobject-server

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https://bugs.launchpad.net/bugs/1015126

Title:
  Wrong permission check with ir.actions.server

Status in OpenERP Server:
  New

Bug description:
  I am using a server-action (ir.actions.server) and want to call it via a menu item.
  The user also has all rights to access the model of the server action

  Expected Result:
  User sees menu item with the server action

  Actual Result:
  User does not see the item with the server action

  Problem:
  Method _filter_visible_menus of ir_ui_menu.py calls modelaccess.check() with a browse_record(model_id, ..) as parameter.
  Then method "check()" in ir_model.py wants to obtain the name of the model with "model.name".
  The correct way would be to use "model.model" instead of model.name because this is the correct model name.

  I am using OpenERP 6.0.3, but the problem should also exist in the
  current trunk.

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