yep but do not reinvent the wheel: use URL(... user_signature) and @auth.requires_signature()
On Tuesday, September 24, 2013 11:03:39 PM UTC+2, Avi A wrote: > > Thanks a lot, I will check it. > I just thought that I could add a args(1) which will be a random number > the controller will create and then check if the args(1) == to the var > random number it created? Is that the "token" method? > > On Tuesday, September 24, 2013 11:55:50 PM UTC+3, Niphlod wrote: > >> check if the t_categories row that the user is loading actually belongs >> to him .... >> >> record = db(db.t_categories.id == >> request.args(0))(db.t_categories.created_by == >> auth.user_id).select().first() >> if not record: >> redirect(URL('not_authorized')) >> >> BTW: with SQLFORM.grid it's easier (really). >> allowed_records = db.t_categories.created_by == auth.user_id >> grid = SQLFORM.grid(allowed_records) >> >> By default, all links to edit, create, etc are all created with the >> signature >> web2py.com/books/default/chapter/29/04/the-core#Digitally-signed-urls , >> so users can't "tamper" playing with request.args >> >> On Tuesday, September 24, 2013 7:47:39 PM UTC+2, Avi A wrote: >>> >>> Hi , >>> i want to display a user a list of item tht belongs to him: >>> my_items= db(db.t_categories.created_by==auth.user.id >>> ).select(db.t_categories.ALL) >>> Then allow him to update each item from the list: >>> item_update=crud.update(db.t_categories,request.args(0)) >>> so far so good. >>> My problem is on the /args(0) page, I still have to make sure that: >>> db.t_categories.created_by==auth.user.id >>> I don't know how to inform the controller the var: >>> db.t_categories.created_by. >>> The ajax function or a different way? >>> Thanks. >>> >>> -- Resources: - http://web2py.com - http://web2py.com/book (Documentation) - http://github.com/web2py/web2py (Source code) - https://code.google.com/p/web2py/issues/list (Report Issues) --- You received this message because you are subscribed to the Google Groups "web2py-users" group. To unsubscribe from this group and stop receiving emails from it, send an email to web2py+unsubscr...@googlegroups.com. For more options, visit https://groups.google.com/groups/opt_out.