[web2py] 'id' on 1.99.3 with SQLFORM.grid and links

[David Watson]

I am running web2py version 1.99.3

I have the following code in my default controller. The view shows the
item. When i run
  links = [lambda row:
A('Edit',_href=URL("default","sweeps",args=[row.id])),
             lambda row:
A('Delete',_href=URL("default","promos",args=[row.id]))]
    grid2 = SQLFORM.grid(db.auth_user, links=links)
    return dict(item=grid2)

but it blows up with a keyerror on id. I thought my lambda links were
correct, but maybe something has changed or I've gotten it wrong?

Thanks,
David

<type 'exceptions.KeyError'> 'id'

Traceback (most recent call last):
  File "/home/watson/latte/web2py/gluon/restricted.py", line 204, in
restricted
    exec ccode in environment
  File "/home/watson/latte/web2py/applications/latte/controllers/
default.py", line 139, in <module>
  File "/home/watson/latte/web2py/gluon/globals.py", line 162, in
<lambda>
    self._caller = lambda f: f()
  File "/home/watson/latte/web2py/applications/latte/controllers/
default.py", line 45, in apps
    (db.portal_keyword.user_id==auth.user_id), links=links,
fields=fields, paginate=10, ui="jquery-ui")
  File "/home/watson/latte/web2py/gluon/sqlhtml.py", line 1810, in
grid
    if link(row):
  File "/home/watson/latte/web2py/applications/latte/controllers/
default.py", line 41, in <lambda>
    links = [lambda row:
A('Edit',_href=URL("default","index",args=[row.id])),
  File "/home/watson/latte/web2py/gluon/dal.py", line 3948, in
__getattr__
    return self[key]
  File "/home/watson/latte/web2py/gluon/dal.py", line 3939, in
__getitem__
    return dict.__getitem__(self, key)
KeyError: 'id'