| Thanks a lot! I would have tried to do something with tuples, but this seems to make easy sense.
Todd On Mar 30, 2006, at 6:06 AM, Ned Batchelder wrote: Sorry, sent the email before the formatting was right:
class K:
def __init__(self, label=None, **kwargs):
assert(len(kwargs) == 1)
for k, v in kwargs.items():
self.id = k
self.v = v
self.label = label or self.id
class Constants:
def __init__(self, *args):
self.klist = args
for k in self.klist:
setattr(self, k.id, k.v)
def choices(self):
return [(k.id, k.label) for k in self.klist]
kBranchKind = Constants(
K(main=1, label='Main branch'),
K(dead=2, label='An ex-branch'),
K(aux=3) # I don't know how to spell 'Auxilliary' anyway!
) Now you can define ids and values, and optionally labels for each choice.
--Ned.
Ned Batchelder wrote: The choices= attribute in the model defines what goes into the select box. In this case, Constants.choices() returns a list based on the values based to the Constants constructor. My example below has it backwards for integers. It should be:
def choices(self):
return [(v,k) for k,v in self.__dict__.items()]
Then for our example, the choices for Branch.kind would be [(1, 'main'), (2, 'aux'), (3, 'dead')], showing 'main', 'aux', 'dead' in the admin interface. Nicer labels than the Python identifiers wouldn't be possible with this code, you'd have to do something more elaborate:
class K: def __init__(self, label=None, **kwargs): assert(len(kwargs) == 1) for k, v in kwargs.items(): self.id = k self.v = v self.label = label or self.id class Constants: def __init__(self, *args): self.klist = args for k in self.klist: setattr(self, k.id, k.v) def choices(self): return [(k.id, k.label) for k in self.klist] kBranchKind = Constants( K(main=1, label='Main branch'), K(dead=2, label='An ex-branch'), K(aux=3) # I don't know how to spell 'Auxilliary' anyway! )
Todd O'Bryan wrote: Wait. How do I define the user-friendly stuff that will show up in the select box for the admin interface?
Todd On Mar 29, 2006, at 9:59 PM, Ned Batchelder wrote: What I've done in these cases is to define a Constants class:
class Constants:
""" Construct one of these with keyword arguments, and you can use the
attributes.
"""
def __init__(self, **kwargs):
for k, v in kwargs.items():
setattr(self, k, v)
def choices(self):
return list(self.__dict__.items()) (I guess Enumeration would be a better name), then I can define a list of constants:
kBranchKind = Constants(
main = 1,
aux = 2,
dead = 3
) Then in the code, you can use kBranchKind.dead, and in your model, you can use:
class Branch(meta.Model):
trunk = meta.ForeignKey(Trunk)
kind = meta.IntegerField(choices=kBranchKind.choices()) It keeps the list of choices in one place, gives you run-time errors if you mistype the constant name (string literals would not), and it works just as well with strings for the values.
--Ned.
Ivan Sagalaev wrote: Todd O'Bryan wrote:
Your comment at the end got me thinking, though. Writing
trunk.get_branch(kind__exact=2)
is not very illuminating, but you're correct that the value 'Dead'
could get changed later. In Java, I'd use constants for the integer
values
public static final int DEAD = 2;
but that seems to violate DRY, because the semantics is already
listed in the choices list. I like using integers for what end up
being enumerated types because they don't take much space in the
database and, as you mentioned, it's easy to change the English
version without having to do anything to the db representation.
Is there a better way to do this kind of thing?
This got me thinking too :-)
Generally when I need a constant in Python I don't hesitate to use
string values for constants which are both values and names. So I'd have
BRANCH_KINDS = (('main', 'Main'), ('aux', 'Auxiliary'), ('dead',
'Dead'),)
I think it won't even hurt performance in DB lookups if you create index
for this field. However this implies changing the field to CharField
which won't become a
--
Ned Batchelder, http://nedbatchelder.com
--
Ned Batchelder, http://nedbatchelder.com
--~--~---------~--~----~------------~-------~--~----~
You received this message because you are subscribed to the Google Groups "Django users" group.
To post to this group, send email to django-users@googlegroups.com
To unsubscribe from this group, send email to [EMAIL PROTECTED]
For more options, visit this group at http://groups.google.com/group/django-users
-~----------~----~----~----~------~----~------~--~---
|
