Keep track of a variable which tells us the action user is currently
doing like "draft,publish,edit "?
the first time the user does a publish you need to trigger an action.
details:
How to know the current action of user?
---------------------------------------------------------------
a) 1. savedraft ----->URL1 --->status = "draft"
save/edit--------------->URL2-->status="publish"
(or)
pass an additional parameter in the url
&status="action" which tell what action a user is doing?
c) Let us categorize his actions in to two
1) draft =>when he does save draft
2) publish => when he does save/edit
d) Also lets have a field in model named status which tells the
current status of a record.
by default associate status "draft" to every record.
status = models.Charfield(default="draft")
e) now instead of save() lets call customesave(param status) for
every object which you want to save
(i am assuming you bound a database object to form whenever such
record already exists in the database)
def customsave(self,status)
{
if(self.status = ="draft" && curstatus == "draft")
{
self.save();
}
elif (self.status == "draft" && curstatus == "publish")
{
self.status = "publish"
self.save()
Trigger the actions which you need to perform
for the first save()
}
elif(self.status == "publish" && curstatus=="publish")
#user is editing
{
self.save()
}
}
Hope the above helps.
Please let me know if you need more clarifications.
On Mar 6, 5:18 pm, Malcolm Tredinnick <malc...@pointy-stick.com>
wrote:
> On Thu, 2009-03-05 at 22:02 -0800, hanks...@gmail.com wrote:
> > Sorry for the unwieldy title, but nothing else strikes me at the
> > moment.
>
> > I have a blog app -- a version of basic.blog, actually. There's a
> > field in the model called "status" with two options: "draft" and
> > "public."
>
> > What I want to do is trigger an action the first time (and /only/ the
> > first time) a particular instance is saved as "public."
>
> > So:
>
> > Write a post in the admin, save as draft --> No action
> > Edit the post some, save as draft again --> No action
> > Edit more, publish --> Action triggered!
> > Edit again --> No action.
>
> > How would I go about this? My thought is to override save(), but I
> > can't figure out how to inspect the instance at that point to
> > determine if the status attribute has changed since it was created.
>
> If the object has a primary key value, assuming you don't set that
> manually yourself, then it already exists in the database. So fetch the
> existing values.
>
> def save(*args, **kwargs):
> if self.pk:
> old_version = self.model.objects.get(pk=self.pk)
> # Work out the changes between "self" and "old_version"
>
> (Before anybody starts complaining about the extra database hit, think
> about how relatively infrequently new posts are saved in the scheme of
> things. Even if it was at the rapid rate of once per minute, you
> wouldn't notice the extra read.)
>
> > I
> > assume that this information must be in there somewhere, since it
> > seems like you'd need it to generate the SQL statement.
>
> No. Django just checks to see if the record is already in the database.
> We can't make any assumptions based on the primary key value for all
> models, since people can assign to that attribute. However, if you know
> that *your* code doesn't assign to that attribute, which is a pretty
> normal case, you'll be able to use the pk test to know if you're saving
> or updating.
>
> One day (maybe 1.2 timeframe?!) we'll add at least optional change
> detection to models and only update changed columns. There's a couple of
> implementation things to be worked out there first, but it's on the
> medium-term feature list.
>
> Regards,
> Malcolm
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