On Jan 5, 2:24 pm, Bill Freeman wrote:
> On Mon, Jan 4, 2010 at 11:00 PM, Marc Aymerich wrote:
> > class user(models.model):
> > person = models.OneToOneField(person)
>
> I presume that you meant:
>
> person = models.OneToOneField(employee)
>
> Is so, then greatlemer's method will work.
On Mon, Jan 4, 2010 at 11:00 PM, Marc Aymerich wrote:
> class user(models.model):
> person = models.OneToOneField(person)
I presume that you meant:
person = models.OneToOneField(employee)
Is so, then greatlemer's method will work.
Or, especially if you will have use for the employee
On Jan 5, 7:09 am, greatlemer wrote:
> On Jan 5, 4:00 am, Marc Aymerich wrote:
> You want to use filter [1] with something like:
> user.objects.filter(employee__company__name='companyname')
>
>
> [1]http://docs.djangoproject.com/en/dev/topics/db/queries/#retrieving-sp...
> --
>
Or simply us
On Tue, Jan 5, 2010 at 8:09 AM, greatlemer wrote:
> On Jan 5, 4:00 am, Marc Aymerich wrote:
> > Hi!
> > I have a model like this:
> >
> > class company(models.model):
> > name = models.CharField(mx_length=20)
> > class employee(models.model):
> > comany = models.ForeignKey(company)
> > class us
On Jan 5, 4:00 am, Marc Aymerich wrote:
> Hi!
> I have a model like this:
>
> class company(models.model):
> name = models.CharField(mx_length=20)
> class employee(models.model):
> comany = models.ForeignKey(company)
> class user(models.model):
> person = models.OneToOneField(person)
>
> One com
Hi!
I have a model like this:
class company(models.model):
name = models.CharField(mx_length=20)
class employee(models.model):
comany = models.ForeignKey(company)
class user(models.model):
person = models.OneToOneField(person)
One company can have multiple employees but one employee only can ha
Hi Guys,
Just wondered if anyone else could help out on this one? To me it
doesn't seem like it's a difficult thing to do, I just can't get it to
work :(
Cheers,
Chris
On Feb 23, 11:48 am, Darthmahon <[EMAIL PROTECTED]> wrote:
> Hmmm ok, can't get this to print anything. When trying to print th
Hmmm ok, can't get this to print anything. When trying to print the
threads array it does this:
[, []]
[, []]
So, it is looping through my friends properly, but it not getting
their threads. To be honest, ideally I just want to print a list of
all the threads my friends are in, and not a list of
If you look at my code, I put "related_name='threads'"
If you don't specify that, then it is automatically set to
"thread_set" (see here:
http://django.tomsk.ru/documentation/models/reverse_lookup/)
Also, in your view, do this instead:
return render_to_response('people/threads.html', { "friends"
Ok, just tried this:
View:
-
# get threads for friends
threads = []
for friend in friends:
threads.append([friend, friend.thread_set.all()])
-
This appears to print something back, but it looks like it is just
printing my friends.
Not sure how to loop through this result properly in
Hi Julien,
Just gave that a go, getting an error:
'User' object has no attribute 'threads'
I'm guessing that because I am referring to the User model, which has
no direct relation to the Thread module, this won't work?
Julien wrote:
> Hi, how about this? (I haven't tested it myself)
>
> Models
Hi, how about this? (I haven't tested it myself)
Models:
---
class UserProfile(models.Model):
user = models.ForeignKey(User, unique=True)
friends = models.ManyToManyField(User, blank=True,
related_name='friends')
class Thread(models.Model):
title = models.CharField(maxlength=200
Hey,
I have three models but I can only handle the relationship between two
at any one time.
Basically I have a page that needs to show a threads the current
user's friends are currently participating in. I.E. something like
this:
Thread: ABC
Participating: Friend 1
Thread: XYZ
Participating:
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