Thank you C. Kirby. This is great, and will work very well for what I am
trying to accomplish. Thanks again.
On Monday, March 17, 2014 2:01:19 PM UTC-4, C. Kirby wrote:
>
> I would do it with a recursive ForeignKey
>
> class Employee(models.Model):
> employee = models.OneToOneField(User)
> supervisor = models.ForeignKey('self', blank=True, null=True)
>
> This keeps the information in one table and pretty clean looking.
> Allows all employees to have a supervisor, but they don't have to have one
> (say, CEO)
> If an employee may have more than one supervisor you can use a ManyToMany
> instead of ForeignKey
>
>
>
> On Sunday, March 16, 2014 5:59:59 PM UTC-5, Gene Coetzee wrote:
>>
>> Hi There!
>>
>> Starting out with Django, and was looking for a bit of guidance. I am
>> trying to establish a Organizational structure for employee/supervisor type
>> relationships. Since each of these individuals will be a user, I am
>> extending trying to also take advantage of django's built-in contrib.auth
>> package.
>>
>> In Short, employees report to supervisors, and in turn supervisors in
>> themselves can report to their respective supervisors... I developed the
>> following models, but was curious if there might be a better way to do this:
>>
>> class Supervisor(models.Model):
>> supervisor = models.OneToOneField(User)
>> def __unicode__(self):
>> return self.supervisor.username
>>
>> class Employee(models.Model):
>> employee = models.OneToOneField(User)
>> supervisor = models.ForeignKey(Supervisor)
>> def __unicode__(self):
>> return self.employee.username
>>
>> Thank you!
>>
>>
>>
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