Konkretne v djangu sa to robi tak ze vramci modulu ked definujes
triedy, tak ako argument do ForeignKey nedavas triedu ale jej nazov.
Django to potom automaticky prevedia na triedu:
class teacher(aModels.User):
telefon = models.IntegerField()
class Trida(models.Model):
name = models.CharField(max_length = 15)
classTeacher = models.OneToOneField('teacher')
students = models.ManyToManyField('student')
class student(aModels.User):
Tridy = models.ManyToManyField('Trida')
2010/10/26 Visgean Skeloru <visg...@gmail.com>:
> Dobrý den, mám tady takovýhle kód:
>
>> from django.db import models
>> from django.contrib.auth import models as aModels
>>
>> class teacher(aModels.User):
>> telefon = models.IntegerField()
>>
>> class Trida(models.Model):
>> name = models.CharField(max_length = 15)
>> classTeacher = models.OneToOneField(teacher)
>>
>> students = models.ManyToManyField(student)
>>
>>
>> class student(aModels.User):
>> Tridy = models.ManyToManyField(Trida)
>
> problém je že jedno je definované druhým, nevíte co s tím?
>
> Zkoušel jsem i následovný kód:
>>
>> from django.db import models
>> from django.contrib.auth import models as aModels
>>
>> class teacher(aModels.User):
>> telefon = models.IntegerField()
>>
>> class Trida(models.Model):
>> name = models.CharField(max_length = 15)
>> classTeacher = models.OneToOneField(teacher)
>>
>> class student(aModels.User):
>> Tridy = models.ManyToManyField(Trida)
>>
>>
>> Trida.students = models.ManyToManyField(student)
>
>
> Ale django nevytvoří patřičný model, nevíte o nějaké metodě jak to řešit?
>
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