子查詢在Django

Question

我有2種型號：

class Professors(models.Model): 
    professors_name = models.CharField('professor', max_length=32, unique=True) 
class Discipline(models.Model): 
    auditorium = models.IntegerField('auditorium') 
    professors_name = models.ForeignKey(Professors) 

在訪問量：

disciplines = Discipline.objects.all() 

所以，我有禮堂和professors_name_id的數量。但我需要完整的名稱，而不是id。怎麼做？

Answer 1

型號：

# models usually named in the singular 
class Professor(models.Model): 
    professors_name = models.CharField('professor', max_length=32, unique=True) 
class Discipline(models.Model): 
    auditorium = models.IntegerField('auditorium') 
    # your pointer is to a professor, not to the name 
    professor = models.ForeignKey(Professor) 

鑑於：

# select_related('professor') to avoid a second query when accessing professor 
disciplines = Discipline.objects.select_related('professor') 

模板：

{% for disc in disciplines %} 
{{ disc.auditorium }}: {{ disc.professor.name }} 
{% endfor %} 

對於值：

Discipline.objects.values('auditorium', 'professor__name') 

Answer 2

Django ORM將始終返回不是ID的對象。你應該有這樣的設計。

class Professor(models.Model): 
    name = models.CharField('professor', max_length=32, unique=True) 
class Discipline(models.Model): 
    auditorium = models.IntegerField('auditorium') 
    professor = models.ForeignKey(Professors) 

並使用discipline.professor.name單獨檢索名稱。