django视图的python代码

Question

模型:

class Pathology(models.Model):
    pathology = models.CharField(max_length=100)

class Publication(models.Model):
    pubtitle = models.TextField()

class Pathpubcombo(models.Model):
    pathology = models.ForeignKey(Pathology)
    publication = models.ForeignKey(Publication) 

1. 发送到HTML模板的病理列表作为下拉菜单

视图:

def search(request):
    pathology_list = Pathology.objects.select_related().order_by('pathology')

2. 用户从下拉菜单中选择一个病理名称,并通过检索ID

视图:

def pathology(request):
    pathology_id = request.POST['pathology_id'] 
    p = get_object_or_404(Pathology, pk=pathology_id)

我被困在哪里 我需要python/django语法来编写以下内容:

pathology_id现在必须从表Pathpubcombo(中间manytomany表)中检索publication_id.检索publication_id后,必须使用它来检索发布表中的所有属性,并将这些属性发送到另一个html模板以显示给用户.

Answer 1

你应该使用这里描述的多对多关系:http: //www.djangoproject.com/documentation/models/many_to_many/

喜欢:

class Pathology(models.Model):
    pathology = models.CharField(max_length=100)
    publications = models.ManyToManyField(Publication)

class Publication(models.Model):
    pubtitle = models.TextField()

然后

def pathology(request):
    pathology_id = request.POST['pathology_id'] 
    p = get_object_or_404(Pathology, pk=pathology_id)
    publications = p.publications.all()
    return render_to_response('my_template.html',
                              {'publications':publications},
                              context_instance=RequestContext(request))

希望这有效,没有测试过,但你明白了.

编辑:

如果不可能重命名表并使用django的buildin支持,也可以使用select_related().

http://docs.djangoproject.com/en/dev/ref/models/querysets/#id4