Question

我在“rlist”中列出药物清单

rlist = []
#dName = disease name and r = resulted disease
rlist.append(Disease.objects.filter(dName = r).values_list('medicine',flat=True))

通过运行此语句，我有标签管理员“医药”的ID

rlist =  [<QuerySet [7]>, <QuerySet [27, 28, 29, 30]>]

但是，我想要药物的名称而不是id。

我怎么能得到???

models.py

class TaggedMedicine(TaggedItemBase):
    content_object = models.ForeignKey("Disease")

    def __str__(self):
       return self.name

class Disease(models.Model):
     did = models.AutoField(verbose_name='Disease Id', primary_key=True,auto_created=True)
     dName = models.CharField(max_length=100,unique=True)

     symptoms = TaggableManager(verbose_name='symptoms list', through=TaggedSymptoms)                #tags with comma or space separated values
     symptoms.rel.related_name = "+"

     medicine = TaggableManager(verbose_name='medicine list',through=TaggedMedicine)
     medicine.rel.related_name = "+"

Answer 1

这只是一种表现形式。但是，如果要更改该表示形式，请在models.py中，对于Medicine类，重写__str__方法。

def __str__(self):
    return self.name #if you have a name attribute for it

如何从Django中的id获取标记名称

1 个答案: