我正在尝试创建一个简单的操作,从数据库中获取一条记录(具有ManyToMany关系),然后显示JSON序列化实例,以下是我现在的处理方式:
服务模式:
class SystemService(models.Model):
name = models.CharField(max_length=35, unique=True, null=False, blank=False)
verion = models.CharField(max_length=35, unique=True, null=False, blank=False)
def __str__(self):
return self.name
服务器型号:
class Server(models.Model):
name = models.CharField(max_length=35, unique=True, null=False, blank=False)
ip_address = models.GenericIPAddressField(protocol='both', unpack_ipv4=True,
null=False, blank=False, unique=True)
operating_system = models.ForeignKey(OperatingSystem, null=False, blank=False)
monitored_services = models.ManyToManyField(SystemService)
info = models.CharField(max_length=250, null=True, blank=True)
pause_monitoring = models.BooleanField(default=False)
created_at = models.DateTimeField(auto_now_add=True)
updated_at = models.DateTimeField(auto_now=True)
def __str__(self):
return self.name
以下是现在使用muj-gabriel的答案:
def get_server(request, server_id):
try:
server_object = Server.objects.get(id=server_id)
data = {
'name': server_object.name,
'ip_address': server_object.ip_address,
'os': server_object.operating_system.name,
'info': server_object.info,
'monitoring_paused': server_object.pause_monitoring,
'created_at': server_object.created_at,
'update_at': server_object.updated_at,
'services': {service['id']: service['name'] for service
in server_object.monitored_services.values('id', 'name')}
}
return JsonResponse(data)
except Server.DoesNotExist:
return JsonResponse({'error': 'Selected object does not exits!'})
我不认为我所做的就足够了,因为每次我需要将一个实例作为JSON时我必须重复相同的事情,所以我想知道是否有一个pythonic和动态的方法来做它?
答案 0 :(得分:1)
如果你只需要价值观' id'和' name'我建议用这个:
'services': {service['id']: service['name']
for service in server_object.monitored_services.values('id', 'name')}
请参阅django docs
此外,您可以将代码移动到Model类的属性中,以便在其他地方重用它。
class Server(models.Model):
.......
@property
def data(self):
return {
'name': self.name,
'ip_address': self.ip_address,
'os': self.operating_system.name,
'info': self.info,
'monitoring_paused': self.pause_monitoring,
'created_at': self.created_at,
'update_at': self.updated_at,
'services': {service['id']: service['name'] for service in self.monitored_services.values('id', 'name')}
}
您的查看功能将是:
def get_server(request, server_id):
try:
server_object = Server.objects.get(id=server_id)
return JsonResponse(server_object.data)
except Server.DoesNotExist:
return JsonResponse({'error': 'Selected object does not exits!'})
答案 1 :(得分:0)
在看了一下Django文档后,我找到了model_to_dict函数,它基本上可以做我需要的东西(Model实例到Dict),但是对于ManyToMany关系,它只返回一个PK列表,所以我自己写了一个函数在它上面:
def db_instance2dict(instance):
from django.db.models.fields.related import ManyToManyField
metas = instance._meta
data = {}
for f in chain(metas.concrete_fields, metas.many_to_many):
if isinstance(f, ManyToManyField):
data[str(f.name)] = {tmp_object.pk: db_instance2dict(tmp_object)
for tmp_object in f.value_from_object(instance)}
else:
data[str(f.name)] = str(getattr(instance, f.name, False))
return data
希望它可以帮助别人。