python - 在django中保存递增字典的数据

Question

我试图将 API 响应的数据保存在我的 django 表中，但我面临的问题是 API 在一个由多个字典组成的列表中进行响应，并且每次调用 API 时字典的数量都会增加。

示例 第一次通话时

A = [{foo:4,
      bar:5}]

第二次通话时

A = [{foo:4,
      bar:5}, {foo:8, bar:10}]

第三次通话时

A = [{foo:4,
      bar:5}, {foo:8, bar:10}, {foo:12, bar:15}]

等等

我的模型如下所示:

class Tracking(models.Model):

    foo = models.CharField(max_length=255, default=0)
    bar = models.CharField(max_length=255, default=0)

我的观点.py

class trackapi(APIView):

    def get(self, request, pk):

       response = requests.request('GET', url, headers=headers, data=payload, auth=user_pass)

       for i in response.text:
           Tracking.objects.create(foo = i["foo"], bar = i["bar"]

但是这会在表中创建很多多个条目，我应该如何执行此操作才能不复制表中的数据？

我尝试过的:

    lr_id = LR.objects.filter(lr_quiz=pk)[0].id
    invoice_id = Invoice.objects.filter(invoice_quiz=pk)[0].id

    for i in track_response:

        Tracking.objects.get_or_create(lr_no=lr_id,invoice_no=invoice_id,
                                       tracking_id=i["_id"],trip_id = i["tripId"],telephone=i["tel"],
                                       loc = i["loc"], address=i["address"],city=i["city"],
                                       created_at=i["createdAt"])

API 数据:

[{"_id":"5dccedadff47e867a2833819","tel":"xxxxxxx","loc":[28.498692,77.095215],"tripId":"5dccedaaff47e867a28337ec","mode":"automated","osm_data":{"distance_remained":10791,"time_remained":1649.5},"distance_remained":10870,"time_remained":1173,"curr_ETA":"2019-11-14T06:43:19.664Z","address":"100,
The National Media Centre, Sector 24, Gurugram, Haryana 122022,
India","city":"Gurugram","createdAt":"2019-11-14T06:01:17.166Z"},{"_id":"5dccedacff47e867a2833801","tel":"xxxxxxx","loc":[28.498692,77.095215],"tripId":"5dccedaaff47e867a28337ec","mode":"automated","osm_data":{"distance_remained":10791,"time_remained":1649.5},"distance_remained":10870,"time_remained":1173,"curr_ETA":"2019-11-14T06:43:18.459Z","address":"100,
The National Media Centre, Sector 24, Gurugram, Haryana 122022,
India","city":"Gurugram","createdAt":"2019-11-14T06:01:16.163Z"}]

Answer 1

虽然 @drd 的答案可能有效，但我认为可能有更好的方法来利用 Django 的广泛工具。

对于非常简单的字典，例如您的示例中使用的 foo 和 bar 字典，很难使用以下方法，但我怀疑您的数据是否那么简单。

所以，您可以使用美妙的 get_or_create 方法!来自 docs :

A convenience method for looking up an object with the given kwargs (may be empty if your model has defaults for all fields), creating one if necessary.

Returns a tuple of (object, created), where object is the retrieved or created object and created is a boolean specifying whether a new object was created.

This is meant to prevent duplicate objects from being created when requests are made in parallel, and as a shortcut to boilerplatish code.

但是，为了确保其正常工作，因为这有点边缘情况，所以您需要对其进行一些技巧。

假设 API 返回的数据集与堆栈结构相当，您可以设计一个处理程序，根据每个条目在 API 返回的数据集中出现的位置为每个条目提供一个 ID，即第一个字典是 1，第二个字典是 1。是 2，等等。

models.py

class Tracking(models.Model):
    id = models.CharField("ID", max_length=255, primary_key=True)
    ...

views.py

def get(...):    
...
# we also need to deserialize the response data into an iterable list
data = json.loads(response.text)

for i, entry in enumerate(data):
    Tracking.objects.get_or_create(pk=i, **entry)

这将在新对象出现时创建它们，而不会因先前调用中的重复内容而使数据库变得困惑。

编辑

如果从 API 获取的数据已经具有某种唯一字段(在本例中为 _id)，您可以只使用该值作为主键，如下所示:

views.py

def get(...):
    ...
    for entry in data:
        pk = entry.pop("_id")
        Tracking.objects.get_or_create(pk=pk, **entry)