python - 从json文件创建Python Active Resource对象

因此，出于测试目的，我尝试从json文件创建python ActiveResource对象（我希望该对象具有json文件中的属性）。更具体地说，我正在使用（https://github.com/Shopify/shopify_python_api）中的ShopifyResource，它扩展了ActiveResource对象。

我浏览了源代码，发现一些我认为有用的功能：
（https://github.com/Shopify/shopify_python_api/blob/master/shopify/base.py）

from pyactiveresource.activeresource import ActiveResource
import shopify.mixins as mixins

class ShopifyResource(ActiveResource, mixins.Countable):
    _format = formats.JSONFormat

    def _load_attributes_from_response(self, response):
        if response.body.strip():
            self._update(self.__class__.format.decode(response.body))

_update来自ActiveResource（https://github.com/Shopify/pyactiveresource/blob/master/pyactiveresource/activeresource.py）

    def _update(self, attributes):
        """Update the object with the given attributes.

        Args:
            attributes: A dictionary of attributes.
        Returns:
            None
        """
        if not isinstance(attributes, dict):
            return
        for key, value in six.iteritems(attributes):
            if isinstance(value, dict):
                klass = self._find_class_for(key)
                attr = klass(value)
            elif isinstance(value, list):
                klass = None
                attr = []
                for child in value:
                    if isinstance(child, dict):
                        if klass is None:
                            klass = self._find_class_for_collection(key)
                        attr.append(klass(child))
                    else:
                        attr.append(child)
            else:
                attr = value
            # Store the actual value in the attributes dictionary
            self.attributes[key] = attr

因此，我尝试执行以下操作：

order = Order()
with open("file.json")) as json_file:
    x = json.loads(json_file.read())
    order._update(x)

顺序扩展ShopifyResource（扩展ActiveResource）的位置。如果没记错的话，x应该是字典，它是_update（）函数的适当参数。

但是我得到以下输出：

raceback (most recent call last):
  File "/home/vineet/Documents/project/tests/test_sync.py", line 137, in testSaveOrder1
    self.getOrder()
  File "/home/vineet/Documents/project/tests/tests/test_sync.py", line 113, in getOrder
    order._update(x)
  File "/home/vineet/Documents/project/venv/lib/python3.6/site-packages/pyactiveresource/activeresource.py", line 962, in _update
    attr.append(klass(child))
  File "/home/vineet/Documents/project/venv/lib/python3.6/site-packages/shopify/base.py", line 126, in __init__
    prefix_options, attributes = self.__class__._split_options(attributes)
  File "/home/vineet/Documents/project/venv/lib/python3.6/site-packages/pyactiveresource/activeresource.py", line 466, in _split_options
    if key in cls._prefix_parameters():
  File "/home/vineet/Documents/project/venv/lib/python3.6/site-packages/pyactiveresource/activeresource.py", line 720, in _prefix_parameters
    for match in template.pattern.finditer(path):
TypeError: cannot use a string pattern on a bytes-like object

我什至尝试了以下方法：

 order._update(order._format.decode(json_file.read()))

但这没有用，因为'str'对象没有属性'decode'。

最佳答案

您似乎担心x是否具有正确的格式。打印并检查。

顺便说一句：并使用

x = json.load(json_file)

代替

x = json.loads(json_file.read())

关于python - 从json文件创建Python Active Resource对象，我们在Stack Overflow上找到一个类似的问题：https://stackoverflow.com/questions/56468796/