API给出了TypeError:不可理解的类型：'dict‘

Question

我正在开发一个django-，在该API中，我对VectorWise数据库进行一些SQL查询和计算，并使用get对一些输入参数创建响应。

一开始我有这样的结构来回应：

    response = {'Date': input_date,
                'Date comparable': date_comparable,
                'CA': {
                    'CA TTC (€)': ca_ttc_n,
                    'Rang magasin': rank,
                    'progression (%)': progression_percentage
                },
                'Nb art / pass caisse': {
                    'nombre': nb_art_pass_caiss,
                    'progression (%)': nb_art_pass_caiss_prog
                },
                'Panier': {
                    '(€)': panier,
                    'progression (%)': prog_panier}
                }

哪个工人很好，给了我一个回复。

然后，我试图改进响应的代码结构和可读性，如下所示：

    response = {'Date': input_date,
                'Comparable date': date_comparable,
                'Indicators': {
                    {"Name": "sales",
                     "Value": ca_ttc_n,
                     "Unit": currency,
                     "Rank": rank},
                    {"Name": "sales_progression",
                     "Value": progression_percentage,
                     "Unit": percentage,
                     "Rank": rank},
                    {"Name": "customer_items_number",
                     "Value": nb_art_pass_caiss,
                     "Unit": "Units",
                     "Rank": rank},
                    {"Name": "customer_items_number_progression",
                     "Value": nb_art_pass_caiss_prog,
                     "Unit": percentage,
                     "Rank": rank},
                    {"Name": "basket",
                     "Value": panier,
                     "Unit": currency,
                     "Rank": rank}
                }

另一方面，这会导致抛出TypeError的错误

Traceback (most recent call last):
  File "C:\ProgramData\Anaconda3\lib\site-packages\django\core\handlers\exception.py", line 34, in inner
    response = get_response(request)
  File "C:\ProgramData\Anaconda3\lib\site-packages\django\core\handlers\base.py", line 115, in _get_response
    response = self.process_exception_by_middleware(e, request)
  File "C:\ProgramData\Anaconda3\lib\site-packages\django\core\handlers\base.py", line 113, in _get_response
    response = wrapped_callback(request, *callback_args, **callback_kwargs)
  File "C:\ProgramData\Anaconda3\lib\site-packages\django\views\decorators\csrf.py", line 54, in wrapped_view
    return view_func(*args, **kwargs)
  File "C:\ProgramData\Anaconda3\lib\site-packages\django\views\generic\base.py", line 71, in view
    return self.dispatch(request, *args, **kwargs)
  File "C:\ProgramData\Anaconda3\lib\site-packages\rest_framework\views.py", line 495, in dispatch
    response = self.handle_exception(exc)
  File "C:\ProgramData\Anaconda3\lib\site-packages\rest_framework\views.py", line 455, in handle_exception
    self.raise_uncaught_exception(exc)
  File "C:\ProgramData\Anaconda3\lib\site-packages\rest_framework\views.py", line 466, in raise_uncaught_exception
    raise exc
  File "C:\ProgramData\Anaconda3\lib\site-packages\rest_framework\views.py", line 492, in dispatch
    response = handler(request, *args, **kwargs)
  File "C:\ProgramData\Anaconda3\lib\site-packages\rest_framework_extensions\cache\decorators.py", line 49, in inner
    kwargs=kwargs,
  File "C:\ProgramData\Anaconda3\lib\site-packages\rest_framework_extensions\cache\decorators.py", line 68, in process_cache_response
    response = view_method(view_instance, request, *args, **kwargs)
  File "C:\Users\10124077\Projects\dataplatform--web--api\wdc\views.py", line 484, in get
    "Rank": rank}
TypeError: unhashable type: 'dict'

为了更好地理解这个问题，我尝试删除了这个级别字段，并做了一些改进，但是我没有做到，因为它一直在响应体的其他字段上提高TypeError！有什么帮助吗？

Answer 1

问题是：

'Indicators': {
                {"Name": "sales",
                 "Value": ca_ttc_n,
                 "Unit": currency,
                 "Rank": rank},
                {"Name": "sales_progression",
                 "Value": progression_percentage,
                 "Unit": percentage,
                 "Rank": rank},
                {"Name": "customer_items_number",
                 "Value": nb_art_pass_caiss,
                 "Unit": "Units",
                 "Rank": rank},
                {"Name": "customer_items_number_progression",
                 "Value": nb_art_pass_caiss_prog,
                 "Unit": percentage,
                 "Rank": rank},
                {"Name": "basket",
                 "Value": panier,
                 "Unit": currency,
                 "Rank": rank}
            }

...attempts来创建一组字典。所以，您的代码基本上是这样做的：

{{"this": "is", "a": "dictionary"}, {"within": "a set"}}

这是不可能的，因为一个人只能构造一组可持续的对象，而dict是不可用的。你可以读到“哈斯可”是什么意思在医生里。不过，您可以使用一个字典列表：

"Indicators": [
    {'Name': ..., 'Value': ...},
    {'Name': ..., 'Value': ...},
    ...
]

Answer 2

你错过了一份清单：

            'Indicators': [
                {"Name": "sales",
                 "Value": ca_ttc_n,
                 "Unit": currency,
                 "Rank": rank},
                  ...
              ]
            }

Answer 3

它的原因是，您试图设置一个字典指示器(具有诸如字典这样的不可访问的值)，而没有任何键值。

response = {'Date': input_date,
                'Comparable date': date_comparable,
                'Indicators': {
                    'key1':{"Name": "sales",
                     "Value": ca_ttc_n,
                     "Unit": currency,
                     "Rank": rank},
                    'key2':{"Name": "sales_progression",
                     "Value": progression_percentage,
                     "Unit": percentage,
                     "Rank": rank},
                    'key3':{"Name": "customer_items_number",
                     "Value": nb_art_pass_caiss,
                     "Unit": "Units",
                     "Rank": rank},
                    'key4':{"Name": "customer_items_number_progression",
                     "Value": nb_art_pass_caiss_prog,
                     "Unit": percentage,
                     "Rank": rank},
                    'key5':{"Name": "basket",
                     "Value": panier,
                     "Unit": currency,
                     "Rank": rank}
                }
}

否则，如果不需要密钥，请将指示器作为列表。