如何将json转换为对象？

I need to convert a json-string to python object. By object I mean "new" python3 object like:

我需要将json-string转换为python对象。按对象我的意思是“新”python3对象，如：

class MyClass(object):

I found several help for example on jsonpickle documentation. But all I found are tutorials which convert object to json first and after this convert backwards.

我在jsonpickle文档中找到了一些帮助。但我发现的所有内容都是将对象首先转换为json并在此转换后向后转换的教程。

I want to convert a json-string from a Rest-API.

我想从Rest-API转换json-string。

Here is what I have done so far:

这是我到目前为止所做的：

import requests
import jsonpickle

class Goal(object):
    def __init__(self):
        self.GoaldID = -1
        self.IsPenalty = False

class Match(object):
    def __init__(self):
        self.Goals = []

headers = {
    "Content-Type": "application/json; charset=utf-8"
}

url = "https://www.openligadb.de/api/getmatchdata/39738"

result = requests.get(url=url, headers=headers)
obj = jsonpickle.decode(result.json)
print (obj)

This results in:

这导致：

TypeError: the JSON object must be str, bytes or bytearray, not 'method'

It's quite clear to me that jsonpickle can't convert this to my classes (Goal, Match), because I don't tell jsonpickle in which class the output should be converted. The problem is I don't know how to tell jsonpickle to convert the JSON in object from type Match? And how can I tell that the list of goals should be of type List<Goal>?

我很清楚，jsonpickle无法将其转换为我的类（目标，匹配），因为我不告诉jsonpickle应在哪个类中转换输出。问题是我不知道如何告诉jsonpickle从对象类型转换对象中的JSON？我怎么能告诉目标列表应该是List 类型？

2 个解决方案

#1

The following lines will give you a dictionary:

以下几行将为您提供一个字典：

obj = jsonpickle.decode(result.content)  # NOTE: `.content`, not `.json`

obj = result.json()

But none of above will give you what you want (python object (not dicitonary)). because the json from the url is not encoded with jsonpickle.encode - whcih add additional information to a generated json (something like {"py/object": "__main__.Goal", ....})

但上面没有一个能给你你想要的东西（python对象（不是dicitonary））。因为来自url的json没有用jsonpickle.encode编码 - whcih向生成的json添加其他信息（类似于{“py / object”：“_ _ main __。Goal”，....}）

>>> import jsonpickle
>>> class Goal(object):
...     def __init__(self):
...         self.GoaldID = -1
...         self.IsPenalty = False
...
>>> jsonpickle.encode(Goal())
'{"py/object": "__main__.Goal", "IsPenalty": false, "GoaldID": -1}'
# ^^^^^^^^^^^^^^^^^^^^^^^^^^^^
# JSON encoded with jsonpickle.encode  (default unpicklable=True)
#   => additional python class information attached
#   => can be decoded back to Python object
>>> jsonpickle.decode(jsonpickle.encode(Goal()))
<__main__.Goal object at 0x10af0e510>


>>> jsonpickle.encode(Goal(), unpicklable=False)
'{"IsPenalty": false, "GoaldID": -1}'
# with unpicklable=False   (similar output with json.dumps(..))
#   => no python class information attached
#   => cannot be decoded back to Python object, but a dict
>>> jsonpickle.decode(jsonpickle.encode(Goal(), unpicklable=False))
{'IsPenalty': False, 'GoaldID': -1}

If you want an actual Python object which is not a dictionary, i.e. you prefer dic.Goals.[0].GoalGetterName to dic["Goals"][0]["GoalGetterName"], use json.loads with object_hook:

如果你想要一个不是字典的实际Python对象，即你更喜欢dic.Goals。[0] .GoalGetterName到dic [“Goals”] [0] [“GoalGetterName”]，请使用带有object_hook的json.loads：

import json
import types    
import requests

url = "https://www.openligadb.de/api/getmatchdata/39738"

result = requests.get(url)
data = json.loads(result.content, object_hook=lambda d: types.SimpleNamespace(**d))
# OR   data = result.json(object_hook=lambda d: types.SimpleNamespace(**d))
goal_getter = data.Goals[0].GoalGetterName
# You get `types.SimpleNamespace` objects in place of dictionaries

2 个解决方案

#1

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