将JSON文件转换为Python对象

Question

我有一个JSON文件，我想把它放到python对象中。它有两个部分，员工和资产，我想把它们分为两部分。以下是JSON文件的示例：

{
"staff": [
    {
        "id": "DA7153",
        "name": [
            "Fran\u00c3\u00a7ois",
            "Ullman"
        ],
        "department": {
            "name": "Admin"
        },
        "server_admin": "true"
    },
    {
        "id": "DA7356",
        "name": [
            "Bob",
            "Johnson"
        ],
        "department": {
            "name": "Admin"
        },
        "server_admin": "false"
    },
],
"assets": [
    {
        "asset_name": "ENGAGED SLOTH",
        "asset_type": "File",
        "owner": "DA8333",
        "details": {
            "security": {
                "cia": [
                    "HIGH",
                    "INTERMEDIATE",
                    "LOW"
                ],
                "data_categories": {
                    "Personal": "true",
                    "Personal Sensitive": "true",
                    "Customer Sensitive": "true"
                }
            },
            "retention": 2
        },
        "file_type": "Document",
        "server": {
            "server_name": "ISOLATED UGUISU",
            "ip": [
                10,
                234,
                148,
                52
            ]
        }
    },
    {
        "asset_name": "ISOLATED VIPER",
        "asset_type": "File",
        "owner": "DA8262",
        "details": {
            "security": {
                "cia": [
                    "LOW",
                    "HIGH",
                    "LOW"
                ],
                "data_categories": {
                    "Personal": "false",
                    "Personal Sensitive": "false",
                    "Customer Sensitive": "true"
                }
            },
            "retention": 2
        },
    },
]

我曾尝试为staff创建一个类，但每次创建时都会出现错误“TypeError:dict最多需要1个参数，得到3个”

我使用的代码如下所示：

import json

with open('Admin_sample.json') as f:
    admin_json = json.load(f)

class staffmem(admin_json):
    def __init__(self, id, name, department, server_admin):
        self.id = id
        self.name = name
        self.deparment = department[name]
        self.server_admin = server_admin

    def staffid(self):
        return self.id

print(staffmem.staffid)

我就是想不出来。任何帮助都将不胜感激

谢谢

Answer 1

以下内容应该是一个很好的起点，但您必须解决一些问题。请注意，如果密钥不存在，我将在任何地方使用get()来提供“安全”默认值：

import json

class StaffMember:
    def __init__(self, json_entry):
        self.name = ",".join(json_entry.get("name"))
        self.id = json_entry.get("id")
        self.dept = json_entry.get("department", {}).get("name")
        self.server_admin = (
            True
            if json_entry.get("server_admin", "false").lower() == "true"
            else False
        )

# Get the data
with open("/tmp/test.data") as f:
    data = json.load(f)

# For every entry in the data["staff"] create object and index them by ID
all_staff = {}
for json_entry in data.get("staff", []):
    tmp = StaffMember(json_entry)
    all_staff[tmp.id] = tmp


print(all_staff)
print(all_staff['DA7153'].name)

输出：

$ python3 /tmp/test.py
{'DA7153': <__main__.StaffMember object at 0x1097b2d50>, 'DA7356': <__main__.StaffMember object at 0x1097b2d90>}
François,Ullman

可能的改进：

• Unicode处理
• 添加getter/setter
<> LI>在Cor中传递JSONDICT，考虑添加^ {< CD2> }静态方法来创建对象
• 丢失值的错误处理
• 如果此对象仅用于/主要用于存储数据，请考虑在py3中使用dataclass
• 如果不打算修改对象（只读），请考虑注释中的namedtuple方法

注:

• 您提供的json不正确-您需要修复它
• 示例中的语法错误，并且命名约定不太符合Python（请阅读更多here）