How to handle "TypeError: init() missing 1 required positional argument: 'path'"

Question

我想在 Python 3 中设置一个新的 tinyDB 对象，但是 TinyDB() 给我这个错误

Traceback (most recent call last):
  File "C:/Users/Utente/Documents/CodingStuff/AlgManager/AlgManager.py", line 4, in <module>
    db = tdb.TinyDB()
  File "C:\Users\Utente\PycharmProjects\prova\venv\lib\site-packages\tinydb\database.py", line 159, in __init__
    self._storage = storage(*args, **kwargs)
TypeError: __init__() missing 1 required positional argument: 'path'

估计是安装的问题，因为我写的代码真的只有

import tinydb as tdb

db = tdb.TinyDB()

我用pip安装了TinyDB。我能做什么？

Answer 1

无论您使用什么 python 版本，TinyDB 在您创建实例时都有一个必需的参数 "path to db"。

像这样：

import tinydb as tdb
db = tdb.TinyDB('path/to/db.json')

How to handle "TypeError: __init__() missing 1 required positional argument: 'path'"

How to handle "TypeError: __init__() missing 1 required positional argument: 'path'"

python

runtime-error

tinydb

How to handle "TypeError: init() missing 1 required positional argument: 'path'"

How to handle "TypeError: init() missing 1 required positional argument: 'path'"