Provided config url is not valid
时间: 2024-01-09 08:04:11 浏览: 97
I'm sorry for the confusion. Can you please provide me with more context on what you're referring to? It would be helpful if you could provide me with some additional information so that I can assist you better.
相关问题
yntaxError: "undefined" is not valid JSON at JSON.parse (
It seems like you have provided an undefined value as an argument to the JSON.parse() function, which is causing a syntax error. JSON.parse() expects a valid JSON string as its argument. Please make sure that the argument you are passing to the function is a valid JSON string, otherwise you will continue to get this error.
Write a function which check if provided variable is a valid python variable
名。(编写一个函数来检查提供的变量名是否是有效的 Python 变量名。))
可以按照以下方式创建一个这样的函数:
```python
import keyword
def is_valid_variable_name(variable_name):
"""
Check if provided variable name is a valid Python variable name.
"""
if not variable_name.isidentifier() or keyword.iskeyword(variable_name):
return False
return True
```
该函数使用 Python 内置的 `isidentifier()` 函数来检查变量名是否是有效的标识符。它还使用 `keyword.iskeyword()` 函数来检查变量名是否是 Python 关键字。如果变量名是有效的 Python 变量名,则该函数将返回 `True`,否则返回 `False`。
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