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What is the purpose of the -m switch?
...s this only works on the subset of modules that are executable i.e. have a __main__.py file. Most don't and will break e.g. python -m sys 'print(sys.version)' fails with python: No code object available for sys. Suggest you make that clear in the answer.
– smci
...
Inheritance and Overriding __init__ in python
...his nowadays:
class FileInfo(dict):
"""store file metadata"""
def __init__(self, filename=None):
super(FileInfo, self).__init__()
self["name"] = filename
Note the following:
We can directly subclass built-in classes, like dict, list, tuple, etc.
The super function handle...
Sibling package imports
...gly hacks to allow imports from siblings modules or parents package from a __main__ module. The issue is detailed in PEP 366. PEP 3122 attempted to handle imports in a more rational way but Guido has rejected it one the account of
The only use case seems to be running scripts that happen
to be...
Set Django IntegerField by choices=… name
...epresents the proper integer.
Like so:
LOW = 0
NORMAL = 1
HIGH = 2
STATUS_CHOICES = (
(LOW, 'Low'),
(NORMAL, 'Normal'),
(HIGH, 'High'),
)
Then they are still integers in the DB.
Usage would be thing.priority = Thing.NORMAL
...
Import a module from a relative path
...
Assuming that both your directories are real Python packages (do have the __init__.py file inside them), here is a safe solution for inclusion of modules relatively to the location of the script.
I assume that you want to do this, because you need to include a set of modules with your script. I us...
How to mock an import
...importing A to get what you want:
test.py:
import sys
sys.modules['B'] = __import__('mock_B')
import A
print(A.B.__name__)
A.py:
import B
Note B.py does not exist, but when running test.py no error is returned and print(A.B.__name__) prints mock_B. You still have to create a mock_B.py where ...
Get absolute path of initially run script
...
The correct solution is to use the get_included_files function:
list($scriptPath) = get_included_files();
This will give you the absolute path of the initial script even if:
This function is placed inside an included file
The current working directory is dif...
Node.js - getting current filename
... is then easy:
var path = require('path');
var scriptName = path.basename(__filename);
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Difference between abstract class and interface in Python
... NotImplementedError("Class %s doesn't implement aMethod()" % (self.__class__.__name__)) is more informative error message :)
– naught101
Sep 3 '14 at 1:43
9
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