flask 当中获得运行 app 脚本的根目录这种写法是否多余

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This topic created in 4125 days ago, the information mentioned may be changed or developed.

run.py

from flask import Flask
app = Flask(__name__)

flask的源码:

def _get_package_path(name):
    try:
        return os.path.abspath(os.path.dirname(sys.modules[name].__file__))
    except (KeyError, AttributeError):
        return os.getcwd()

class Flask:
    def __init__(self, name):
        self.name = name
        self.root_path = _get_package_path(self.name)

这其中 _get_package_path 这么写也能获得run.py的根目录:

def _get_package_path(name):
    try:
        return os.path.abspath(os.path.dirname(name))
    except (KeyError, AttributeError):
        return os.getcwd()

为什么要加上sys.modules[name].__file__

flask

根目录

def

1 replies • 2015-06-15 14:42:28 +08:00

yuankui

Jun 15, 2015

因为函数命名的节操: do what you say,say what you mean

根据函数命名: 这里是要根据模块名,取得模块文件路径

而不是仅仅实现一个功能

另外
如果出现
.
└── hello
├── __init__.py
└── kitty
├── __init__.py
└── real_module.py

$ cat hello/__init__.py
import kitty.real_module as real_module

的情况hello中的real_module虽然叫做hello.real_module
但是,他的文件路径却在