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How to retrieve a module's path?

...tually give you the path to the .pyc file that was loaded, at least on Mac OS X. So I guess you can do: import os path = os.path.abspath(a_module.__file__) You can also try: path = os.path.dirname(a_module.__file__) To get the module's directory. ...
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Using Python's os.path, how do I go up one directory?

... os.path.abspath(os.path.join(os.path.dirname( __file__ ), '..', 'templates')) As far as where the templates folder should go, I don't know since Django 1.4 just came out and I haven't looked at it yet. You should probably a...
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来自微软的一手内幕:Windows 10是怎么出炉的 - 创意 - 清泛网 - 专注C/C++及内核技术

...iPad并未碾压笔记本,消费者们对触屏电脑兴趣缺缺。另一个Windows 8.1版本试图挽回败局,但是为时已晚。市场已有定论,就像之前的Vista和Windows ME版本一样,Windows 8再次惨遭用户们跳过。 现在,在CEO 萨提亚·纳德拉(Satya Nade...
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How to open every file in a folder?

... Os You can list all files in the current directory using os.listdir: import os for filename in os.listdir(os.getcwd()): with open(os.path.join(os.cwd(), filename), 'r') as f: # open in readonly mode # do your stuff...
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php 获取操作系统、浏览器版本信息(持续更新) - 更多技术 - 清泛网 - 专...

...版本,以及用户使用何种浏览器等信息,本文主要提供getOS、getBrowser、getBrowserVer三个方法,对网上各方法进行测试汇总整理,持续更新希望能成为最佳实现,欢迎大家多提意见。一、获取操作系统信息: /** * 获取os信息 * ...
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How to get the parent dir location

...can only go as far as the root package, however. If this is a problem, use os.path.abspath: dirname(dirname(abspath(file))). share | improve this answer | follow ...
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How can I safely create a nested directory?

What is the most elegant way to check if the directory a file is going to be written to exists, and if not, create the directory using Python? Here is what I tried: ...
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任何组件块 · App Inventor 2 中文网

... « 返回首页 编写应用程序时的一个常见概念是“不要重复自己”(或 DRY)。与其制作大量重复代码,例如通过复制和粘贴,不如使用称为“任何组件”块的特殊块。 “任何组件”块到底是什么? 每...
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C++中智能指针的设计和使用 - C/C++ - 清泛网 - 专注C/C++及内核技术

...用实现技术是使用引用计数(reference count)。智能指针类将一个计数器与类指向的对象相关联,引用计数跟踪该类有多少个对象共享同一指针。每次创建类的新对象时,初始化指针并将引用计数置为1;当对象作为另一对象的副本而...
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What is the difference between os.path.basename() and os.path.dirname()?

What is the difference between os.path.basename() and os.path.dirname() ? 2 Answers ...