在Python中接受文件参数(从“发送到"上下文菜单中) [英] Accepting File Argument in Python (from Send To context menu)

问题描述

我首先要指出我几乎没有python经验.

I'm going to start of by noting that I have next to no python experience.

替代文字http://www.aquate.us/u/9986423875612301299.jpg

您可能知道，只需在Windows PC上的发送到"文件夹中放置一个快捷方式，就可以允许程序将文件作为参数.

As you may know, by simply dropping a shortcut in the Send To folder on your Windows PC, you can allow a program to take a file as an argument.

我将如何编写一个以该文件为参数的python程序?

How would I write a python program that takes this file as an argument?

而且，如果有人有机会，这是一种奖励- 如何将其与urllib2集成以将文件发布到服务器上的PHP脚本中?

And, as a bonus if anyone gets a chance -- How would I integrate that with a urllib2 to POST the file to a PHP script on my server?

谢谢.

编辑-同样，如何使某些内容显示在发送至"菜单中?我的印象是，您只需将快捷方式放到SendTo文件夹中，它会自动在菜单中添加一个选项... 没关系.我发现自己在做什么错了:)

Edit-- also, how do I make something show up in the Sendto menu? I was under the impression that you just drop a shortcut into the SendTo folder and it automatically adds an option in the menu... Never mind. I figured out what I was doing wrong :)

推荐答案

1. 找出所拖动的文件是什么: http://docs.python. org/library/sys.html#sys.argv
2. 打开它: http://docs.python.org/library/functions.html #open
3. 在以下位置阅读: http://docs.python.org/library/stdtypes.html#file.read
4. 发布它: http://docs.python.org/library/urllib2 .html#urllib2.urlopen

1. Find out what the dragged file was: http://docs.python.org/library/sys.html#sys.argv
2. Open it: http://docs.python.org/library/functions.html#open
3. Read it in: http://docs.python.org/library/stdtypes.html#file.read
4. Post it: http://docs.python.org/library/urllib2.html#urllib2.urlopen

这篇关于在Python中接受文件参数(从“发送到"上下文菜单中)的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！