如何在bash从zippyshare下载? [英] How to download in bash from zippyshare?
问题描述
谁能帮助我如何使用bash(服务器在CentOS)下载从zippyshare.com文件?
他们的下载按钮,闪光灯做,所以我不能获取下载链接。
Can anyone help me how to download files from zippyshare.com using bash (server with CentOS)? Their download button made with flash, so i can't retrieve download link.
推荐答案
这是绝对有可能的,但它是一个有点哈克。我一直在使用的wget
做我自己相当一段时间。诀窍在于使服务器认为请求被从浏览器制成,要做到这一点,你需要几件事情:
It's definitely possible, but it's a little hacky. I've been doing it myself for quite some time using wget
. The trick is to make the server think that the request is being made from a browser and to make that happen, you need a couple of things:
- 的下载链接(实际的链接文件)
- 链接访问来源(使用的下载按钮网页)
- Zippyshare会话ID(发现在饼干)
下面是截图说明那里你可以得到每个项目:
Here's a screenshot explaining where you can get each item:
现在打开你的终端,并使用下面的命令(更换必要的项目):
wget <download_link> \
--referer='<referrer>' \
--cookies=off --header "Cookie: JSESSIONID=<session_id>" \
--user-agent='Mozilla/5.0 (Windows NT 6.0) Gecko/20100101 Firefox/14.0.1'
示例:
wget http://www16.zippyshare.com/d/29887835/8895183/hello.txt \
--referer='http://www16.zippyshare.com/v/29887835/file.html' \
--cookies=off --header "Cookie: JSESSIONID=26458C0893BF69F88EB5743D74FE0F8C" \
--user-agent='Mozilla/5.0 (Windows NT 6.0) Gecko/20100101 Firefox/14.0.1'
原来的答复:如何使用wget从托管网站下载
的注:在命令中,它实际上是引用者'而不是'引用'的
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