URL-spli的Python脚本

Question

我是python新手，正在学习基础知识。在

我的查询： 我有多个页面作为请求从如下日志文件访问

"GET /img/home/search-user-ico.jpg HTTP/1.1"  
"GET /SpellCheck/am.tlx HTTP/1.1"
"GET /img/plan-comp-nav.jpg HTTP/1.1" 
"GET /ie6.css HTTP/1.1"
"GET /img/portlet/portlet-content-bg.jpg HTTP/1.1"
"GET /SpellCheck/am100k2.clx HTTP/1.1" 
"GET /SpellCheck/am.tlx HTTP/1.1" 

我的问题是我只想要页面上的文件部分， 例如， 让我们把"GET /img/home/search-user-ico.jpg HTTP/1.1" ,"GET /ie6.css HTTP/1.1"看作一个页面，然后从上面我要拆分search-user-ico.jpg HTTP，ie6.css HTTP。在

所以请专家们帮我写python脚本，让上面的代码分开。在

Answer 1

data = [
"GET /img/home/search-user-ico.jpg HTTP/1.1",
"GET /SpellCheck/am.tlx HTTP/1.1",
"GET /img/plan-comp-nav.jpg HTTP/1.1" ,
"GET /ie6.css HTTP/1.1",
"GET /img/portlet/portlet-content-bg.jpg HTTP/1.1",
"GET /SpellCheck/am100k2.clx HTTP/1.1" ,
"GET /SpellCheck/am.tlx HTTP/1.1" 
]

for url in data:
    print url.split(' ')[1].split('/')[-2]

Answer 2

data = [
"GET /img/home/search-user-ico.jpg HTTP/1.1",
"GET /SpellCheck/am.tlx HTTP/1.1",
"GET /img/plan-comp-nav.jpg HTTP/1.1" ,
"GET /ie6.css HTTP/1.1",
"GET /img/portlet/portlet-content-bg.jpg HTTP/1.1",
"GET /SpellCheck/am100k2.clx HTTP/1.1" ,
"GET /SpellCheck/am.tlx HTTP/1.1" 
]

for url in data:
    print url.split(' ')[1].split('/')[-1]

Answer 3

假设您的文件名中没有空格，并且您不希望在末尾使用“HTTP”。在

你可以按空格分隔这条线。在

parts = line.split(" ")

然后使用os模块从路径中获取文件名。在

^{pr2}$

例如。在

>>> line = "GET /img/home/search-user-ico.jpg HTTP/1.1"
>>> parts = line.split(" ")
>>> parts[1]
'/img/home/search-user-ico.jpg'
>>> os.path.basename(parts[1])
'search-user-ico.jpg'