Welcome to OGeek Q&A Community for programmer and developer-Open, Learning and Share
Welcome To Ask or Share your Answers For Others

Categories

0 votes
121 views
in Technique[技术] by (71.8m points)

python - Parsing apache log files

I just started learning Python and would like to read an Apache log file and put parts of each line into different lists.

line from the file

172.16.0.3 - - [25/Sep/2002:14:04:19 +0200] "GET / HTTP/1.1" 401 - "" "Mozilla/5.0 (X11; U; Linux i686; en-US; rv:1.1) Gecko/20020827"

according to Apache website the format is

%h %l %u %t "%r" %>s %b "%{Referer}i" "%{User-Agent}i

I'm able to open the file and just read it as it is but I don't know how to make it read in that format so I can put each part in a list.

See Question&Answers more detail:os

与恶龙缠斗过久,自身亦成为恶龙;凝视深渊过久,深渊将回以凝视…
Welcome To Ask or Share your Answers For Others

1 Reply

0 votes
by (71.8m points)

This is a job for regular expressions.

For example:

line = '172.16.0.3 - - [25/Sep/2002:14:04:19 +0200] "GET / HTTP/1.1" 401 - "" "Mozilla/5.0 (X11; U; Linux i686; en-US; rv:1.1) Gecko/20020827"'
regex = '([(d.)]+) - - [(.*?)] "(.*?)" (d+) - "(.*?)" "(.*?)"'

import re
print re.match(regex, line).groups()

The output would be a tuple with 6 pieces of information from the line (specifically, the groups within parentheses in that pattern):

('172.16.0.3', '25/Sep/2002:14:04:19 +0200', 'GET / HTTP/1.1', '401', '', 'Mozilla/5.0 (X11; U; Linux i686; en-US; rv:1.1) Gecko/20020827')

与恶龙缠斗过久,自身亦成为恶龙;凝视深渊过久,深渊将回以凝视…
OGeek|极客中国-欢迎来到极客的世界,一个免费开放的程序员编程交流平台!开放,进步,分享!让技术改变生活,让极客改变未来! Welcome to OGeek Q&A Community for programmer and developer-Open, Learning and Share
Click Here to Ask a Question

...