python利用正则表达式提取字符串

import re
str = "a123b"
print re.findall(r"a(.+?)b",str)#
输出['123']

import re
str = "a123b456b"

print re.findall(r"a(.+?)b", str)
#输出['123']#?控制只匹配0或1个,所以只会输出和最近的b之间的匹配情况

print re.findall(r"a(.+)b", str)
#输出['123b456']

print re.findall(r"a(.*)b", str)
#输出['123b456']

str = "a23b\na34b"

re.findall(r"a(\d+)b.+a(\d+)b", str)
#输出[]
#因为不能处理str中间有\n换行的情况

re.findall(r"a(\d+)b.+a(\d+)b", str, re.S)
#s输出[('23', '34')]

str = "a23b\na34b"

re.findall(r"^a(\d+)b", str)
#输出['23']

re.findall(r"^a(\d+)b", str, re.M)
#输出['23', '34']

import re
line ='192.168.0.1 25/Oct/2012:14:46:34 "GET /api HTTP/1.1" 200 44 "http://abc.com/search" 
"Mozilla/5.0"'
reg = re.compile('^(?P<remote_ip>[^ ]*) (?P<date>[^ ]*) "(?P<request>[^"]*)" 
(?P<status>[^ ]*) (?P<size>[^ ]*) "(?P<referrer>[^"]*)" "(?P<user_agent>[^"]*)"')
regMatch = reg.match(line)
linebits = regMatch.groupdict()
print linebits
for k, v in linebits.items() :
 print k+": "+v

status: 200
referrer: 
request: GET /api HTTP/1.1
user_agent: Mozilla/5.0
date: 25/Oct/2012:14:46:34size: 44
remote_ip: 192.168.0.1