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= 如何从后向前匹配正则？ =
{{{Mr Shore <shore.cloud@gmail.com>
reply-to	python-cn@googlegroups.com,
to	python-cn@googlegroups.com,
date	Wed, Mar 19, 2008 at 3:37 PM
subject	[CPyUG:/] Re: 如何指定正则匹配从后往前？
}}}	
算了 直接贴代码了 很简单的东西 越说越糊涂了

{{{#!python
import time
import re
import os
import stat
#pattern = "(.*),\'([^,]+)\'\.split\(\'\|\'\),0,{}\)\)"
#pattern = "(.*)(\')([^\']+)(\'\.split\(\'\|\'\),0,{}\)\))\t+$"
pattern = "\'([^\']+)\'\.split\(\'\|\'\),0,{}\)\)\t*$"
regex = re.compile(pattern, re.DOTALL|re.VERBOSE|re.IGNORECASE)


def mapper(match):
    tmp = ''

    if 'IRN' in match.group(3):
        tmp = match.group(3).replace('IRN','test')
    else:
        tmp = match.group(3)
    return "%s%s%s%s" %(match.group(1), match.group(2), tmp, match.group(4))



s = open(r'D:/work/IR/python/compressjs/files.txt').read().split('\n')
p = r'D:/work/IR/php/src'
r = []

start = time.time()
for each in s:
    print each
    fp = p + each
    ss = open(fp).read()
    ssl = len(ss)
    #ss = regex.sub(mapper,ss)
    ss_sub = regex.findall(ss)
    if len(ss_sub) == 0:
        continue
    ss_sub = ss_sub[0]
    l = len(ss_sub)
    post = ss_sub.replace('IRN','test')
    ss_sub = ss[:ssl - l -2 - len("'.split('|'),0,{}))")] + "'" + post + "'.split('|'),0,{}))"
   
    print ss_sub
    #sos.chmod(fp,stat.S_IWRITE)
    #file(fp,'w').write(ss)
    #r.append(ss)
print '%s' %(str(time.time() - start),)

"""
s = open(r'D:/work/IR/php/src/Client/Javascript/Body_min.js').read()
r = regex.sub(mapper,s)
"""
}}}


== 反馈 ==
是这样 最后没用re.sub，而是用了re.findall
 * 用re.sub的话正则是`pattern = "(.*)(\')([^\']+)(\'\.split\(\'\|\'\),0,{}\)\))\t+$"`
 * 而用re.findall的话`pattern = "\'([^\']+)\'\.split\(\'\|\'\),0,{}\)\)\t*$"`
 * 少了个`.*` 而这个正是杀手
也可能是re.sub用的不熟 绕了远