I think you can do this:
a=[]
b=redata.split('::')
for e in b:
if e.find('@') != -1:
a.append(e.split('@')[1])
list a includes all the domain
在 2012年4月9日 上午5:26，Wayne Werner <[email protected]>写道：
> On Fri, 6 Apr 2012, Khalid Al-Ghamdi wrote:
>
> hi all,
>> I'm trying to extract the domain in the following string. Why doesn't my
>> pattern (patt) work:
>>
>> >>> redata
>> 'Tue Jan 14 00:43:21 2020::[email protected]::**1578951801-6-10 Sat
>> Jul 31 15:17:39 1993::[email protected]::**744121059-5-6 Mon Sep 21
>> 20:22:37
>> 1987::[email protected]::**559243357-6-7 Fri Aug 2 07:15:23
>> 1991::[email protected]::**681106523-4-9 Mon Mar 18 19:59:47
>> 2024::[email protected]::**1710781187-6-7 '
>> >>> patt=r'\w+\.\w{3}(?<=@)'
>> >>> re.findall(patt,redata)
>> []
>>
>> This pattern works but the first should, too. shouldn't it?
>>
>
> The all too familiar quote looks like it applies here: "Often programmers,
> when faced with a problem, think 'Aha! I'll use a regex!'. Now you have two
> problems."
>
> It looks like you could easily split this string with redata.split('::')
> and then look at every second element in the list and split *that* element
> on the last '.' in the string.
>
> With data as well-formed as this, regex is probably overkill.
>
> HTH,
> Wayne
>
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