Extracting patterns after matching a regex

Mart. mdekauwe at gmail.com
Tue Sep 8 16:24:08 CEST 2009


On Sep 8, 3:14 pm, "Andreas Tawn" <andreas.t... at ubisoft.com> wrote:
> > > > Hi,
>
> > > > I need to extract a string after a matching a regular expression. For
> > > > example I have the string...
>
> > > > s = "FTPHOST: e4ftl01u.ecs.nasa.gov"
>
> > > > and once I match "FTPHOST" I would like to extract
> > > > "e4ftl01u.ecs.nasa.gov". I am not sure as to the best approach to the
> > > > problem, I had been trying to match the string using something like
> > > > this:
>
> > > > m = re.findall(r"FTPHOST", s)
>
> > > > But I couldn't then work out how to return the "e4ftl01u.ecs.nasa.gov"
> > > > part. Perhaps I need to find the string and then split it? I had some
> > > > help with a similar problem, but now I don't seem to be able to
> > > > transfer that to this problem!
>
> > > > Thanks in advance for the help,
>
> > > > Martin
>
> > > No need for regex.
>
> > > s = "FTPHOST: e4ftl01u.ecs.nasa.gov"
> > > If "FTPHOST" in s:
> > >     return s[9:]
>
> > > Cheers,
>
> > > Drea
>
> > Sorry perhaps I didn't make it clear enough, so apologies. I only
> > presented the example  s = "FTPHOST: e4ftl01u.ecs.nasa.gov" as I
> > thought this easily encompassed the problem. The solution presented
> > works fine for this i.e. re.search(r'FTPHOST: (.*)',s).group(1). But
> > when I used this on the actual file I am trying to parse I realised it
> > is slightly more complicated as this also pulls out other information,
> > for example it prints
>
> > e4ftl01u.ecs.nasa.gov\r\n', 'FTPDIR: /PullDir/0301872638CySfQB\r\n',
> > 'Ftp Pull Download Links: \r\n', 'ftp://e4ftl01u.ecs.nasa.gov/PullDir/
> > 0301872638CySfQB\r\n', 'Down load ZIP file of packaged order:\r\n',
>
> > etc. So I need to find a way to stop it before the \r
>
> > slicing the string wouldn't work in this scenario as I can envisage a
> > situation where the string lenght increases and I would prefer not to
> > keep having to change the string.
>
> If, as Terry suggested, you do have a tuple of strings and the first element has FTPHOST, then s[0].split(":")[1].strip() will work.

It is an email which contains information before and after the main
section I am interested in, namely...

FINISHED: 09/07/2009 08:42:31

MEDIATYPE: FtpPull
MEDIAFORMAT: FILEFORMAT
FTPHOST: e4ftl01u.ecs.nasa.gov
FTPDIR: /PullDir/0301872638CySfQB
Ftp Pull Download Links:
ftp://e4ftl01u.ecs.nasa.gov/PullDir/0301872638CySfQB
Down load ZIP file of packaged order:
ftp://e4ftl01u.ecs.nasa.gov/PullDir/0301872638CySfQB.zip
FTPEXPR: 09/12/2009 08:42:31
MEDIA 1 of 1
MEDIAID:

I have been doing this to turn the email into a string

email = sys.argv[1]
f = open(email, 'r')
s = str(f.readlines())

so FTPHOST isn't the first element, it is just part of a larger
string. When I turn the email into a string it looks like...

'FINISHED: 09/07/2009 08:42:31\r\n', '\r\n', 'MEDIATYPE: FtpPull\r\n',
'MEDIAFORMAT: FILEFORMAT\r\n', 'FTPHOST: e4ftl01u.ecs.nasa.gov\r\n',
'FTPDIR: /PullDir/0301872638CySfQB\r\n', 'Ftp Pull Download Links: \r
\n', 'ftp://e4ftl01u.ecs.nasa.gov/PullDir/0301872638CySfQB\r\n', 'Down
load ZIP file of packaged order:\r\n',

So not sure splitting it like you suggested works in this case.

Thanks



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