On Sep 8, 2:16 pm, "Andreas Tawn" <andreas.t...@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. Many thanks -- http://mail.python.org/mailman/listinfo/python-list
