Extracting patterns after matching a regex

[nn]

On Sep 8, 11:19 am, Dave Angel <da... at ieee.org> wrote:
> Mart. wrote:
> > <snip>
> > I have been doing this to turn the email into a string
>
> > email =ys.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',
> > <snip>
>
> The mistake I see is trying to turn a list into a string, just so you
> can try to parse it back again.  Just write a loop that iterates through
> the list that readlines() returns.
>
> DaveA

No kidding.

Instead of this:
s = str(f.readlines())

ftphost = re.search(r'FTPHOST: (.*?)\\r',s).group(1)
ftpdir  = re.search(r'FTPDIR: (.*?)\\r',s).group(1)
url = 'ftp://' + ftphost + ftpdir


I would have possibly done something like this (not tested):
lines = f.readlines()
header={}
for row in lines:
    key,sep,value = row.partition(':')[2].rstrip()
    header[key.lower()]=value
url = 'ftp://' + header['ftphost'] + header['ftpdir']