<[EMAIL PROTECTED]> wrote: ... > 1511156 As Requested 2000 adv server -2000AdvSrv.vol001+02.PAR2 (1/4) - > 21/27 > 1511157 As Requested 2000 adv server -2000AdvSrv.vol001+02.PAR2 (2/4) - > 21/27 ... > would be to look for (1/ in the subject string then find the > denominator and loop thru as many times as the denominator to create > the <segment> part of the nzb file. Is this the best way or is there an > easier method? Also what would be the best way to search for the (1/ > using string searches or RegExp? If REgExp could someone provide me > with the RegExp for searching for this string?
If the only thing that identifies these posts as part of one logical post is that (n/m) in the subject then I guess there's nothing for it but analysis of the subject, and a re seems appropriate. import re re_n_of_m = re.compile(r'(.*)\s*\((\d+)/(\d+)\)') might be one approach. Then, somewhere in your loop, mo = re_n_of_m.search(subject) sets mo to None if it doesn't match the pattern; if it does match, then mo is a match object with three groups. mo.group(1) is the part of the subject before the (n/m) marker; mo.group(2) is n as a string; mo.group(3) is n as a string. You can use mo.group(1) as the key into a dictionary where you collect (n,m) pairs as values, so that once you've collected all posts you can tell which one are multipart and also check for inconsistency (different m values), duplicates, missing parts. What you need to do in each case, I don't know... Alex -- http://mail.python.org/mailman/listinfo/python-list
