On Mon, Apr 11, 2011 at 10:20 PM, Yuri Slobodyanyuk <yuri.slobodyan...@gmail.com> wrote: > Good day everyone, > I am trying to make this pretty simple regex to work but got stuck, > I'd appreciate your help . > Task: Get current date , then read file of format below, find the line that > matches > the current date of month,month and year and extract the number from such > line. > Here is what I did , but if i run it at 11 April 2011 ... > - regex pattern_match matches nothing > - regex pattern_test matches the line "4141411 Fri 11 11 2011" , logical > as it is the last matching line in year 2011 with the date of 11th. > My question - why regex pattern_match does not match anything and how to > make it match the exact line i want. > Thanks > Yuri > > from datetime import datetime, date, time > import re > today_day = datetime.now() > time_tuple= today_day.timetuple() > pattern_match = re.compile("([0-9])+ +" + "Fri +" + str(time_tuple[1]) + " > +" + str(time_tuple[2]) + " +" + str(time_tuple[0]) + " +") ^^^^^^
This trailing " +" *requires* that the lines have trailing spaces. Do they? Such files typically don't, and your example input doesn't either (although that may be due to email formatting lossage). Cheers, Chris -- http://blog.rebertia.com > hhh = open("file_with_data.data",'r') > pattern_test = re.compile("([0-9]+)" + ".*" + " +" + > str(time_tuple[2]).strip(' \t\n\r') + " +" + str(time_tuple[0]).strip(' > \t\n\r') ) > for nnn in hhh: > if (re.search(pattern_test,nnn)): > print nnn.split()[0] > > 1111111 Fri 4 8 2011 > 2323232 Fri 4 15 2011 > 4343434 Fri 4 22 2011 > 8522298 Fri 4 29 2011 > ......... > 5456678 Fri 10 28 2011 > 5633333 Fri 11 4 2011 > 4141411 Fri 11 11 2011 > 3324444 Fri 11 18 2011 -- http://mail.python.org/mailman/listinfo/python-list