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How do I Compare Dates

by Anonymous Monk
on Jun 26, 2000 at 21:13 UTC ( #19878=categorized question: print w/replies, xml ) Need Help??
Contributed by Anonymous Monk on Jun 26, 2000 at 21:13 UTC
Q&A  > dates and times


given string representation of dates in the format dd-MMM-yyyy, coming from a database.

Answer: How do I Compare Dates
contributed by httptech

Date::Manip is particularly good at this. Look at the DateCalc() method.

Alternatively, since you're getting the values from a database, see if you can get it to return the values in the format yyyy-mm-dd; then a simple string comparison will do. Or - heck - get it in nnnnnnnnn format, a number representing seconds since the epoch. Then a simple numeric comparison will do!

Answer: How do I Compare Dates
contributed by tenatious

If the date you want to compare is in the same format as that you are pulling from the database, you might be able to perform a SELECT statement on it and let the database tell you which date is younger (or older).

This method requires a connection to a database, and might not be the fastest way, but if you are designing for portability, would require less packages be installed than going whole hog and installing Date::Manip.

You could also just compare the dates manually. I.e:

($d,$m,$y) = split /-/, $date; $dv1 = $d+31*$m+365*$y; ($d,$m,$y) = split /-/, $date_from_db; $dv2 = $d+31*$m+365*$y; if ($dv1 < $dv2) { something; ... ... done; }
Answer: How do I Compare Dates
contributed by Anonymous Monk

Hello. For what it's worth, I wrote the following to be able to determine whether date1 is within 3 days of date2 etc. I didn't know about Date::Manip.

sub abs_day { # this subroutine accepts a month, day, and two-digit year in # mm/dd/yy numerical form, e.g., 3/23/04. # it then calculates the "absolute day", # i.e., the number of days # since day 0, which is the turn of the 21st century. # so, 1/10/2000 is absolute day 10; # 1/10/2001 is absolute day 376 # (because year 2000 was a leap year). # the routine deals with leap years by # multiplying years by 365.25, and then rounding down. my $month = shift(@_); my $day = shift(@_); my $year = shift(@_); if ($debug) { print "i'm in abs_day and month is $month,", " day is $day, year is $year\n"; } my $abs_day; undef($abs_day); # so each year has 365.25 days # years are of the form 00, 01, 02, etc. $abs_day = $year * 365.25; # add .75 to account for fact that year 00 is a leap year $abs_day = $abs_day + .75; if ($debug) { print "abs_day is $abs_day after adding .75 to yr*365\n"; } if ($month == 1) #jan { $abs_day = $abs_day + 0; } if ($month == 2) #feb { $abs_day = $abs_day + 31; } # starting in march, we add .25 to number of days # to account for leap year if ($month == 3) #mar { $abs_day = $abs_day + 59.25; } if ($month == 4) #apr { $abs_day = $abs_day + 90.25; } if ($month == 5) #may { $abs_day = $abs_day + 120.25; } if ($month == 6) #jun { $abs_day = $abs_day + 151.25; } if ($month == 7) #jul { $abs_day = $abs_day + 181.25; } if ($month == 8) #aug { $abs_day = $abs_day + 212.25; } if ($month == 9) #sep { } if ($month == 11) #nov { $abs_day = $abs_day + 304.25; } if ($month == 12) #dec { $abs_day = $abs_day + 334.25; } $abs_day = $abs_day + $day; if ($debug) { print "now that we have added year, month, and", " day, abs_day is $abs_day\n"; } $abs_day = sprintf "%d", $abs_day; if ($debug) { print "now that we have converted from float to integer,", " abs_day is $abs_day\n"; } return $abs_day; }

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