Increment date based on time - date

Question

I need to increment a date by +1d based on the time. This seems to be a bit over my head but I suspect it can be done with perl or maybe even awk. For every timestamp with 0000, I need to increment the day by +1.
input:
15,07/31/2020 1600,1048,31,0
15,07/31/2020 1700,1086,45,0
15,07/31/2020 1800,1092,42,0
15,07/31/2020 1900,1017,20,1
15,07/31/2020 2000,945,0,28
15,07/31/2020 2100,897,0,43
15,07/31/2020 2200,869,0,55
15,07/31/2020 2300,812,0,72
15,07/31/2020 0000,778,0,82
expected output:
15,07/31/2020 1600,1048,31,0
15,07/31/2020 1700,1086,45,0
15,07/31/2020 1800,1092,42,0
15,07/31/2020 1900,1017,20,1
15,07/31/2020 2000,945,0,28
15,07/31/2020 2100,897,0,43
15,07/31/2020 2200,869,0,55
15,07/31/2020 2300,812,0,72
15,08/01/2020 0000,778,0,82
It looks like awk can increment date but my question is how do I only increment the date on the lines with 0000 while still printing out the rest of the lines unaltered?

Answer 1

This Perl solution uses Time::Piece (that is included in the standard distribution).
#!/usr/bin/perl
use strict;
use warnings;
use Time::Piece;
use Time::Seconds 'ONE_DAY';
my $regex = '(\d\d/\d\d/\d\d\d\d) (\d{4})';
while (<>) { # reads from the input file supplied on the CLI
my ($mdy, $time) = /$regex/;
if ($time eq '0000') {
# parse the date ($mdy) and add 1 day
my $date = Time::Piece->strptime($mdy, "%m/%d/%Y") + ONE_DAY;
# convert the Time::Piece object to a string rep
$date = $date->strftime("%m/%d/%Y");
# substitute the new date for the date 1 day earlier
s/$regex/$date $time/;
}
print;
}
Prints
C:\Old_Data\perlp>perl test3.pl f0.txt
15,07/31/2020 1600,1048,31,0
15,07/31/2020 1700,1086,45,0
15,07/31/2020 1800,1092,42,0
15,07/31/2020 1900,1017,20,1
15,07/31/2020 2000,945,0,28
15,07/31/2020 2100,897,0,43
15,07/31/2020 2200,869,0,55
15,07/31/2020 2300,812,0,72
15,08/01/2020 0000,778,0,82

Answer 2

Note: This is prefaced by my top comment. Since this question has been reopened, I can post the answer here.
In order to increment the day, we also need to roll over the month if necessary.
For that, we need to know the number of days in the month.
To know that, we also need to know if the current year is a leap year (i.e. february has either 28 or 29 days). The formula is readily available with a web search.
I'm not sure about the format of your input:
15,07/31/2020 0000,778,0,82
What is the 15? I assume you just pass it through.
What is the 0000,778,0,82? I assume you can just compare the first number [as text] with 0000.
Here's the code in perl. I've coded the entire algorithm without any reliance on installing extra perl packages or use of functions such as mktime, etc. I've tried to annotate it as much as possible:
#!/usr/bin/perl
# code to increment day
master(#ARGV);
exit(0);
sub master
{
my(#days_in_month) = (31,28,31,30,31,30,31,31,30,31,30,31);
# input format:
# 15,07/31/2020 0000,778,0,82
while ($buf = <STDIN>) {
chomp($buf);
# split off date and time
my($date,$time) = split(" ",$buf);
# split up date
my($pre,$mon,$day,$year) = split(m%[,/]%,$date);
# decide if current year is leap year
my($isleap) = 0;
{
# year must be divisible by 4
last if (($year % 4) != 0);
# if year is divisible by 100, it is _not_ a leap year unless ...
if (($year % 100) == 0) {
# ... it is also divisible by 400
last if (($year % 400) != 0);
}
$isleap = 1;
}
# get number of days in february
if ($isleap) {
$days_in_month[1] = 29;
}
else {
$days_in_month[1] = 28;
}
# change only if time is 0000
if ($time =~ /^0000,/) {
# increment day
$day += 1;
# advance month if necessary
if ($day > $days_in_month[$mon - 1]) {
$day = 1;
$mon += 1;
}
# advance year if necessary
if ($mon > 12) {
$mon = 1;
$year += 1;
}
}
$date = sprintf("%2.2d,%2.2d/%2.2d/%4.4d",$pre,$mon,$day,$year);
printf("%s %s\n",$date,$time);
}
}

Answer 3

With GNU awk:
awk '$5=="0000"{
# convert columns to unixtime
unixtime=mktime($4 " " $2 " " $3 " " "00 00 00")
# add one day to unixtime and convert back to month/day/year
nextday=strftime("%m/%d/%Y",unixtime+60*60*24)
# rebuild current row
$0=$1 OFS nextday " " $5 OFS $6 OFS $7 OFS $8
}
{print}' FS='[,/ ]' OFS=',' file
As one line:
awk '$5=="0000"{unixtime=mktime($4 " " $2 " " $3 " " "00 00 00"); nextday=strftime("%m/%d/%Y",unixtime+60*60*24); print $1,nextday " " $5,$6,$7,$8; next}{print}' FS='[,/ ]' OFS=',' file
Output:
15,07/31/2020 1600,1048,31,0
15,07/31/2020 1700,1086,45,0
15,07/31/2020 1800,1092,42,0
15,07/31/2020 1900,1017,20,1
15,07/31/2020 2000,945,0,28
15,07/31/2020 2100,897,0,43
15,07/31/2020 2200,869,0,55
15,07/31/2020 2300,812,0,72
15,08/01/2020 0000,778,0,82
See: 8 Powerful Awk Built-in Variables – FS, OFS, RS, ORS, NR, NF, FILENAME, FNR