Solaris 10无date -d选项时，实现Unix时间加减n天的方法求助

Since you can't use the -d flag with Solaris 10's date (even /usr/xpg4/bin/date) and can't install any extra Perl modules or tools, here are two reliable, pre-installed-only methods:

Method 1: Use Perl with Core Modules #

Solaris 10's pre-installed Perl includes Time::Local (a core, always-present module), which we can leverage to handle date arithmetic without extra dependencies.

Example 1: Add/Subtract days to the current date #

Run these one-liners directly in your shell:

# Add 5 days to today
perl -MTime::Local -e '@t = localtime(time + 5*86400); printf "%04d-%02d-%02d\n", $t[5]+1900, $t[4]+1, $t[3]'

# Subtract 3 days from today
perl -MTime::Local -e '@t = localtime(time - 3*86400); printf "%04d-%02d-%02d\n", $t[5]+1900, $t[4]+1, $t[3]'

Example 2: Add/Subtract days to a specific date (format: YYYY-MM-DD) #

Save this as a reusable script (name it date_adjust.pl):

#!/usr/bin/perl
use Time::Local;

die "Usage: $0 <YYYY-MM-DD> <days_to_add_or_subtract>" unless @ARGV == 2;

my ($date_str, $days) = @ARGV;
my ($year, $mon, $mday) = split /-/, $date_str;

# Convert target date to epoch (note: months are 0-based in Time::Local)
my $epoch = timelocal(0, 0, 12, $mday, $mon-1, $year-1900);

# Calculate new epoch by adding/subtracting days (1 day = 86400 seconds)
my $new_epoch = $epoch + $days * 86400;

# Convert back to human-readable date
my ($sec, $min, $hour, $new_mday, $new_mon, $new_year) = localtime($new_epoch);
printf "%04d-%02d-%02d\n", $new_year+1900, $new_mon+1, $new_mday;

Make it executable with chmod +x date_adjust.pl, then use it like so:

# Add 10 days to 2024-05-15
./date_adjust.pl 2024-05-15 10

# Subtract 7 days from 2024-05-15
./date_adjust.pl 2024-05-15 -7

Method 2: Use /usr/xpg4/bin/date with Epoch Arithmetic #

Solaris 10's XPG4 date supports the %s format to output epoch time (seconds since 1970-01-01). We can calculate the epoch for our target date and convert it back to a readable format.

Example 1: Add/Subtract days to today #

# Add 5 days to today
new_epoch=$(( $(/usr/xpg4/bin/date +%s) + 5*86400 ))
/usr/xpg4/bin/date -r $new_epoch +"%Y-%m-%d"

# Subtract 3 days from today
new_epoch=$(( $(/usr/xpg4/bin/date +%s) - 3*86400 ))
/usr/xpg4/bin/date -r $new_epoch +"%Y-%m-%d"

Example 2: Add/Subtract days to a specific date #

To handle a custom input date, combine the epoch approach with the Perl core module trick to convert the input date to epoch first:

# Example: Add 7 days to 2024-05-15
input_date="2024-05-15"
days=7

# Convert input date to epoch using Perl's core Time::Local
epoch=$(perl -MTime::Local -e '($y,$m,$d)=split/-/,shift;print timelocal(0,0,12,$d,$m-1,$y-1900)' "$input_date")
new_epoch=$(( epoch + days*86400 ))
/usr/xpg4/bin/date -r $new_epoch +"%Y-%m-%d"

Both methods stick strictly to pre-installed tools and avoid any external dependencies, so they’ll work within your system’s strict constraints.

内容的提问来源于stack exchange，提问作者JimR