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Programming-Idioms

# 184 Tomorrow
Assign to t a string representing the day, month and year of the day after the current date.
New implementation

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Please try to avoid dependencies to third-party libraries and frameworks.

Other implementations
var t = DateTime.Today.AddDays(1).ToShortDateString();
var now = new DateTime.now();
var t = now.add(new Duration(days: 1));
def main() do
  Date.utc_today()
  |> Date.add(1)
  |> Date.to_string()
end
import "time"
t := time.Now().Add(24 * time.Hour).Format("2006-01-02")
import java.time.LocalDate
LocalDate t = LocalDate.now() + 1
var nextDate = new Date(new Date().getTime() + 24 * 60 * 60 * 1000);
var day = nextDate.getDate()
var month = nextDate.getMonth() + 1
var year = nextDate.getFullYear()
var t = `${day}/${month}/${year}`;
var tomorrow = new Date();
tomorrow.setDate(tomorrow.getDate() + 1);
let t = new Date();
t.setDate(t.getDate() + 1)
t = t.toString()
var now = new Date()
var year = now.getFullYear()
var month = now.getMonth()
var day = now.getDate()

var tomorrow = new Date(0)
tomorrow.setFullYear(year, month, day + 1)
tomorrow.setHours(0, 0, 0, 0)

var shortDateFormat = Intl.DateTimeFormat(undefined, { dateStyle: "short" })
var t = shortDateFormat.format(tomorrow)
import java.time.LocalDate;
String t = LocalDate.now().plusDays(1).toString();
import java.time.LocalDate
val t = LocalDate.now().plusDays(1).toString()
(ql:quickload :chronicity)
(setf _t (chronicity:parse "tomorrow"))
$t = date('Y-m-d',strtotime('+1 day'));
$d = new DateTime('tomorrow');
$t = $d->format('Y-m-d');
uses SysUtils;
t := DateToStr(Now+1.0);
use DateTime qw();
my $dt = DateTime->today;
$dt->add(days => 1);
my $t = $dt->strftime('%F');
use DateTime;
$t = DateTime->today->add(days => 1)->ymd;
from datetime import date, timedelta
t = str(date.today() + timedelta(days=1))
from datetime import date, timedelta
t = str(date.today() + timedelta(1))
require "date"
t = (Date.today + 1).to_s
require "date"
t = Date.tomorrow.to_s
require 'active_suport'
t = 1.day.since.to_s
let t = chrono::Utc::now().date().succ().to_string();
t := Date tomorrow asString.
Dim t As Date = Date.Today.AddDays(1)