Date after 5 days from current date in YYYYMMDD format

Shell Programming and Scripting
1

Hello Experts,

How do i get date after 5 days from current date in YYYYMMDD format?

How do you compare date in YYYYMMDD format?

Thanks

2

You would have to write a script which takes in account for instance like today:
we are month=09 day=29
If you add 5 to 29, you are a bit out of day range... ( you would have to reset day (+4) and add 1 to month value...)

---------- Post updated at 17:50 ---------- Previous update was at 17:31 ----------

Some clues:

# with this date format month position is 99 to 9999 and day 9 to 99 so:

ant:/home/vbe $ IN5DAYS=$(expr $(date +%Y%m%d) + 105);echo $IN5DAYS 
20101034      # 34 days sep had 30 so:
ant:/home/vbe $ IN5DAYS=$(expr $IN5DAYS - 30);echo $IN5DAYS        
20101004
octant:/home/vbe $
3

If you have GNU date

date -d "5 days"
4

Waiting to see the script you will have to write! (nice algorithm...)

5

The following DIBOL program is a re hash of a program published in Computer World Newspaper in the early 1970's. The original was written in Fortran
It converts a date in the format yyyymmdd into the number days that have elapsed since Jan 1 1900.
It also converts a number representing the number of days since Jan 1 1900 to a date.

All arithmetic is integer, so 8/3 = 2

#cat xdatex.dbl
.SUBROUTINE XDATEX
DATEXREF,    A
RECORD DATEREC
           CLDRDATE, D8
               CLDRYEAR, D4 @CLDRDATE 
               CLDRMTH,  D2 @CLDRDATE +4    
               CLDRDAY , D2 @CLDRDATE +6
           DAYS1901, D10            
           DAYOFWK,  D1
           NAMOFDAY, A9      
           NAMOFMTH, A9      
RECORD 
        CALCYEAR , D10 
        CALCMTH ,D10
        CALCDAY , D10 
        DYYR , D10 
        DYWK , D10
        DYMO , D10 
        TEMPYEAR , D10 
        TEMPDAY , D10 
        LEAPYEAR , D10
        HOLDDATE , D8   
        M, A8
        DAYDATA, 7A9, 'SUNDAY   ','MONDAY   ','TUESDAY  ','WEDNESDAY'
&,'THURSDAY ','FRIDAY   ','SATURDAY '
        MTHDATA, 12A9, 'JANUARY  ','FEBRUARY ','MARCH   ','APRIL   ',
&'MAY     ','JUNE    ','JULY    ','AUGUST  ',
&'SEPTEMBER','OCTOBER  ','NOVEMBER ','DECEMBER '
PROC
SOJ,
DATEREC=DATEXREF
HOLDDATE=CLDRDATE
IF (DAYS1901 .NE. 0) GO TO CENTOCLD
LEAPYEAR=2
CALCYEAR= CLDRYEAR - 1900
CALCMTH= CLDRMTH
CALCDAY= CLDRDAY
TEMPYEAR= CALCYEAR/4
TEMPYEAR= TEMPYEAR * 4
IF (TEMPYEAR .EQ. CALCYEAR)  LEAPYEAR=1
DYYR= (CALCMTH * 275)/9 + CALCDAY - 30
IF (CALCMTH .GT. 2) DYYR= DYYR -LEAPYEAR
DAYS1901= CALCYEAR - 1
DAYS1901= (DAYS1901 * 1461) / 4 + DYYR
CALL    CENTOCLD
IF (CLDRDATE .NE. HOLDDATE)
BEGIN
DAYS1901= 0
CLDRDATE= 0
DAYOFWK = 0
NAMOFDAY=  
NAMOFMTH=
DATEXREF=DATEREC
RETURN
END
RETURN
CENTOCLD,
CALCYEAR=(DAYS1901/1461) 
CALCYEAR=(DAYS1901 - CALCYEAR + 364)/365
DYYR=((CALCYEAR - 1)*1461)/4
DYYR=DAYS1901-DYYR
LEAPYEAR= 2
        TEMPYEAR= CALCYEAR/4
        TEMPYEAR= TEMPYEAR * 4
IF (TEMPYEAR .EQ. CALCYEAR) LEAPYEAR=1
TEMPDAY= DYYR
TEMPYEAR= 61 - LEAPYEAR 
IF (TEMPDAY .GT. TEMPYEAR) TEMPDAY= TEMPDAY + LEAPYEAR
CALCMTH= (TEMPDAY * 9 + 269) / 275
DYMO= ((CALCMTH * 275) / 9) - 30
CALCDAY= TEMPDAY - DYMO
DYMO=CALCDAY
DYWK=DAYS1901 + 1
DYWK=DYWK-((DYWK/7)*7)+1
DAYOFWK=DYWK
CLDRYEAR = CALCYEAR + 1900
CLDRMTH= CALCMTH
CLDRDAY= CALCDAY
NAMOFDAY=  DAYDATA(DYWK)
NAMOFMTH=MTHDATA(CALCMTH)
DATEXREF=DATEREC
RETURN
#
6

I think this is what you're looking for.

date -d "5 days" +%Y%m%d

What type of comparison are you doing? If you just want to know which dates are before or after, a simple date1 > date2 (or vice versa) will work if the date is formatted in YYYYMMDD format. If you need to know specific amount of time between dates you might have to write a shell script. Let me know which you're trying to do and I'll to help out.

7

Basically, I would like to find last date of next month in YYYYMMDD format and want to find number of days between today and last date of next month.

Thanks

8

Here's one way to do it with Perl -

$
$
$ perl -MDateTime -le '$d = DateTime->now;
                       $start = $d->clone;
                       $d1 = $d->add(months=>1);
                       $end = DateTime->last_day_of_month(year=>$d1->year, month=>$d1->month);
                       print "The date today          = ", $start->ymd("");
                       print "Last date of next month = ", $end->ymd("");
                       print "Number of days left     = ", $start->delta_days($end)->delta_days;
                      '
The date today          = 20100930
Last date of next month = 20101031
Number of days left     = 31
$
$

tyler_durden

9

That sounds different than the original post. This probably isn't the most elegant way to do this, but it works and keeps the dependencies to a minimum. (Just GNU date)

#!/bin/bash

#Get last date
dt=$(($(date +%-m) + 2))
yr=$(date +%Y)
if [ $dt -gt 12 ]; then
        dt=$((1 + ($dt - 12)))
        yr=$(($yr + 1))
fi
value=$(date -d "$yr-$dt-01 -1 day" +%Y%m%d)

echo "Last day of next month is $value"

#Get days between

seconds1=$(date -d $value +%s)
seconds2=$(date +%s)
elapsed=$(($seconds1 - $seconds2))
secondsperday=86400
numdays=$(($elapsed / $secondsperday))
echo "Number of days between is $numdays"