I want to add one day to a date and store it in a variable.
From GUI we are passing value (last day of the month)to $t_date.
This $t_date will give me the value like this %Y%m%d 20150531.
Now I want to add one day to this value and store it in a variable "datein".
datein should give me the first day of the next month i.e. 20150601.
I tried the below script it 's not working
Code:
#!/bin/ksh
echo "Initialization"
echo "str Date : $t_date"
echo "str Date = First day of the month"
datein=`$t_date '+%Y%m%d'`
datein=$((datein +1))
echo "$datein"
Hi,
I have a date input in MMDDYYYY format..
I have to give the day (whether that DD is sunday/monday...)
Is there any command for it...
Or do I have to write a script for that...
Thanks in Advance
Yeheya (1 Reply)
Hi, I have been trying just about every unix command to come up with yesterday's date (today's date - 1). I have seen all of the help on this forum, and none of it seems to work for me here. We are using Sun Solaris 9 Unix. I am using this script to create a .txt file with ftp commands that I will... (2 Replies)
It's easy as pie to get the date minus one day on opensolaris:
date -d "-1 day" +"%Y%m%d"run this command on our crappy Solaris 10 machines however (which I'm guessing doesn't have GNU date running on it) and you get:
date: illegal option -- d
date: illegal option -- 1
date: illegal option --... (5 Replies)
Hi guys,
I had a scenario...
1. I had to get the previous days date in yyyymmdd format
2. i had to create a file with Date inthe format yyyymmdd.txt format
both are different
thanks guys in advance.. (4 Replies)
Hi! I am trying to read a file and every line has a specific date as one of its fields.
I want to take that date and compare it to the date today plus 6 days.
while read line
do
date=substr($line, $datepos, 8) #date is expected to be YYYYMMDD
if ; then
...proceed commands
... (1 Reply)
Hi,
I want to add some hours and minutes to the current date. For example, if the current date is "July 16, 2012 15:20", i want to add 5 hours 30 minutes to "July 16, 2012 00:00" not to "July 16, 2012 15:20". Please help.
Thanks! (4 Replies)
I need to get the next day's date of the user entered date
for example:
Enter date (yyyy/mm/yy):
2013/10/08I need to get the next day's date of the user entered date
Desired Output:
2013/10/09Though there are ways to achieve this is Linux or Unix environment (date command) ,I need to... (1 Reply)
Hi,
I have a scenario like this.
I get a file on anyday of the week, so the file name is like this.
FILE_NAME_YYYYMMDD
I want to get the tuesday date before this day.
For example
FILE_NAME_20180413 I should get the date as 20180410
FILE_NAME_20180404 I should get the date as 20180403... (5 Replies)
I Have text like
XXX_20190908.csv.gz need to replace Only date in this format with current date every day
Thanks! (1 Reply)
Discussion started by: yamasani1991
1 Replies
LEARN ABOUT PHP
cal_from_jd
CAL_FROM_JD(3) 1 CAL_FROM_JD(3)cal_from_jd - Converts from Julian Day Count to a supported calendarSYNOPSIS
array cal_from_jd (int $jd, int $calendar)
DESCRIPTION cal_from_jd(3) converts the Julian day given in $jd into a date of the specified $calendar. Supported $calendar values are CAL_GREGORIAN,
CAL_JULIAN, CAL_JEWISH and CAL_FRENCH.
PARAMETERS
o $jd
- Julian day as integer
o $calendar
- Calendar to convert to
RETURN VALUES
Returns an array containing calendar information like month, day, year, day of week, abbreviated and full names of weekday and month and
the date in string form "month/day/year".
EXAMPLES
Example #1
cal_from_jd(3) example
<?php
$today = unixtojd(mktime(0, 0, 0, 8, 16, 2003));
print_r(cal_from_jd($today, CAL_GREGORIAN));
?>
The above example will output:
Array
(
[date] => 8/16/2003
[month] => 8
[day] => 16
[year] => 2003
[dow] => 6
[abbrevdayname] => Sat
[dayname] => Saturday
[abbrevmonth] => Aug
[monthname] => August
)
SEE ALSO cal_to_jd(3), jdtofrench(3), jdtogregorian(3), jdtojewish(3), jdtojulian(3), jdtounix(3).
PHP Documentation Group CAL_FROM_JD(3)