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Top Forums UNIX for Beginners Questions & Answers Add parameter in UNIX from date Post 303005234 by Nileshta on Monday 16th of October 2017 05:31:56 AM
Old 10-16-2017
Hi Rudic...Thanks for your help. I am using below code to get the number of days.
Code:
today_date=$(date +%d/%m/%Y) mm=$5 dd=$(date -d"$mm" +%d/%m/%Y) echo [[ $(( ($today_date - $dd) / (60 * 60 * 24) ))]]

. So I am getting below error " value too great for base (error token is "08")" I am having Todays Date: 16/10/2017 and Date From : 11/08/2017. I tried with adding 10# to the both the dates but still getting same error. Thanks in advance

Last edited by Scott; 10-16-2017 at 06:41 AM.. Reason: Added code tags (did not reformat your code)
 

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DATETIME.ADD(3) 							 1							   DATETIME.ADD(3)

DateTime::add - Adds an amount of days, months, years, hours, minutes and seconds to a DateTime object

       Object oriented style

SYNOPSIS
public DateTime DateTime::add (DateInterval $interval) DESCRIPTION
Procedural style DateTime date_add (DateTime $object, DateInterval $interval) Adds the specified DateInterval object to the specified DateTime object. PARAMETERS
o $object -Procedural style only: A DateTime object returned by date_create(3). The function modifies this object. o $interval - A DateInterval object RETURN VALUES
Returns the DateTime object for method chaining or FALSE on failure. EXAMPLES
Example #1 DateTime.add(3) example Object oriented style <?php $date = new DateTime('2000-01-01'); $date->add(new DateInterval('P10D')); echo $date->format('Y-m-d') . " "; ?> Procedural style <?php $date = date_create('2000-01-01'); date_add($date, date_interval_create_from_date_string('10 days')); echo date_format($date, 'Y-m-d'); ?> The above examples will output: 2000-01-11 Example #2 Further DateTime.add(3) examples <?php $date = new DateTime('2000-01-01'); $date->add(new DateInterval('PT10H30S')); echo $date->format('Y-m-d H:i:s') . " "; $date = new DateTime('2000-01-01'); $date->add(new DateInterval('P7Y5M4DT4H3M2S')); echo $date->format('Y-m-d H:i:s') . " "; ?> The above example will output: 2000-01-01 10:00:30 2007-06-05 04:03:02 Example #3 Beware when adding months <?php $date = new DateTime('2000-12-31'); $interval = new DateInterval('P1M'); $date->add($interval); echo $date->format('Y-m-d') . " "; $date->add($interval); echo $date->format('Y-m-d') . " "; ?> The above example will output: 2001-01-31 2001-03-03 NOTES
DateTime.modify(3) is an alternative when using PHP 5.2. SEE ALSO
DateTime.sub(3), DateTime.diff(3), DateTime.modify(3). PHP Documentation Group DATETIME.ADD(3)
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