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Full Discussion: Date issue
Top Forums UNIX for Beginners Questions & Answers Date issue Post 303039296 by vedanta on Sunday 29th of September 2019 05:25:07 PM
Old 09-29-2019
I have made some correction. Please let me know if there is a better way to do it.


Code:
unset date_to
unset date_from
unset counter=15
date_to=$(date -d 2018-06-15 +"%Y%m%d")    
todate=$(date -d 2018-08-18 +"%Y%m%d")  
counter=0
while [ $date_to -lt $todate ]; #put the loop where you need it
do
 date_from=$(date -d "$date_to +1 days" '+%Y%m%d') 
 date_to=$(date -d "$date_from +$counter days" '+%Y%m%d')
 echo "date from"$date_from
 echo "date to" $date_to 
 ls -lrt $(find . -type f -newermt $date_from \! -newermt $date_to )
done

 

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DATETIME.SUB(3) 							 1							   DATETIME.SUB(3)

DateTime::sub - Subtracts an amount of days, months, years, hours, minutes and seconds from a DateTime object

       Object oriented style

SYNOPSIS
public DateTime DateTime::sub (DateInterval $interval) DESCRIPTION
Procedural style DateTime date_sub (DateTime $object, DateInterval $interval) Subtracts the specified DateInterval object from 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.sub(3) example Object oriented style <?php $date = new DateTime('2000-01-20'); $date->sub(new DateInterval('P10D')); echo $date->format('Y-m-d') . " "; ?> Procedural style <?php $date = date_create('2000-01-20'); date_sub($date, date_interval_create_from_date_string('10 days')); echo date_format($date, 'Y-m-d'); ?> The above examples will output: 2000-01-10 Example #2 Further DateTime.sub(3) examples <?php $date = new DateTime('2000-01-20'); $date->sub(new DateInterval('PT10H30S')); echo $date->format('Y-m-d H:i:s') . " "; $date = new DateTime('2000-01-20'); $date->sub(new DateInterval('P7Y5M4DT4H3M2S')); echo $date->format('Y-m-d H:i:s') . " "; ?> The above example will output: 2000-01-19 13:59:30 1992-08-15 19:56:58 Example #3 Beware when subtracting months <?php $date = new DateTime('2001-04-30'); $interval = new DateInterval('P1M'); $date->sub($interval); echo $date->format('Y-m-d') . " "; $date->sub($interval); echo $date->format('Y-m-d') . " "; ?> The above example will output: 2001-03-30 2001-03-02 NOTES
DateTime.modify(3) is an alternative when using PHP 5.2. SEE ALSO
DateTime.add(3), DateTime.diff(3), DateTime.modify(3). PHP Documentation Group DATETIME.SUB(3)
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