Hi All,
I want to find the previous day's date and store that in a variable, which will be usuful for further processing.
Any help please.
Regards,
raju (4 Replies)
Hello,
I wanted to display the month for previous day date. Like, today date is 18-Nov-2008. So the previous date is 17-Nov-2008. The output should be November.
If the today date is 1-DEC-2008, then output should be NOVEMBER.
If the today date is 1-JAN-2008, then output should be DECEMBER.... (4 Replies)
I have a problem of Finding Day of the week from date, but i need to do it within awk On SOLARIS
Input:20101007(YYYYMMDD)
Output:Thursday
kindly provide suggestions.
Thanks in advance (8 Replies)
hi there
I have file names in different format as below
triss_20111117_fxcb.csv
triss_fxcb_20111117.csv
xpnl_hypo_reu_miplvdone_11172011.csv
xpnl_hypo_reu_miplvdone_11-17-2011.csv
xpnl_hypo_reu_miplvdone_20111117.csv
xpnl_hypo_reu_miplvdone_20111117xfb.csv... (10 Replies)
Dear all,
I have 2 questions.
I have a file with many rows which has date of the format YYYYMMDD.
1. I need to change the date to that weeks friday date(Ex: 20120716(monday) to 20120720). Satuday/Sunday has to be changed to next week friday date too.
2. After converting the date to... (10 Replies)
Ok, the title is confusing i know, but it is a weird question.
I have a bash script running on Centos5.8 and want to find a better way to do some date manipulation. What i am trying to do is get 3 days of files (day before, that day, and day after), concatenate the three files and pass them on... (2 Replies)
I have requirment to get last date of previous month and the first date of previous 4th month:
Example:
Current date: 20130320 (yyyymmdd)
Last date of previous month: 20130228 (yyyymmdd)
First date of previous 4th month: 20121101 (yyyymmdd)
In my shell --date, -d, -v switches are not... (3 Replies)
Hi Experts,
i am using the below code get the date of previous day.
#!/usr/bin/ksh
datestamp=`date '+%Y%m%d'`
yest=$((datestamp -1))
echo $yest
When i execute the code i am getting output as:
20130715
What i am trying here is, based on the date passed i am fetching previus day's... (0 Replies)
Legends,
i need to get previous day using date command.
Can you please help.
sdosanjh:/home> date +%m%d%y
011514
i tried -d '-1 day' but it is not working (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
datetime.sub
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)