I am getting a date from environment variable and want to do some processing by subtracting 2 months from the date passed through the environment variable.
Hi i am trying to subtract days from current date. For example todays date is 10/03/2006. If i subtract 2 days it should give 8/03/2006. I am also trying to find the access date of a file in dd/mm/yyyy format. Can any one please help in how to do this.
Ramesh (1 Reply)
I have looked through the forums and found many date / time manipulation tools, but cannot seem to find something that fits my needs for the following.
I have a log file with date time stamps like this:
Jun 21 17:21:52
Jun 21 17:24:56
Jun 21 17:27:59
Jun 21 17:31:03
Jun 21 17:34:07
Jun... (0 Replies)
how can we add or subtract days from the output of date command in unix...
like if i want to subtract a day from the result of date command like this..
v_date=`date +%Y%m%d`
this wud give me 20080519
now i want to subtract one day from this.. so tht it wud give me 20080518..
how do i do... (1 Reply)
Hi All,
I have a requirement in my project where a batch runs on any day of a week. The input files will land to the server every Sunday. I need to read these files based on the current BUS_DAY (which falls any day of the week).
For e.g : if the BUS_DAY is 20120308, we need to derive the... (3 Replies)
Hi Gurus!
I have a static date in a YYYYMMDD format; and I want get the date 2 years in the past and 2 years in the future.
static_date=20010203
old_date=$static_date - 3 years
future_date=$static_date + 2 years
I was only able to research on dates that are current and not on static... (3 Replies)
I got a statement like below to subtract 1 from given date using teradata. I am looking for a one line unix command to perform the same.
select 'parse_this_record', (DATE '${FILE_DATE}' - 1) (FORMAT 'YYYY-MM-DD');
Input: 2012-02-21
Expected Output: 2012-02-20
PS: One liner because I am... (2 Replies)
Hi,
Is there a way to subtract a month or at least 30 days from an arbitrary or user inputted date without the GNU date?
Example:
please Input date: 2011-05-11
then the answer will be 2011-04-11
This code doesn't work:
$date -d "2011-05-11 - 1 month" "+%Y%m%d"
Thanks (7 Replies)
I am trying to achieve to get only the month and the day. Example Feb 5 (as you can see if it is feb 1-9) the space is 2. If it is feb 10-28, the space is only 1. I am trying to right a script that will list a directory and shoot an email if there is an activity in last 7 days. I dont really trust... (5 Replies)
Hi All,
I have a CSV file which is as below. Basically I need to take the year column in it and find if the year is >= 20152 . If that is then I should subtract all values by 6. In the below example in description I am having number mentioned as YYWW so I need to subtract those by -5. Whereever... (8 Replies)
Discussion started by: arunkumar_mca
8 Replies
LEARN ABOUT PHP
datetime.modify
DATETIME.MODIFY(3) 1 DATETIME.MODIFY(3)DateTime::modify - Alters the timestamp
Object oriented style
SYNOPSIS
public DateTime DateTime::modify (string $modify)
DESCRIPTION
Procedural style
DateTime date_modify (DateTime $object, string $modify)
Alter the timestamp of a DateTime object by incrementing or decrementing in a format accepted by strtotime(3).
PARAMETERS
o $object
-Procedural style only: A DateTime object returned by date_create(3). The function modifies this object.
o $modify
-A date/time string. Valid formats are explained in Date and Time Formats.
RETURN VALUES
Returns the DateTime object for method chaining or FALSE on failure.
CHANGELOG
+--------+---------------------------------------------------+
|Version | |
| | |
| | Description |
| | |
+--------+---------------------------------------------------+
| 5.3.6 | |
| | |
| | Absolute date/time statements now take effect. |
| | Previously, only relative parts were used. |
| | |
| 5.3.0 | |
| | |
| | Changed the return value on success from NULL to |
| | DateTime. |
| | |
+--------+---------------------------------------------------+
EXAMPLES
Example #1
DateTime.modify(3) example
Object oriented style
<?php
$date = new DateTime('2006-12-12');
$date->modify('+1 day');
echo $date->format('Y-m-d');
?>
Procedural style
<?php
$date = date_create('2006-12-12');
date_modify($date, '+1 day');
echo date_format($date, 'Y-m-d');
?>
The above examples will output:
2006-12-13
Example #2
Beware when adding or subtracting months
<?php
$date = new DateTime('2000-12-31');
$date->modify('+1 month');
echo $date->format('Y-m-d') . "
";
$date->modify('+1 month');
echo $date->format('Y-m-d') . "
";
?>
The above example will output:
2001-01-31
2001-03-03
SEE ALSO strtotime(3), DateTime.add(3), DateTime.sub(3), DateTime.setDate(3), DateTime.setISODate(3), DateTime.setTime(3), DateTime.setTimes-
tamp(3).
PHP Documentation Group DATETIME.MODIFY(3)