I have the date of the file passed into a variable also current date formatted same passed into a separate variable and compare the two with an if statement and statement always comes up false. Even though I verified the dates. Any help would be awesome.
Last edited by coderanger; 09-15-2010 at 01:22 PM..
hi ,
I have two variables both containg dates,
x= `date`
and
y= `date'
their format being -> Fri Nov 12 22:59:50 MST 2004
how do I compare which one is greater.
->Can dates be converted into integer and then compared?
( one lengthy way would be to compare the words one by... (7 Replies)
Hi guys
I have a a variable called check_ts which holds a date value. this date value keeps refreshing every 15 minutes.
I am going to start a cron job 5 minutes after the refresh. I have to check if the current date > 20 min of check_ts. how do i do that.
thanks
ragha (17 Replies)
Hello friends,
I am looking for a script or method that can display all the dates between any 2 given dates.
Input:
Date 1
290109
Date 2
010209
Output:
300109
310109
Please help me. Thanks. :):confused: (2 Replies)
Hi I have yesterday date and todays date stored in two variables.
Today date is stored in variable -- testdate=`date +%m/%d/%Y`
I found the yesterday date and stored in variable -- ydate=$month'/'$day1'/'$year
Now i am trying to find out whether $testdate is less that $ydate.
I am... (6 Replies)
Hi,
I have written the following script which will provide the first date of the 3rd previous month and last date of the 2nd previous month but when I change the month for 03 (March) then I am getting the incorrect results.
Ex : - If today is 1 Jan 2010 then the script should provide the... (1 Reply)
Hi,
I want to compare today's date(DDMMYYYY) with yesterday(DDMMYYYY) from system date,if (today month = yesterday month) then execute alter query else do nothing.One more condition is change of year also i.e today is Jan1 2012 and yesterday is Dec 31 2011.
The above rek i want in Shell... (4 Replies)
Hi
I'm trying to compare the current date (dd-Mmm-yyyy) against a variable that is an extracted date from an sql script. Below is the code:
datenow=`date '+%d-%h-%Y'`
#datenow is the current date in the format dd-Mmm-yyyy
sqlplus $dbuserid/$dbpassword @ $SCRIPT_PATH/business-date.sql >... (3 Replies)
Hi
I am failing to write a script which compares a list of dates in a file with today's date.
OS: Solaris 10
I have a file which has server names & dates, i need to compare the date in this file with today's date, if it is less than today it should print the server name.
code i tried is ... (3 Replies)
Discussion started by: nanz143
3 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)