You would have to write a script which takes in account for instance like today:
we are month=09 day=29
If you add 5 to 29, you are a bit out of day range... ( you would have to reset day (+4) and add 1 to month value...)
---------- Post updated at 17:50 ---------- Previous update was at 17:31 ----------
Some clues:
# with this date format month position is 99 to 9999 and day 9 to 99 so:
hii all.
I have to get the date of the 7th day past from the current date.
if i give the current date as sep 3 then i must get the date as 27th of august.
can we get the values from the "cal" command.
cal | awk '{print $2}' will this type of command work.
actually my need is
if today is... (17 Replies)
Hi,
Anybody knows how to get what date was 28 days ago of the current system date through UNIX script.
Ex : - If today is 28th Mar 2010 then I have to delete the files which arrived on 1st Mar 2010, (15 Replies)
Hi i am writing a cron job.
so for it i need the 60 days old date form current date in variable.
Like today date is 27 jan 2011 then output value will be stote in variable in formet Nov 27.
i am using EST date, and tried lot of solution and see lot of post but it did not helpful for me. so... (3 Replies)
I am trying to find out the number of days between the current date and user defined date.
I took reference from here for the date2jd() function.
Modified the function according to my requirement. But its not working properly.
Original code from here is working fine.
#!/bin/sh... (1 Reply)
Well guys,
I know the right syntax for displaying the current date is $(date). However, I am planning to send emails to some customers which displays their subscription date, and then the expiry. The expiry being 30 days from the current date.
What would the right syntax be? (6 Replies)
hi all..
i want 2 know how 2 find 7days past date from current date..
when i used set datetime = `date '+%m%d%y'` i got 060613..
i just want to know hw to get 053013..
i tried using date functions but couldnt get it :( i use c shell and there is no chance that i can change that ..... (3 Replies)
I have to display only those subscribers which are in "unconnected state" and the date is 90 days older than today's date.
Below command is used for this purpose:
cat vfsubscriber_20170817.csv | sed -e 's/^"//' -e '1d' | nawk -F '",' '{if ( (substr($11,2,4) == 2017) && ( substr($11,2,8) -lt... (1 Reply)
current date command runs well
awk -v t="$(date +%Y-%m-%d)" -F "'" '$1 < t' myname.dat
subtract 30 days fails
awk -v t="$(date --date="-30days" +%Y-%m-%d)" -F "'" '$1 < t' myname.dat
awk command in hp unix subtract 30 days automatically from current date without date illegal option error... (20 Replies)
Discussion started by: kmarcus
20 Replies
LEARN ABOUT PHP
gmmktime
GMMKTIME(3) 1 GMMKTIME(3)gmmktime - Get Unix timestamp for a GMT dateSYNOPSIS
int gmmktime ([int $hour = gmdate("H")], [int $minute = gmdate("i")], [int $second = gmdate("s")], [int $month = gmdate("n")], [int
$day = gmdate("j")], [int $year = gmdate("Y")], [int $is_dst = -1])
DESCRIPTION
Identical to mktime(3) except the passed parameters represents a GMT date. gmmktime(3) internally uses mktime(3) so only times valid in
derived local time can be used.
Like mktime(3), arguments may be left out in order from right to left, with any omitted arguments being set to the current corresponding
GMT value.
PARAMETERS
o $hour
- The number of the hour relative to the start of the day determined by $month, $day and $year. Negative values reference the hour
before midnight of the day in question. Values greater than 23 reference the appropriate hour in the following day(s).
o $minute
- The number of the minute relative to the start of the $hour. Negative values reference the minute in the previous hour. Values
greater than 59 reference the appropriate minute in the following hour(s).
o $second
- The number of seconds relative to the start of the $minute. Negative values reference the second in the previous minute. Values
greater than 59 reference the appropriate second in the following minute(s).
o $month
- The number of the month relative to the end of the previous year. Values 1 to 12 reference the normal calendar months of the
year in question. Values less than 1 (including negative values) reference the months in the previous year in reverse order, so 0
is December, -1 is November, etc. Values greater than 12 reference the appropriate month in the following year(s).
o $day
- The number of the day relative to the end of the previous month. Values 1 to 28, 29, 30 or 31 (depending upon the month) refer-
ence the normal days in the relevant month. Values less than 1 (including negative values) reference the days in the previous
month, so 0 is the last day of the previous month, -1 is the day before that, etc. Values greater than the number of days in the
relevant month reference the appropriate day in the following month(s).
o $year
- The year
o $is_dst
- Parameters always represent a GMT date so $is_dst doesn't influence the result.
Note
This parameter has been removed in PHP 7.0.0.
RETURN VALUES
Returns a integer Unix timestamp.
CHANGELOG
+--------+---------------------------------------------------+
|Version | |
| | |
| | Description |
| | |
+--------+---------------------------------------------------+
| 7.0.0 | |
| | |
| | $is_dst parameter has been removed. |
| | |
| 5.1.0 | |
| | |
| | As of PHP 5.1.0, the $is_dst parameter became |
| | deprecated. As a result, the new timezone han- |
| | dling features should be used instead. |
| | |
+--------+---------------------------------------------------+
EXAMPLES
Example #1
gmmktime(3) basic example
<?php
// Prints: July 1, 2000 is on a Saturday
echo "July 1, 2000 is on a " . date("l", gmmktime(0, 0, 0, 7, 1, 2000));
?>
SEE ALSO mktime(3), date(3), time(3).
PHP Documentation Group GMMKTIME(3)