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 $2}' will this type of command work.
actually my need is
if today is... (17 Replies)
Hi,
Anybody knows how to get what date was 28 days ago of the current system date through UNIX script.
Ex : - If today is 28th Mar 2010 then I have to delete the files which arrived on 1st Mar 2010, (15 Replies)
Hi i am writing a cron job.
so for it i need the 60 days old date form current date in variable.
Like today date is 27 jan 2011 then output value will be stote in variable in formet Nov 27.
i am using EST date, and tried lot of solution and see lot of post but it did not helpful for me. so... (3 Replies)
I am trying to find out the number of days between the current date and user defined date.
I took reference from here for the date2jd() function.
Modified the function according to my requirement. But its not working properly.
Original code from here is working fine.
#!/bin/sh... (1 Reply)
Hi! I am trying to read a file and every line has a specific date as one of its fields.
I want to take that date and compare it to the date today plus 6 days.
while read line
do
date=substr($line, $datepos, 8) #date is expected to be YYYYMMDD
if ; then
...proceed commands
... (1 Reply)
Hi,
One of my Unix scripts needs to look for files coming in on Fridays. This script runs on Mondays. $date +"%y%m%d" will give me today's date. How can I get previous Friday's date.. can I do "today's date minus 3 days" to get Friday's date? If not, then any other way?? Name of the files is... (4 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
intlcalendar.setfirstdayofweek
INTLCALENDAR.SETFIRSTDAYOFWEEK(3) 1 INTLCALENDAR.SETFIRSTDAYOFWEEK(3)IntlCalendar::setFirstDayOfWeek - Set the day on which the week is deemed to start
Object oriented style
SYNOPSIS
public bool IntlCalendar::setFirstDayOfWeek (int $dayOfWeek)
DESCRIPTION
Procedural style
bool intlcal_set_first_day_of_week (IntlCalendar $cal, int $dayOfWeek)
Defines the day of week deemed to start the week. This affects the behavior of fields that depend on the concept of week start and end
such as IntlCalendar::FIELD_WEEK_OF_YEAR and IntlCalendar::FIELD_YEAR_WOY.
PARAMETERS
o $cal
- The IntlCalendar resource.
o $dayOfWeek
- One of the constants IntlCalendar::DOW_SUNDAY, IntlCalendar::DOW_MONDAY, , IntlCalendar::DOW_SATURDAY.
RETURN VALUES
Returns TRUE on success. Failure can only happen due to invalid parameters.
EXAMPLES
Example #1
IntlCalendar.setFirstDayOfWeek(3)
<?php
ini_set('date.timezone', 'Europe/Lisbon');
ini_set('intl.default_locale', 'es_ES');
$cal = IntlCalendar::createInstance();
$cal->set(2013, 5 /* June */, 30); // A Sunday
var_dump($cal->getFirstDayOfWeek()); // 2 (Monday)
echo IntlDateFormatter::formatObject($cal, <<<EOD
week of month : 'W'
week of year : 'ww
EOD
), "
";
$cal->setFirstDayOfWeek(IntlCalendar::DOW_SUNDAY);
echo IntlDateFormatter::formatObject($cal, <<<EOD
week of month : 'W'
week of year : 'ww
EOD
), "
";
The above example will output:
int(2)
local day of week: 7
week of month : 4
week of year : 26
local day of week: 1
week of month : 5
week of year : 27
PHP Documentation Group INTLCALENDAR.SETFIRSTDAYOFWEEK(3)