You have variables YYYY DD MM, but try to use DD1, YYYY1, MM1.
And in function you have more than one echo. If you need test echo in function, then redirect it ex. to the stderr.
Now you echo your all values to the jd1 or jd2.
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)
I need to increment a date value through shell script.
Input value consist of start date and end date in DATE format of unix.
For eg.
I need increment a date value of 1/1/09 to 31/12/09 i.e for a whole yr.
The output must look like
1/1/09
2/2/09
.
.
.
31/1/09
.
.
1/2/09
.
28/2/09... (1 Reply)
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)
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
datefmt_get_pattern
DATEFMT_GET_PATTERN(3) 1 DATEFMT_GET_PATTERN(3)IntlDateFormatter::getPattern - Get the pattern used for the IntlDateFormatter
Object oriented style
SYNOPSIS
public string IntlDateFormatter::getPattern (void )
DESCRIPTION
Procedural style
string datefmt_get_pattern (IntlDateFormatter $fmt)
Get pattern used by the formatter.
PARAMETERS
o $fmt
- The formatter resource.
RETURN VALUES
The pattern string being used to format/parse.
EXAMPLES
Example #1
datefmt_get_pattern(3) example
<?php
$fmt = datefmt_create(
'en_US',
IntlDateFormatter::FULL,
IntlDateFormatter::FULL,
'America/Los_Angeles',
IntlDateFormatter::GREGORIAN,
'MM/dd/yyyy'
);
echo 'pattern of the formatter is : ' . datefmt_get_pattern($fmt);
echo 'First Formatted output with pattern is ' . datefmt_format($fmt, 0);
datefmt_set_pattern($fmt,'yyyymmdd hh:mm:ss z');
echo 'Now pattern of the formatter is : ' . datefmt_get_pattern($fmt);
echo 'Second Formatted output with pattern is ' . datefmt_format($fmt, 0);
?>
Example #2
OO example
<?php
$fmt = new IntlDateFormatter(
'en_US',
IntlDateFormatter::FULL,
IntlDateFormatter::FULL,
'America/Los_Angeles',
IntlDateFormatter::GREGORIAN,
'MM/dd/yyyy'
);
echo 'pattern of the formatter is : ' . $fmt->getPattern();
echo 'First Formatted output is ' . $fmt->format(0);
$fmt->setPattern('yyyymmdd hh:mm:ss z');
echo 'Now pattern of the formatter is : ' . $fmt->getPattern();
echo 'Second Formatted output is ' . $fmt->format(0);
?>
The above example will output:
pattern of the formatter is : MM/dd/yyyy
First Formatted output is 12/31/1969
Now pattern of the formatter is : yyyymmdd hh:mm:ss z
Second Formatted output is 19690031 04:00:00 PST
SEE ALSO datefmt_set_pattern(3), datefmt_create(3).
PHP Documentation Group DATEFMT_GET_PATTERN(3)