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Full Discussion: Error when executing script
Operating Systems Linux Red Hat Error when executing script Post 302996992 by anaigini45 on Friday 5th of May 2017 05:03:25 AM
Old 05-05-2017
I have another question on command substitution.
The same script, but different command substitution.

I simply run this command from the terminal :

Code:
[root@L28tstream1 emokheng]# stat -c %z * | awk '{print $1}'
2015-08-18
2015-08-18
[root@L28tstream1 emokheng]#

As you see, I get the dates of the file generated in the directory.

However, when I attempt to use this command as a substitute, like this, I get a diferent output :

Code:
[root@L28tstream1 emokheng]# D=$(stat -c %z * | awk '{print $1}')
[root@L28tstream1 emokheng]# echo $D
2015-08-18 2015-08-18

It all displays in one line. Why is this, and how do I get it to display in two separate lines like the command above this?

Actually the main output I want is to get ONLY the day section of the whole date. Means I want to extract only '18' from 2015-08-18. This is for all the files in the directory.
Therefore, I use this command :

Code:
[root@L28tstream1 emokheng]# FDAY=$(date -d "$D" '+%d')
date: invalid date `2015-08-18\n2015-08-18'
[root@L28tstream1 emokheng]#

However, as you can see, the error is shown. Perhaps if
Code:
echo $D

displays two separate lines, the error above will not happen?

How do I display ONLY the day part of the whole date for a file in a directory?

Last edited by anaigini45; 05-05-2017 at 06:12 AM..
 
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INTLCALENDAR.GETMINIMALDAYSINFIRSTWEEK(3)				 1				 INTLCALENDAR.GETMINIMALDAYSINFIRSTWEEK(3)

IntlCalendar::getMinimalDaysInFirstWeek - Get minimal number of days the first week in a year or month can have

	Object oriented style

SYNOPSIS
public int IntlCalendar::getMinimalDaysInFirstWeek (void ) DESCRIPTION
Procedural style int intlcal_get_minimal_days_in_first_week (IntlCalendar $cal) Returns the smallest number of days the first week of a year or month must have in the new year or month. For instance, in the Gregorian calendar, if this value is 1, then the first week of the year will necessarily include January 1st, while if this value is 7, then the week with January 1st will be the first week of the year only if the day of the week for January 1st matches the day of the week returned by IntlCalendar.getFirstDayOfWeek(3); otherwise it will be the previous years last week. PARAMETERS
o $cal - The IntlCalendar resource. RETURN VALUES
An int representing a number of days or FALSE on failure. EXAMPLES
Example #1 IntlCalendar.getMinimalDaysInFirstWeek(3) <?php ini_set('date.timezone', 'UTC'); ini_set('intl.default_locale', 'en_US'); $cal = new IntlGregorianCalendar(2013, 0 /* January */, 2); var_dump(IntlDateFormatter::formatObject($cal, 'cccc')); // Wednesday var_dump($cal->getMinimalDaysInFirstWeek(), // 1 $cal->getFirstDayofWeek()); // 1 (Sunday) // Week 1 of 2013 var_dump(IntlDateFormatter::formatObject($cal, "'Week 'w' of 'Y")); $cal->setMinimalDaysInFirstWeek(4); // Still Week 1 of 2013 (1st week has 5 days in the new year) var_dump(IntlDateFormatter::formatObject($cal, "'Week 'w' of 'Y")); $cal->setMinimalDaysInFirstWeek(6); // Week 53 of 2012 var_dump(IntlDateFormatter::formatObject($cal, "'Week 'w' of 'Y")); The above example will output: string(9) "Wednesday" int(1) int(1) string(14) "Week 1 of 2013" string(14) "Week 1 of 2013" string(15) "Week 53 of 2012" PHP Documentation Group INTLCALENDAR.GETMINIMALDAYSINFIRSTWEEK(3)

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