Hi,
haven't found anything about this through searching, so may be a new topic:
when doing this:
set -o nounset
set -o errexit
find . -name "*.lib" | while read library; do
echo ${libary}
done
echo "after while"
I expect the script to exit within the while loop (because of nounset and... (6 Replies)
Hi all,
Im using a script which contains read command.. the script works perfectly but when I alias the script it gave "undifined variable" after I enter the read command input (variable)...
Does any one know why ? (4 Replies)
Below script cuts date part from the date entered by the user.
#!/bin/csh
echo 'Enter date in the format dd/mm/yyyy'
read DATE
DD=`echo $DATE | cut -c1-2`
echo $DD;
when debug with -x option , it works perfectly but without -x doesnot.:confused:
$ sh -x unix_12.sh
+ echo Enter... (2 Replies)
Hello group,
Still fairly new at the whole scripting thing so be gentle. I'm trying to write a simple script that archives my log files into a master log broken into weeks of the year.
My script runs fine up till the "cat" lines which I get a undefined or illegal variable name error. But... (2 Replies)
hi there
I have this C shell script that was migrated from AIX to Linux, could someone please help me, I checked the syntax numerous times but I can't find out where the error is. The script is meant to find files older than 27 days and delete it
#!/usr/bin/csh
#
... (22 Replies)
Hi,
i am a beginner in ubuntu. my default shell is bash. everytime i try to change the shell with command "csh", i get a message (probably an error message). after i get into c-shell, when i try to execute a c shellscript, then it showed the same message. any idea about what is this about or any... (1 Reply)
Hello,
I'm new here. I test these expressions's value in my script :
(in centOS 6 )
#!/bin/bash
array='something'
echo "############"
echo ${array}
echo ${array}
echo ${array}
echo "############"
The output result is :
#################
something
something
#################... (5 Replies)
I am getting the error undefined variable even after following these steps
#read name
abcd
#echo $name
na: undefined variable (6 Replies)
Discussion started by: Vishawdeep
6 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)