I should have included some of my code for you to understand what Im on about..
if I use your method, and I have no arguments it skips the first if statement.
basically i want the first if statement to be "if no searchMonth or searchyear, then"
Cheers for your help
Code:
#if no arguments entered
if [ $# = 0 ]; then
echo "Invalid Arguments Entered."
echo
echo "Usage of this program defined with either of the following inputs:"
echo "sh programName Month Year"
echo "sh programName Month"
echo "sh programName Year"
echo
echo "Full month and year inputs (eg: March 2004) can be used, abbreviation is also a valid input (eg: Mar 04)"
echo
exit
elif [ -z $searchMonth ]; then
for (( i=0; i<12; i++))
do
#string comparison
if [ "$2" = ${month[$i]} -a "$3" = "$searchYear" ]; then
monthCount[$i]=`expr ${monthCount[$i]} + 1`
fi
done
print=1
#if no searchYear entered
elif [ -z "$searchYear" ]; then
for (( i=0; i<"${#year[*]}"; i++))
do
#string comparison
if [ "$2" = "$searchMonth" -a "$3" = "${year[$i]}" ]; then
yearCount[$i]=`expr ${yearCount[$i]} + 1`
fi
done
print=2
In my ksh script, if the conditions of a if statement are true, then do nothing; otherwise, execute some commands.
How do I write the "do nothing" statement in the following example?
Example:
if (( "$x"="1" && "$y"="a" && "$z"="happy" ))
then
do nothing
else
command
command
fi... (3 Replies)
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I need to insert values to a table using the o/p from a slelect statement.
Can anybody Help!
My script looks like tihs.
---`sqlplus -s username/password@SID << EOF
set heading off
set feedback off
set pages 0
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I can't find anything wrong with this line of code, it works when there is one file in the directory but more than one i get a "too many arguements2 error
if ; then
am i missing something? (3 Replies)
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I have a boubt passing different arguments at a time for any one option in below code.
I would also like to check which option has been selected (any one of i, r, u ) so that whether or not matching argument passed can be verified.
for i and r - install and re-install -... (4 Replies)
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Hi
Am pretty new to C..
Am trying to pass the arguments from command line and use them in switch case statement..
i have tried the following
#include <stdlib.h>
main(int argc, char* argv)
{
int num=0;
if ( argc == 2 )
num = argv;
printf("%d is the num value",num);
switch ( num )
... (2 Replies)
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I have a question regarding the usage statement of a script.
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How this would be printed out within a usage statement?
My suggestion would be:
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Discussion started by: Eric7giants
6 Replies
LEARN ABOUT PHP
datetime.setdate
DATETIME.SETDATE(3) 1 DATETIME.SETDATE(3)DateTime::setDate - Sets the date
Object oriented style
SYNOPSIS
public DateTime DateTime::setDate (int $year, int $month, int $day)
DESCRIPTION
Procedural style
DateTime date_date_set (DateTime $object, int $year, int $month, int $day)
Resets the current date of the DateTime object to a different date.
PARAMETERS
o $object
-Procedural style only: A DateTime object returned by date_create(3). The function modifies this object.
o $year
- Year of the date.
o $month
- Month of the date.
o $day
- Day of the date.
RETURN VALUES
Returns the DateTime object for method chaining or FALSE on failure.
CHANGELOG
+--------+---------------------------------------------------+
|Version | |
| | |
| | Description |
| | |
+--------+---------------------------------------------------+
| 5.3.0 | |
| | |
| | Changed the return value on success from NULL to |
| | DateTime. |
| | |
+--------+---------------------------------------------------+
EXAMPLES
Example #1
DateTime.setDate(3) example
Object oriented style
<?php
$date = new DateTime();
$date->setDate(2001, 2, 3);
echo $date->format('Y-m-d');
?>
Procedural style
<?php
$date = date_create();
date_date_set($date, 2001, 2, 3);
echo date_format($date, 'Y-m-d');
?>
The above examples will output:
2001-02-03
Example #2
Values exceeding ranges are added to their parent values
<?php
$date = new DateTime();
$date->setDate(2001, 2, 28);
echo $date->format('Y-m-d') . "
";
$date->setDate(2001, 2, 29);
echo $date->format('Y-m-d') . "
";
$date->setDate(2001, 14, 3);
echo $date->format('Y-m-d') . "
";
?>
The above example will output:
2001-02-28
2001-03-01
2002-02-03
SEE ALSO DateTime.setISODate(3), DateTime.setTime(3).
PHP Documentation Group DATETIME.SETDATE(3)