I keep getting an error at line 21, it doesn't like my if statement. Previously I have tried using (( )), but still get errors. The current error is that server_busy is not found. This is the script:
#! /bin/ksh
server_busy="na"
for file in $1 $2 $3 $4 $5 $6
do
echo " ${file}\t\c"
... (1 Reply)
I need an IF statement that will compare the contents of the variable CX with the actual string "CP". ie. If the contents of $CX are NOT equal to the actual string "CP" then blah blah blah.
I have tried a number of things including the following.......
if ]; then
if ]; then
if ];... (2 Replies)
hi all. i just have a very small problem. i have a menu of 7 choices. i want an if statement so that if the user chooses anything except inside the 1 to 7 range, i can handle the error for it.
i tried this:
if ]
then
.......
fi
(but it dont work)
...any suggestions?
... (4 Replies)
The problem I am having here is that only the 1st option is executed, no matter if I pick yes or no. What am I doing wrong? How can I get this working right without resorting to a case statement?
echo "This is the max size your lvol can be:"
echo $MAXSIZE
echo
echo Do you want to max out... (2 Replies)
Hi I have a bash script like this
if
then
echo "A"
else
echo "B"
fi
$1 is something like 02350 (there is always a trailing '0')
and I would like to have an if based on the value of the digits after the 0.
Can anybody help?
Thanks,
Sarah (3 Replies)
Could someone help me out with this if statement? It's supposed to get a person's website, but it isn't working when I run it.
website=""
echo "Would you like to enter a website? Enter Yes/No"
read choice
if
then
while
do
echo "Please enter a website:";
read... (4 Replies)
Writing my script and I'm banging my head on the desk right now ...
My biggest problem is the 3rd IF statement where I check if the username exists. Doing the grep command on it's own in the shell gives me a 1 or 0 value. Running the script, it always returns a false value (runs the ELSE... (4 Replies)
echo "Enter the variable: " "
read var1
echo " "
for i in ib eb atm
do
if ; then
mv properties environment.properties
break
else
echo "No changes to $var1 "
fi
done
When i run and enter the eb it's not working.Any suggestions please.. (7 Replies)
Hi All,
I am writing an if statement to check multiple conditions, but when I try to execute the script it is breaking at the point of if statement by showing the issue below.
Code I am using is given below.
if -a ]
then
....
else
...
fi
I am not understanding... (3 Replies)
Discussion started by: ginrkf
3 Replies
LEARN ABOUT PHP
datetime.add
DATETIME.ADD(3) 1 DATETIME.ADD(3)DateTime::add - Adds an amount of days, months, years, hours, minutes and seconds to a DateTime object
Object oriented style
SYNOPSIS
public DateTime DateTime::add (DateInterval $interval)
DESCRIPTION
Procedural style
DateTime date_add (DateTime $object, DateInterval $interval)
Adds the specified DateInterval object to the specified DateTime object.
PARAMETERS
o $object
-Procedural style only: A DateTime object returned by date_create(3). The function modifies this object.
o $interval
- A DateInterval object
RETURN VALUES
Returns the DateTime object for method chaining or FALSE on failure.
EXAMPLES
Example #1
DateTime.add(3) example
Object oriented style
<?php
$date = new DateTime('2000-01-01');
$date->add(new DateInterval('P10D'));
echo $date->format('Y-m-d') . "
";
?>
Procedural style
<?php
$date = date_create('2000-01-01');
date_add($date, date_interval_create_from_date_string('10 days'));
echo date_format($date, 'Y-m-d');
?>
The above examples will output:
2000-01-11
Example #2
Further DateTime.add(3) examples
<?php
$date = new DateTime('2000-01-01');
$date->add(new DateInterval('PT10H30S'));
echo $date->format('Y-m-d H:i:s') . "
";
$date = new DateTime('2000-01-01');
$date->add(new DateInterval('P7Y5M4DT4H3M2S'));
echo $date->format('Y-m-d H:i:s') . "
";
?>
The above example will output:
2000-01-01 10:00:30
2007-06-05 04:03:02
Example #3
Beware when adding months
<?php
$date = new DateTime('2000-12-31');
$interval = new DateInterval('P1M');
$date->add($interval);
echo $date->format('Y-m-d') . "
";
$date->add($interval);
echo $date->format('Y-m-d') . "
";
?>
The above example will output:
2001-01-31
2001-03-03
NOTES DateTime.modify(3) is an alternative when using PHP 5.2.
SEE ALSO DateTime.sub(3), DateTime.diff(3), DateTime.modify(3).
PHP Documentation Group DATETIME.ADD(3)