Hi ...
I am trying to calculate the time needed for a command to execute..
but the resulting value is getting as string..
so i am not able to use "expr " command..
please help me to convert the value to integer so that i can proceed with my script..
Regards
esham (1 Reply)
Hi all,
I'm trying to convert a decimal number into an integer number; I'm doing this:
n=`echo |awk '{ print "'"$mem"'"*10}'`
where the variable mem is equal to 3.7
I'd like to obtain 37, but the expression above gives me 30.
Help please!!!!
thx a lot (4 Replies)
Hi,
I am passing a variable to a unix function.
However when I try to assign the value to another variable like
typeset -i I_CACHE_VAL=$2
Is this because of String to Integer conversion?
I get an error.
Please help me with thsi.
Thanks (2 Replies)
Hi everyone,
I would like to know how to convert an integer to a string. for instance if i=1 i would like to creat a variable called constant1. i want to do this in a for loop so for each value of i, i create a new variable such as constant2, constant3,... and so on.
for i in 1 2 3
do ... (1 Reply)
The shell mentioned below will show a warning if the page takes more than 6 seconds to load.
The problem is that myduration variable is not an integer. How do I convert it to integer?
myduration=$(curl http://192.168.50.1/mantisbt/view.php?id=1 -w %{time_total}) > /dev/null ; ] && echo... (3 Replies)
Dear community,
i got a problem to get "date +%j" as the right value.
Today is the 10th day of the year.
#./script.sh 2
#!/bin/bash/
Var1=$(date +%j)
Var2=$1
let result=$Var1+$Var2
echo $Var1 plus $Var2 equals $result
The output of the script is:
010 plus 2 equals 10... (9 Replies)
I have a function that is supposed to check for user processes and wait for 0 count before exiting the function. I am sure I have more than one issue in my code, but the stumbling block right now is that I am trying to convert the value of my variable from a string to integer.
process_count... (10 Replies)
Hi guys,
I'm new here. I have a problem at work. One of our scripts was eventually having a bug and only detected recently. Here's the issue and background:
Bash Script which calls AWK script
Awk script returns a string as per below (example):var1='00000-123'So, when we convert it, the... (18 Replies)
Hello,
How can we convert date like format 20181004171050 in seconds ?
I can able to convert till date but failing for HHMMSS.
date -d "20181004" "+%s" output as 1538596800 .
But when i add hhmmss it is failing date -d "20181004172000" "+%s" result Invalid date
Kindly guide.
Regards (16 Replies)
Hi Folks -
Linux Version = Linux 2.6.39-400.128.17.el5uek x86_64
I have a process that determines the start and end load periods for an Oracle data load process.
The variables used are as follows follows:
They are populated like such:
However, the load requires the month to be the... (11 Replies)
Discussion started by: SIMMS7400
11 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)