I need help on arithmetic
Now I subtract this result by 1 or 01 I get "3" as the answer. I need "03" as the answer, ie last two significant numbers should be there.
10 - 8 = 02
15 - 10 = 05
25 - 10 = 15
30 - 12 - 18
Like above
Last edited by anilcliff; 11-02-2011 at 08:14 AM..
I am begining to learn bourne shell and as a practice I have written a script which when given the purchase price and percentage of discount calculates the savings.
I somehow cannot figure out why my script fails to do arthimatic calculation on real numbers.
Could anyone look at the script... (5 Replies)
Discussion started by: Tirmazi
5 Replies
5. Post Here to Contact Site Administrators and Moderators
page
unix com/answers-frequently-asked-questions/13785-yesterdays-date-date-arithmetic.html
Date Arithmetic with the Shell
has link of
www samag com/documents/s=8284/sam0307b/0307b.htm
which is no longer.
Is this the correct place to post this?:confused:
and I got message... (1 Reply)
Hello everybody,
I decided to take a Unix Introduction class and have never had experience with programming. Everything was fine until recently when the Prof. started shell scripting and he wants us to make a small script to add unlimited numbers from arguments and from standard input.
I... (1 Reply)
Hello everybody,
I decided to take a Unix Introduction class and have never had experience with programming. Everything was fine until recently when the Prof. started shell scripting and he wants us to make a small script to add unlimited numbers from arguments and from standard input.
I... (8 Replies)
Hi,
I need a help with arithmetic calculations in my script. I have two variables: a=17; b=1712
I want to perform ($a/$b)*100 with two decimals in the result.
I tried with following:
res=$((100*a/b))
res=`echo "scale=2; $a / $b" | bc`
But I am not getting the decimal values.... (4 Replies)
i am having a varialbe a , which is input to my file
i want to multiply this input with value .43, and assign it to variable b.
i tried it as below:
#!/bin/sh
a=$1
b=`expr $1\*0.43`
echo b=$b
error : expr: non-integer argument
Please tell me , how to do this.
Thanks (10 Replies)
Discussion started by: rishifrnds
10 Replies
LEARN ABOUT PHP
datetime.settime
DATETIME.SETTIME(3) 1 DATETIME.SETTIME(3)DateTime::setTime - Sets the time
Object oriented style
SYNOPSIS
public DateTime DateTime::setTime (int $hour, int $minute, [int $second])
DESCRIPTION
Procedural style
DateTime date_time_set (DateTime $object, int $hour, int $minute, [int $second])
Resets the current time of the DateTime object to a different time.
PARAMETERS
o $object
-Procedural style only: A DateTime object returned by date_create(3). The function modifies this object.
o $hour
- Hour of the time.
o $minute
- Minute of the time.
o $second
- Second of the time.
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.setTime(3) example
Object oriented style
<?php
$date = new DateTime('2001-01-01');
$date->setTime(14, 55);
echo $date->format('Y-m-d H:i:s') . "
";
$date->setTime(14, 55, 24);
echo $date->format('Y-m-d H:i:s') . "
";
?>
Procedural style
<?php
$date = date_create('2001-01-01');
date_time_set($date, 14, 55);
echo date_format($date, 'Y-m-d H:i:s') . "
";
date_time_set($date, 14, 55, 24);
echo date_format($date, 'Y-m-d H:i:s') . "
";
?>
The above examples will output something similar to:
2001-01-01 14:55:00
2001-01-01 14:55:24
Example #2
Values exceeding ranges are added to their parent values
<?php
$date = new DateTime('2001-01-01');
$date->setTime(14, 55, 24);
echo $date->format('Y-m-d H:i:s') . "
";
$date->setTime(14, 55, 65);
echo $date->format('Y-m-d H:i:s') . "
";
$date->setTime(14, 65, 24);
echo $date->format('Y-m-d H:i:s') . "
";
$date->setTime(25, 55, 24);
echo $date->format('Y-m-d H:i:s') . "
";
?>
The above example will output:
2001-01-01 14:55:24
2001-01-01 14:56:05
2001-01-01 15:05:24
2001-01-02 01:55:24
SEE ALSO DateTime.setDate(3), DateTime.setISODate(3).
PHP Documentation Group DATETIME.SETTIME(3)