I have a script like this--
#!/bin/ksh
echo "To pad a 0 before digits from 1-9"
for i in $*
do
echo $i | sed 's//'0'/g'
done
I run this script as
ksh name 1 2 23 34
The output should be
01 02 23 34
Help me in modifying this script.
Thanks
Namish (2 Replies)
Folks,
Is there a simple way to replace one digit by two digit using sed.
Example,
mydigit1918_2006_8_8_lag1.csv should be
mydigit1918_2006_08_08_lag01.csv.
I tried this way, but doesn't work.
echo mydigit1989_2006_8_8_lag1.csv|sed 's/]/0]/'
Thank you, (5 Replies)
Hi All,
How can i convert a number 24 to 0024
In the same way how can i convert 123 to 0123?
All this has to be done inside a script
Thanks in advance
JS (6 Replies)
hi all,
how do i format the date command so it displays day and month in single digits i.e 8 instead of 08 ??
am using the command (in a ksh) : date +%D
output i get is 10/08/08
thanks in advance. (5 Replies)
hi all,
how do i get the month in single digit in ksh script ?? my command for date is :
/usr/bin/date +%Om/%Oe/%y | sed 's/ //g'
which returns "01/16/09" but i need the output to be "1/16/09" i.e the month without leading zero.
thanks in advance. (2 Replies)
Hi ,
how to add the single digit to front of the word and front of the lines in the one file with compare pattern file and get digit. like example
pattern file pattern.txt
pattern num
bala 2
raja 3
muthu 4
File Name: chennai.dat
muthu is good boy
raja is bad boy
selvam in super... (6 Replies)
Hi,
I'm trying to acheive the following, I have a dat file in which i have several addresses, If the address starts with a single digit then i have to delete the line,
if it starts with 2 or more digits then i have to keep the line
Here is a sample of my file:
377 CARRER DE LA... (5 Replies)
Hi All,
I have a file which keeps count based on completion of a certain activity. I am using the following grep command to return a '1' in case the count is zero
grep -ic "0" abc_count.txt
Now the issue happens when the count is '10', '20' etc .. in these cases as well it returns a... (5 Replies)
Discussion started by: dev.devil.1983
5 Replies
LEARN ABOUT PHP
datetime.sub
DATETIME.SUB(3) 1 DATETIME.SUB(3)DateTime::sub - Subtracts an amount of days, months, years, hours, minutes and seconds from a DateTime object
Object oriented style
SYNOPSIS
public DateTime DateTime::sub (DateInterval $interval)
DESCRIPTION
Procedural style
DateTime date_sub (DateTime $object, DateInterval $interval)
Subtracts the specified DateInterval object from 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.sub(3) example
Object oriented style
<?php
$date = new DateTime('2000-01-20');
$date->sub(new DateInterval('P10D'));
echo $date->format('Y-m-d') . "
";
?>
Procedural style
<?php
$date = date_create('2000-01-20');
date_sub($date, date_interval_create_from_date_string('10 days'));
echo date_format($date, 'Y-m-d');
?>
The above examples will output:
2000-01-10
Example #2
Further DateTime.sub(3) examples
<?php
$date = new DateTime('2000-01-20');
$date->sub(new DateInterval('PT10H30S'));
echo $date->format('Y-m-d H:i:s') . "
";
$date = new DateTime('2000-01-20');
$date->sub(new DateInterval('P7Y5M4DT4H3M2S'));
echo $date->format('Y-m-d H:i:s') . "
";
?>
The above example will output:
2000-01-19 13:59:30
1992-08-15 19:56:58
Example #3
Beware when subtracting months
<?php
$date = new DateTime('2001-04-30');
$interval = new DateInterval('P1M');
$date->sub($interval);
echo $date->format('Y-m-d') . "
";
$date->sub($interval);
echo $date->format('Y-m-d') . "
";
?>
The above example will output:
2001-03-30
2001-03-02
NOTES DateTime.modify(3) is an alternative when using PHP 5.2.
SEE ALSO DateTime.add(3), DateTime.diff(3), DateTime.modify(3).
PHP Documentation Group DATETIME.SUB(3)