Numbers statrting with 0 are evaluated as octal numbers : 070820 is an invalid octal number.
A solution is to force decimal base evaluation :
This script will not give you any error, but i am not sure that the result is what you are expected.
Incrementng the date will give you invalid dates, for example :
070831+1=070832 <= Invalid date
Jean-Pierre.
Hi ,
I am trying to increment the nested for loops parellely,but i cant ,i used continue 2 but the second loop not getting increment.
no1="1 6 5 4 8"
no2="4 7 8 0 1"
for var1 in $no1 ; do
for var2 in $no2 ; do
line1
line 2
line 3
continue 2
done
done
Please help on this (4 Replies)
I need to increment a date value through shell script.
Input value consist of start date and end date in DATE format of unix.
For eg.
I need increment a date value of 1/1/09 to 31/12/09 i.e for a whole yr.
The output must look like
1/1/09
2/2/09
.
.
.
31/1/09
.
.
1/2/09
.
28/2/09... (1 Reply)
Hi,
Iam new to scripting language.:o
can someone help me out solving this thread?hopingly ya....:)
I want to write a script which connects to db and searches the count in a table which has date column and id column.
If the count is not equal to 0 then it should increment the date with the one... (4 Replies)
Hi,
I have a variable lets say DATA_DATE.
I have to pass some value to this variable in YYYYMMDD format.
lets say today I have passed this variable as :
DATA_DATE=20100107
Then pls help me how to calculate another variable DATA_DATE1 (which is DATA_DATE+1).
The code should work... (3 Replies)
hi Friends,
Today_Dt=`date "+%Y-%m-%d"`
So the Today date is 2010-05-03
I have a file which has date values as below
2010-04-27
2010-04-02
2010-04-18
2010-04-28
2010-04-29
.. (1 Reply)
hi experts,
my requirement is like this i need to develop a shell script to update date part with new incremental date time in file some 'X' which is kept at some server location incrementing every two hours.as i am new to this scripting i need support from u people,thanx in advance (1 Reply)
Hello men.
How can i build a simple increment for $a by Xquery such as ?
let $a := 0
for $i in (1 to 10)
let $a := $a + 1
return $a
why a in this loop always is '1'
Thank you for reading, its will really helpful for my job if i can solve this problem :D:D (3 Replies)
Daily stupid question. I want to increment the file name everytime the script is run. So for example if the filename is manager.log and I run the script, I want the next sequence to be manager.log1. So to be clear I only want it to increment when the script is executed. So
./script... (10 Replies)
Hi,
I am using solaris 5.10 environment and need help on doing parallel increment of nested for loop.
Samples
#inside the code the values assigned to a variable by another awk command will be like
a=/xyz/pg/as
/xyz/pg/as2
/xyz/pg/as3
b=/xyz/sd/fd1
/xyz/sd/fd2
/xyz/sd/fd3
for q in... (1 Reply)
Discussion started by: ananan
1 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)