Hi all,
I am getting following output by using commands like sort, uniq and awk to the standard output.
110 d
40 a
59 c
9 b
3 e
Now at the end I would like to add all the numbers in column 1 and display the count of all numbers i.e. (110 + 40 + 59 + 9 + 3).
Also the output may... (3 Replies)
Hi all,
I am very new to shell programming and trying to learn out the basics.
I tried this:
$ echo `expr 20 + 30`
and it worked. But when i tried this,it does not work.
$ a=20
$ b=30
$ echo `expr a + b`
The error is:
expr: non-numeric argument
I cant understand why its... (3 Replies)
Hi,
I want to add 3 new fields in the existing file.Please find the example below.
input:
UID: ABCD
UNAME: XYZ
Desired Output
Tmiestamp: 20101208
UID: ABCD
UNAME: XYZ
DEPTNO:40
ModifyTImestamp:20101209
If you see the above i have added the 3 columns manually in the output section... (2 Replies)
Hi I need to do the following substitution
I have to look for line starting with ABC and add 4 ":" before the first occurence of "+"in that line
Input
ABC:12:Lambert:C278472:1357:0:0:0:0:2:N::::N:9045123:NAP::+DEF
output
ABC:12:Lambert:C278472:1357.00:0.00:0:0:0:2:N::::N:9045123:NAP::::::+DEF... (5 Replies)
How would I print out the total amount through awk? I tried using
print "Total Amount: " $4+$4;
Would I have to do a for loop to get through everything? (2 Replies)
by the script, two files Q1 and Q2 will be generated on the system. Q1 will contain an integer number and Q2 also contain an integer number. i would like to add those numbers and put into new file.
excerpt from my script
22 subcount=`echo $dir/Q$qid.txt` + `echo $dir/Q$qid.txt`
23 echo... (1 Reply)
whats wrong with this addition?
Whats the maximum number of digits can be handled?
pandeeswaran@ubuntu:~/Downloads$ const=201234454654768979799999
pandeeswaran@ubuntu:~/Downloads$ let new+=const
pandeeswaran@ubuntu:~/Downloads$ echo $new
-2152890657037557890
pandeeswaran@ubuntu:~/Downloads$ (4 Replies)
Need assistance . Below code gets me julian date . I wanted to add hour/24 to julian date and output it. Is there a way to do the calculation?
use Time::Local;
use POSIX qw(strftime);
my $time=timelocal(1,2,3,9,11,2013);
printf strftime "%j", localtime($time);
343 (3 Replies)
Hi
I have a file whose contents are as follows:
sorce1 LEN assumption 695 3570 0.770047 - . ID=f000001.1;source_id=A.off_LEN_10008424;
sorce1 LEN descriptive 3334 3570 . - 0 Parent=f000001.1;
sorce1 LEN ... (8 Replies)
Discussion started by: sa@@
8 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)