Hi,
My objective is to achieve from HB.txt
into adding a day column by using the first column - date.
So far, I'm in progress to try this,
However, the generated output for the new column is after a breakline.
System Specifications:
macOS High Sierra
Terminal - Bash
HI guys,
I have created a script to read 1 column in a csv file and then place it in text file.
However, when i checked out the text file, it is not in a column format...
Example:
CSV file contains
name,age
aa,11
bb,22
cc,33
After using awk to get first column
TXT file... (1 Reply)
I have an assignment in a Linux class I am taking. It has multiple scripts. Basicly when it runs it asks the user name and shows information about the user from the /etc/passwd and /etc/shadow files. The one I need help with is the one that reads the /etc/shadow file. I need to format the date into... (1 Reply)
The date format in the delimited file for one column '6/27/2011 12:00:00 AM' Is it possible o change it to '2011-06-27 12:00:00 AM' for all the records..
Thanks in advance..... (8 Replies)
I have this code to compare columns 1 and 10 between file1 and file 2 and give me all records that match column 1 but dont match column 10
However column 10 is date format mm/dd/yy and awk cant read it and compare ...i tried
awk < file1 -F~ '{print $10}'
and it gave blank screen
Is... (1 Reply)
I have below date format in a CSV file. (dd/mm/yyyy)
Ex Input:
9/8/2013
Need to convert it into below format (yyyymmdd ) and redirect to new file.
Ex Output:
20130809
How do I use awk here to change the format and if leading 0 (zero) is not then add it.
Please help. Thanks. (8 Replies)
Could you tell me how to convert the following dates?
If I have m/d/yyyy, I want to have
0m/0d/yyyy. I want my dates to always be 8 digits.
In other words, I want a 0 inserted whenever the month or day is a single digit.
My issue is first I need to use FS="," to get field $4 for the... (7 Replies)
Hi All,
I have file like
“April 10, 2013”,”raj”
“April 29, 2013”,”raj1”
Output :
“2013/04/10”,”raj”
“2013/04/29”,”raj1”
Please help me how to do... (9 Replies)
Hi,
i have a flat file namely temp.txt with this data below
ID|name|contact_date
101|Kay|2013-12-26
102|let|2013-12-26
I need to modify the date data in the flat file into MM/DD/YYYY HH24:MI:SS format
let me know the code for this.
Thank you! (5 Replies)
Hi have a large spreadsheet which has 4 columns
APM00111803814 server_2 96085 Corp IT Desktop and Apps
APM00111803814 server_2 96085 Corp IT Desktop and Apps
APM00111803814 server_2 96034 Storage Mgmt Team
APM00111803814 server_2 96152 GWP... (6 Replies)
Discussion started by: kieranfoley
6 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)