I have ran the command and i am getting following error in the output
Code:
bash-3.00$ awk 'NR==1{print $1,$2,$3}' SystemOut.log_bak14122010
************ Start Display
awk: record `[12/14/10 15:54:22:9...' too long
record number 6014
---------- Post updated at 09:58 AM ---------- Previous update was at 09:56 AM ----------
Hello zaxxon,
I have ran the command and i am getting following output.
Hi all,
I want to increment the date which is in the format "YYYYMMDD".
ex: If the date is 20010601
Increment should be: 20010602, 20010603 etc.,
Any help will be much appreciated.
Many Thanks and Regards,
Ganapati (4 Replies)
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Discussion started by: jcdole
1 Replies
LEARN ABOUT PHP
datetime.format
DATETIME.FORMAT(3) 1 DATETIME.FORMAT(3)DateTime::format - Returns date formatted according to given format
Object oriented style
SYNOPSIS
public string DateTime::format (string $format)
DESCRIPTION
string DateTimeImmutable::format (string $format) string DateTimeInterface::format (string $format)
Procedural style
string date_format (DateTimeInterface $object, string $format)
Returns date formatted according to given format.
PARAMETERS
o $object
-Procedural style only: A DateTime object returned by date_create(3)
o $format
- Format accepted by date(3).
RETURN VALUES
Returns the formatted date string on success or FALSE on failure.
EXAMPLES
Example #1
DateTime.format(3) example
Object oriented style
<?php
$date = new DateTime('2000-01-01');
echo $date->format('Y-m-d H:i:s');
?>
Procedural style
<?php
$date = date_create('2000-01-01');
echo date_format($date, 'Y-m-d H:i:s');
?>
The above example will output:
2000-01-01 00:00:00
NOTES
This method does not use locales. All output is in English.
SEE ALSO date(3).
PHP Documentation Group DATETIME.FORMAT(3)