Compare the system date with date from a text file
I get the date that's inside a text file and assigned it to a variable. When I grep the date from the file, I get this,
Code:
Not After : Jul 28 14:09:57 2017 GMT
So I only crop out the date, with this command
Code:
echo $dateFile | cut -d ':' -f 2,4
The result would be
Code:
Jul 28 14:57 2017 GMT
How do I convert this date to the number of seconds, so I can compare it to the system date? If it is over 2 days old.
I have this code but it doesn't work. I'm getting an error message when I ran it. I think its because $dateFile is a text file and it doesn't know how to convert it.
Any help would be appreciated.
Code:
#!/bin/bash
$dateFile=grep "After :" myfile.txt | cut -d ':' -f 2,4
AGE_OF_MONTH="172800" # 172800 seconds = 2 Days
NOW=$( date +%s )
NEW_DATE=$(( NOW - AGE_OF_MONTH ))
if [ $( stat -c %Y "$dateFile" ) -gt ${NEW_DATE} ]; then
echo Date Less then 2 days
else
echo Date Greater then 2 days
fi
Last edited by rbatte1; 02-21-2017 at 08:31 AM..
Reason: Changed bold text to CODE for file contents
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Discussion started by: pradeepp
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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)