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Top Forums Shell Programming and Scripting Ksh: Read line parse characters into variable and remove the line if the date is older than 50 days Post 302976791 by RudiC on Wednesday 6th of July 2016 09:57:36 AM
Old 07-06-2016
Not sure what you mean with "then remove this line", but with this, you'll have the desired LUSER variable if the date portion is longer than 50 days back:
Code:
while IFS="_ " read LUSER LDATE
  do    [ $(( $(date +%s) - $(date +%s -d"${LDATE:2:2}/${LDATE:0:2}/${LDATE: -4}") - 4320000 )) -ge 0 ] && echo $LUSER, $LDATE
  done < file

This is not ksh tested. You might be able to use ksh's printf "%(fmt)T" builtin should your system's date not allow for the -d option...
 

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DATETIMEINTERFACE.FORMAT(3)						 1					       DATETIMEINTERFACE.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 DateTimeInterface.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 DATETIMEINTERFACE.FORMAT(3)
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