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Top Forums Shell Programming and Scripting Rename multiple file names in a directory Post 302524248 by methyl on Sunday 22nd of May 2011 04:14:58 PM
Old 05-22-2011
To match your requirement, the 'date' command should be:
Code:
date +'%d%m%Y%H%M%S'

The %Y gives a 4-digit year.

Statements like "which is not working" tell us nothing. What went wrong?

Maybe your script should be something like:

Code:
#!/bin/sh
DDMMYYYYHHMMSS=`date +'%d%m%Y%H%M%S'` # Evaluate the date once only
cd /vol/best/srcfiles
for filename in "Payor" "customer" "vendor" Employee"
do
       if [ -f "${filename} ]
       then
              mv "${filename}" "${filename}.${DDMMYYYYHHMMSS}"
       fi
done


Btw. The date format YYYYMMDD etc. is much better than DDMMYYYY etc. in filenames because the files will come out in chronogical order in a default "ls" directory list.

Last edited by methyl; 05-22-2011 at 05:50 PM..
 

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DATE_PARSE_FROM_FORMAT(3)						 1						 DATE_PARSE_FROM_FORMAT(3)

date_parse_from_format - Get info about given date formatted according to the specified format

SYNOPSIS
array date_parse_from_format (string $format, string $date) DESCRIPTION
Returns associative array with detailed info about given date. PARAMETERS
o $format - Format accepted by DateTime.createFromFormat(3). o $date - String representing the date. RETURN VALUES
Returns associative array with detailed info about given date. EXAMPLES
Example #1 date_parse_from_format(3) example <?php $date = "6.1.2009 13:00+01:00"; print_r(date_parse_from_format("j.n.Y H:iP", $date)); ?> The above example will output: Array ( [year] => 2009 [month] => 1 [day] => 6 [hour] => 13 [minute] => 0 [second] => 0 [fraction] => [warning_count] => 0 [warnings] => Array ( ) [error_count] => 0 [errors] => Array ( ) [is_localtime] => 1 [zone_type] => 1 [zone] => -60 [is_dst] => ) SEE ALSO
DateTime.createFromFormat(3), checkdate(3). PHP Documentation Group DATE_PARSE_FROM_FORMAT(3)
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