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Top Forums Shell Programming and Scripting Getting filenames from a directory in single line Post 302459502 by Raamc on Monday 4th of October 2010 09:38:17 PM
Old 10-04-2010
Hi Agama,

Thank you very much, that worked.

I need one more small clue, I got the output like this
/home/RAAM/work/test1.xls /home/RAAM/work/test2.xls

I need to get only filenames with out directory structure, can you please help.


Thank you very much for quick response.

--Thanks,
--Raam.
 

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GET_INCLUDED_FILES(3)							 1						     GET_INCLUDED_FILES(3)

get_included_files - Returns an array with the names of included or required files

SYNOPSIS
array get_included_files (void ) DESCRIPTION
Gets the names of all files that have been included using include(3), include_once(3), require(3) or require_once(3). RETURN VALUES
Returns an array of the names of all files. The script originally called is considered an "included file," so it will be listed together with the files referenced by include(3) and family. Files that are included or required multiple times only show up once in the returned array. EXAMPLES
Example #1 get_included_files(3) example <?php // This file is abc.php include 'test1.php'; include_once 'test2.php'; require 'test3.php'; require_once 'test4.php'; $included_files = get_included_files(); foreach ($included_files as $filename) { echo "$filename "; } ?> The above example will output: abc.php test1.php test2.php test3.php test4.php NOTES
Note Files included using the auto_prepend_file configuration directive are not included in the returned array. SEE ALSO
include(3), include_once(3), require(3), require_once(3), get_required_files(3). PHP Documentation Group GET_INCLUDED_FILES(3)
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