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Full Discussion: passing file extension using external variable
Top Forums Shell Programming and Scripting passing file extension using external variable Post 302706261 by Skrynesaver on Wednesday 26th of September 2012 11:14:18 AM
Old 09-26-2012
If I understand your intent correctly the following should work.

Code:
#!/bin/bash
FROM=$1 
TO=$2 
FILETYPE=$3 

find $FROM -name \*.$FILETYPE -exec mv {} $TO \;

 

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LEARN ABOUT PHP

filetype

FILETYPE(3)								 1							       FILETYPE(3)

filetype - Gets file type

SYNOPSIS

       string filetype (string	$filename)

DESCRIPTION

	Returns the type of the given file.

PARAMETERS

	      o $filename
		- Path to the file.

RETURN VALUES

	Returns the type of the file. Possible values are fifo, char, dir, block, link, file, socket and unknown.

	Returns FALSE if an error occurs. filetype(3) will also produce an E_NOTICE message if the stat call fails or if the file type is unknown.

EXAMPLES

       Example #1

	      filetype(3) example

	      <?php

	      echo filetype('/etc/passwd');  // file
	      echo filetype('/etc/');	     // dir

	      ?>

ERRORS
/EXCEPTIONS
	Upon failure, an E_WARNING is emitted.

NOTES

       Note

	      The results of this function are cached. See clearstatcache(3) for more details.

       Tip

	      As  of  PHP  5.0.0,  this function can also be used with some URL wrappers. Refer to "Supported Protocols and Wrappers" to determine
	      which wrappers support stat(3) family of functionality.

SEE ALSO

       is_dir(3), is_file(3), is_link(3), file_exists(3), mime_content_type(3), pathinfo(3), stat(3).

PHP Documentation Group 													       FILETYPE(3)
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