Create folder from filename and copy file to the folder

[RenegadeXO]

FIrst of all, I am new to AutoIt and scripting in general. I can read some code but not and expert by know means. What I am needing, hopefully is simple.

Scenario: I have two servers (old and new Listserv servers). On the old server I have several files that will need to be copied to the new server. The problem is the new server file structure is slightly different. Each file, on the old server, will need a Directory with the same name as the file on the new server. Example...(abc123.list (old server file) ---> ABC123\abc123.list on the new server. Hopefully there is an easy way to do this. If you notice, the directory name will need to also drop the extension.

[water]

Something like this:

$sSourceFilename = "abc123.list"
$sTargetDrive = "C:"
$aTargetDirectory = StringSplit($sSourceFilename, ".")
DirCreate($sTargetDrive & "" & $aTargetDirectory[1])
FileCopy($sSourceFilename, $sTargetDrive & "" & $aTargetDirectory[1] & "")

What's missing is some error checking code.

[Marlo]

I just threw this together:

To use it just save the script to the directory where the files are and run it.

#include <File.au3>

$Files = _FileListToArray(@ScriptDir, "*", 1)

For $I = 1 To $Files[0]

$Name = StringLeft($Files[$I], StringInStr($Files[$I], ".")-1)
If Not FileExists(@ScriptDir & "" & $Name) Then DirCreate(@ScriptDir & "" & $Name)

FileCopy(@ScriptDir & "" & $Files[$I], @ScriptDir & "" & $Name)

Next

MsgBox(0, "", "Done!")

It copies the file opposed to moving them just in-case something goes wrong and it also lacks recursion (but you didn't mention that so...)

Edited September 12, 2012 by Marlo

[RenegadeXO]

Thank you both for the response, I will try this today to see which works best for me. I will also post later the results. Love this forum!

[water]

Glad to be of service

[guinness]

Here is my take on the output(s) you provided.

Local $sSourceFilePath = @ScriptDir & 'abc123.list'
Local $sDestinationFilePath = 'C:'

_FileCopyToFolder($sSourceFilePath, $sDestinationFilePath, 1 + 8) ; See FileCopy parameters.

Func _FileCopyToFolder($sSource, $sDestination, $iFlag = 0)
    Local $sFilePath = StringTrimLeft($sSource, StringInStr($sSource, '', 2, -1))
    $sDestination = StringRegExpReplace($sDestination, '[/]*$', '') ; Remove trailing slashes.
    $sDestination = $sDestination & '' & StringUpper(StringLeft($sFilePath, StringInStr($sFilePath, '.', 2, -1) - 1)) & '' & $sFilePath
    MsgBox(4096, '', $sDestination)
;~  Return FileCopy($sSource, $sDestination, $iFlag)
EndFunc   ;==>_FileCopyToFolder

Edited September 13, 2012 by guinness