Batch command to move files to a new directory

I want to write a batch job that when executed will grab all the files in the C:\Test\Log folder and move them to a new directory in the C:\Test. This new directory will have a name called "Backup-" and CURRENT DATE. 开发者_开发百科

So once completed, the log folder should be empty with all the files now located in the new folder.

I know I would have to use the MOVE command, but have no idea how to dynamically create a new folder, and use the date to name it.

Something like this might help:

SET Today=%Date:~10,4%%Date:~4,2%%Date:~7,2%
mkdir C:\Test\Backup-%Today%
move C:\Test\Log\*.* C:\Test\Backup-%Today%\
SET Today=

The important part is the first line. It takes the output of the internal DATE value and parses it into an environmental variable named Today, in the format CCYYMMDD, as in '20110407`.

The %Date:~10,4% says to extract a *substring of the Date environmental variable 'Thu 04/07/2011' (built in - type echo %Date% at a command prompt) starting at position 10 for 4 characters (2011). It then concatenates another substring of Date: starting at position 4 for 2 chars (04), and then concats two additional characters starting at position 7 (07).

*The substring value starting points are 0-based.

You may need to adjust these values depending on the date format in your locale, but this should give you a starting point.

this will also work, if you like

 xcopy  C:\Test\Log "c:\Test\Backup-%date:~4,2%-%date:~7,2%-%date:~10,4%_%time:~0,2%%time:~3,2%" /s /i
 del C:\Test\Log