I want to create a few batch files to automate a program.
My question is when I create the batch file, what is the current directory? Is it the directory where the file is located or is it the same directory that appears in the command prompt, or something else?
From within your batch file:
%cd% refers to the current working directory (variable)
%~dp0 refers to the full path to the batch file's directory (static)
%~dpnx0 and %~f0 both refer to the full path to the batch directory and file name (static).
See also: What does %~dp0 mean, and how does it work?
It usually is the directory from which the batch file is started, but if you start the batch file from a shortcut, a different starting directory could be given. Also, when you'r in cmd, and your current directory is c:\dir3
, you can still start the batch file using c:\dir1\dir2\batch.bat
in which case, the current directory will be c:\dir3
.
In a batch file, %cd% is the most commonly used command for the current directory, although you can set your own variable:
set mypath=%cd%
echo %mypath% (where %mypath% is the current directory that the batch file is sitting in)
So say you were wanting to open Myprog.exe. If it was in the same folder, you would use the command:
start %mypath%\Myprog.exe
That would open Myprog from the current folder.
The other option is to make a directory in C: called AutomatePrograms. Then, you transfer your files to that folder then you can open them using the following command:
start "" "C:\AutomatePrograms\Myprog1.exe"
start "" "C:\AutomatePrograms\Myprog2.exe"
start "" "C:\AutomatePrograms\Myprog3.exe"
%cd%
is variable, so running the batch file from Windows Explorer as admin will give C:\WINDOWS\system32
which is almost certainly not what you want. %~dp0
is more consistent.
Say you were opening a file in your current directory. The command would be:
start %cd%\filename.filetype
I hope I answered your question.
It is the directory from where you run the command to execute your batch file.
As mentioned in the above answers you can add the below command to your script to verify:
> set current_dir=%cd%
> echo %current_dir%
It is the directory from where you start the batch file. E.g. if your batch is in c:\dir1\dir2
and you do cd c:\dir3
, then run the batch, the current directory will be c:\dir3
.
%__CD__% , %CD% , %=C:%
There's also another dynamic variable %__CD__%
which points to the current directory but unlike %CD%
it has a backslash at the end. This can be useful if you want to append files to the current directory. Also %CD%
does not work under disabled extensions environment ,but %__CD__%
always works.
With %=C:%
%=D:%
you can access the last accessed directory for the corresponding drive. If the variable is not defined you haven't accessed the drive on the current cmd session.
And %__APPDIR__%
expands to the executable that runs the current script a.k.a. cmd.exe
directory.
Just my 2 cents. The following command fails if called from batch file (Windows 7) placed on a pendrive:
%SystemRoot%\System32\xcopy.exe /e /i "%cd%Ala" "C:\KS\Ala\"
But this does the job:
%SystemRoot%\System32\xcopy.exe /e /i "%~dp0Ala" "C:\KS\Ala\"
%~dp0
Your bat file should be in the directory that the bat file is/was in when you opened it. However if you want to put it into a different directory you can do so with cd [whatever directory]
Success story sharing
.cmd
file and running as administrator,%cd%
givesC:\WINDOWS\system32
and%~dp0
gives the batch file directory with trailing slash.%~dp0
is referencing drive and path of argument 0 of the batch file arguments. It is NOT an environment variable reference. It is a batch file argument reference.%~dp0
expands to full path of currently processed batch file. A batch fileC:\Temp\Devlopement & Test()!\Main.bat
with%~dp0
expanding toC:\Temp\Devlopement & Test()!\
on execution cancall
the batch fileC:\Temp\SubBatch.cmd
, for example withcall "%~dp0..\SubBatch.cmd"
, containing also%~dp0
expanding on execution toC:\Temp\
. Runcall /?
in a cmd window for help.