Windows – Finding the path of the program that will execute from the command line in Windows

command linepathwindows

Say I have a program X.EXE installed in folder c:\abcd\happy\ on the system. The folder is on the system path. Now suppose there is another program on the system that's also called X.EXE but is installed in folder c:\windows\.

Is it possible to quickly find out from the command line that if I type in X.EXE which of the two X.EXE's will get launched? (but without having to dir search or look at the process details in Task Manager).

Maybe some sort of in-built command, or some program out there that can do something like this? :

detect_program_path X.EXE

Best Answer

Use the where command. The first result in the list is the one that will execute.

C:\> where notepad
C:\Windows\System32\notepad.exe
C:\Windows\notepad.exe

According to this blog post, where.exe is included with Windows Server 2003 and later, so this should just work with Vista, Win 7, et al.

On Linux, the equivalent is the which command, e.g. which ssh.

Related Solutions

Windows – n equivalent of ‘which’ on the Windows command line?

Windows Server 2003 and later (i.e. anything after Windows XP 32 bit) provide the where.exe program which does some of what which does, though it matches all types of files, not just executable commands. (It does not match built-in shell commands like cd.) It will even accept wildcards, so where nt* finds all files in your %PATH% and current directory whose names start with nt.

Try where /? for help.

Note that Windows PowerShell defines where as an alias for the Where-Object cmdlet, so if you want where.exe, you need to type the full name instead of omitting the .exe extension.

Windows – How to get the application exit code from a Windows command line

A pseudo environment variable named errorlevel stores the exit code:

echo Exit Code is %errorlevel%

Also, the if command has a special syntax:

if errorlevel

See if /? for details.

Example

@echo off
my_nify_exe.exe
if errorlevel 1 (
   echo Failure Reason Given is %errorlevel%
   exit /b %errorlevel%
)

Warning: If you set an environment variable name errorlevel, %errorlevel% will return that value and not the exit code. Use (set errorlevel=) to clear the environment variable, allowing access to the true value of errorlevel via the %errorlevel% environment variable.

Best Answer

Related Solutions

Windows – n equivalent of ‘which’ on the Windows command line?

Windows – How to get the application exit code from a Windows command line

Example

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