determine version of .exe thru batch file
Posted: 02 Jun 2009, 16:00
I have been researching a way to check the version of a file thru a
batch file; i found the filever.exe utility, but then discovered it is
not compatable with Vista.
Is there a way to do this in Vista? My goal is to create a batch file
that compares the version of an .exe that is locally installed to the
"current" version on a server, replace the local version if it doesn't
match, and then open the local version.
in searching i found a command in one of the forums here (can't find
that post now to save my life!) that i believe was intended to check the
version of a dll, but in the example that was given they used an .exe -
so i was hoping to get it to work for me, but i'm getting errors.
Here's what i'm trying:
$env:"C:\Program Files\ProgramFolder\Program.exe".VersionInfo |
fl *
and here's what i'm getting:
The filename, directory name, or volume label syntax is
incorrect.
Help with this method or the correct method would be GREATLY
appreciated!
thanks in advance!
Rebekah
Ft. Rucker
--
woodrg
batch file; i found the filever.exe utility, but then discovered it is
not compatable with Vista.
Is there a way to do this in Vista? My goal is to create a batch file
that compares the version of an .exe that is locally installed to the
"current" version on a server, replace the local version if it doesn't
match, and then open the local version.
in searching i found a command in one of the forums here (can't find
that post now to save my life!) that i believe was intended to check the
version of a dll, but in the example that was given they used an .exe -
so i was hoping to get it to work for me, but i'm getting errors.
Here's what i'm trying:
$env:"C:\Program Files\ProgramFolder\Program.exe".VersionInfo |
fl *
and here's what i'm getting:
The filename, directory name, or volume label syntax is
incorrect.
Help with this method or the correct method would be GREATLY
appreciated!
thanks in advance!
Rebekah
Ft. Rucker
--
woodrg