From a DOS batch script under (64bit) Win7 I want to execute the following command
with the well known archiver tool 7zip:
start /LOW /B /WAIT D:\tool\7zip\7z.exe e "D:\work 2013\myarchive.rar"
This works fine.
Now I decided to install the 7zip archiver in another directory with a path that contains a blank.
Because it contains a blank I have to quote the program:
start /LOW /B /WAIT "D:\Program Files\7zip\7z.exe" e "D:\work 2013\myarchive.rar"
When I run the DOS script now an error popup appears telling me:
"The system cannot find the file e."
So
start /LOW /B /WAIT does not like a quoted path.
How can I execute otherwise the program with the mentioned parameters (and with start command) ?
Resp: how do I quote the path with blank otherwise?
Peter
Problems with start /LOW /B /WAIT and "quoted" program path
Moderator: DosItHelp
Re: Problems with start /LOW /B /WAIT and "quoted" program p
Have a look at the Help Message. The first quoted string is interpreted as window title.
Try
Regards
aGerman
Try
Code: Select all
start "" /LOW /B /WAIT "D:\Program Files\7zip\7z.exe" e "D:\work 2013\myarchive.rar"
Regards
aGerman
Re: Problems with start /LOW /B /WAIT and "quoted" program p
It works 
Thank you

Thank you