I do a search :
set keypath=HKEY_LOCAL_MACHINE\SOFTWARE\Microsoft\Windows\CurrentVersion\Installer\Folders
reg query "!keypath!" /s /f "%mysearch%"
and i want to delete all the result that match with mysearch
But i have a problem with some results (!myresult!) like this :
"c:\Program Files\mysearch\"
"c:\Program Data\mysearch\"
when i do :
reg delete "!keypath!" /v "!myresult!" /f
it's not working (i think because !myresult! ending with \ and "), asking me confirmation before deleting and don't work because the system no find key
when i do :
set myresult=!myresult:\=\\!
reg delete "!keypath!" /v "!myresult!" /f
it's not working, not asking me confirmation but the system no find key too
when i do :
reg delete "!keypath!" /v !myresult! /f
it's not working i have a syntax error (because of space in !myresult!), but with no space in !myresult! it works
If somebody have an idea.
Thanks.
Your ending backslash is escaping the double quote (more here).
You can check if the value ends in a slash and double it.
if "!myresult:~-1!"=="\" set "myresult=!myresult!\"
reg delete "!keypath!" /v "!myresult!" /f
But this will fail when you have a double backslash ended value, so probably the better option could be to directly use
reg delete "!keypath!" /f /v "!myresult!
The /f has been moved AND the closing quote has been removed. To use this, ensure there is not any space at the end of the line. As the quote has not been closed, any space will be included inside the "quoted" value.