The parameter 'exePath' is invalid in ConfigurationManager.OpenExe

[fran]

I have this code:
System.Configuration.Configuration config =
ConfigurationManager.OpenExeConfiguration("SAS.Int .ComunServicios.ComunProcesos.dll");
The filepath is not useful for me.

If I want to use an absolute path like:
System.Configuration.Configuration config =
ConfigurationManager.OpenExeConfiguration("C:\\SAS .Int.ComunServicios.ComunProcesos.dll");

Then I have the exception:
"An error occurred loading a configuration file: The parameter 'exePath' is
invalid.
Parameter name: exePath"

How can I use absolute path?

[chanmm]

System.Configuration.Configuration is used to open your .config file and not
dll perhaps.
http://msdn2.microsoft.com/en-us/lib...iguration.aspx

chanmm

"fran" <fr**@discussions.microsoft.com> wrote in message
news:79**********************************@microsof t.com...

I have this code:
System.Configuration.Configuration config =
ConfigurationManager.OpenExeConfiguration("SAS.Int .ComunServicios.ComunProcesos.dll");
The filepath is not useful for me.

If I want to use an absolute path like:
config =
ConfigurationManager.OpenExeConfiguration("C:\\SAS .Int.ComunServicios.ComunProcesos.dll");

Then I have the exception:
"An error occurred loading a configuration file: The parameter 'exePath'
is
invalid.
Parameter name: exePath"

How can I use absolute path?