For some odd reaseon this code fails:
p.StartInfo.FileName = @\"%temp%\\SSCERuntime_x86-ENU.msi\";
and this code succes:
p.
The Process class does not expand strings with environment variables (i.e. %temp%
).
If you want to use environment variables to build the FileName property then you'll have to get the environment variables (using the GetEnvironmentVariable method on the Environment class) and perform the substitution yourself, like so:
// Construct the path.
string temp = Environment.GetEnvironmentVariable("temp");
string path = Path.Combine(temp, "SSCERuntime_x86-ENU.msi");
// Launch the process.
Process p = new Process();
p.StartInfo.FileName = path;
p.StartInfo.Arguments = "/passive";
p.Start();
Additionally, you can use the ExpandEnvironmentVariables method with your original string like so:
p.StartInfo.FileName =
Environment.ExpandEnvironmentVariables(@"%temp%\SSCERuntime_x86-ENU.msi");
%TEMP% is parsed and evaluated by Command Shell. You could use use Path.GetTempPath() and Path.Combine for this purpose.
p.StartInfo.FileName = Path.Combine(Path.GetTempPath(), @"SSCERuntime_x86-ENU.msi");
Try this:
string tempPath = Environment.GetEnvironmentVariable("Temp");
Then concat it in:
p.StartInfo.FileName = Path.Combine(tempPath, "SSCERuntime_x86-ENU.msi");
Casper beat me to the punch on the explaination, but the Process.Start method basically treats it literally instead of intrperting it as the shell would.
You can use the Environment.ExpandEnvironmentVariables
to expand environment variables within a string, then pass that to the Process
class:
p.StartInfo.FileName = Environment.ExpandEnvironmentVariables(@"%temp%\SSCERuntime_x86-ENU.msi");
This has the added benefits of
The %temp%
portion of the string is being interpreted literally instead of being replaced with the appropriate environment variable. You'll need to manually expand it
var temp = Environment.GetEnvironmentVariable("temp");
var fileName = Path.Combine(temp, "SSCERuntime_x86-ENU.msi");
p.StartInfo.FileName = fileName;