Process.Start filename using %temp%

问题

For some odd reaseon this code fails:

p.StartInfo.FileName = @"%temp%\SSCERuntime_x86-ENU.msi";

and this code succes:

p.StartInfo.FileName = @"C:\Users\USERNAME\AppData\Local\Temp\SSCERuntime_x86-ENU.msi";

Is there any reason I am missing?

Note I just copied the path, I don't think the rest of the code is needed but I'll put it anyway:

Process p = new Process();
p.StartInfo.FileName = @"%temp%\SSCERuntime_x86-ENU.msi";
p.StartInfo.Arguments = "/passive";
p.Start();

回答1:

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");

回答2:

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;

回答3:

%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");

回答4:

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

1. Working for any environment variable (%APPDATA%, %COMMONPROGRAMFILES%, etc), and
2. Working anywhere in the string (e.g. "%temp%\%username%\foo.txt")

回答5:

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.

来源：https://stackoverflow.com/questions/9675430/process-start-filename-using-temp