Process.StartInfo.FileName是否接受长文件名？

看起来不是.

如果我将文件名转换为其短值,则Process.Start()可以正常工作.

Process runScripts = new Process();
runScripts.StartInfo.FileName = @"C:\long file path\run.cmd";
runScripts.StartInfo.WindowStyle = ProcessWindowStyle.Hidden;
runScripts.StartInfo.UseShellExecute = true;
runScripts.StartInfo.RedirectStandardOutput = false;
runScripts.Start();

上面的代码失败了.但...

Process runScripts = new Process();
runScripts.StartInfo.FileName = @"C:\short\file\path\run.cmd";
runScripts.StartInfo.WindowStyle = ProcessWindowStyle.Hidden;
runScripts.StartInfo.UseShellExecute = true;
runScripts.StartInfo.RedirectStandardOutput = false;
runScripts.Start();

成功.

我设法通过将长路径名称转换为短路径名称来解决这个问题.但我发现这一点有点惊讶.有关于此的任何原因或背景信息？

谢谢.

更新1
Microsoft .NET Framework版本2.0.50727

这很好奇,我重现了这个行为,实际上,它没有像你说的那样执行,但修改它,它起作用了:

    System.Diagnostics.Process runScripts = new System.Diagnostics.Process();
    runScripts.StartInfo.FileName = @"run.cmd";
    // new
    runScripts.StartInfo.WorkingDirectory = @"C:\long file path";
    runScripts.StartInfo.WindowStyle = System.Diagnostics.ProcessWindowStyle.Hidden;
    runScripts.StartInfo.UseShellExecute = true;
    runScripts.StartInfo.RedirectStandardOutput = false;
    runScripts.Start();