Question

        var ssh = new SshClient("ip", "user", "pass");
        var input = new MemoryStream(Encoding.ASCII.GetBytes("exit\r\n"));

        var shell = ssh.CreateShell(input, Console.Out, Console.Out, "xterm", 80, 24, 800, 600, "");
        shell.Stopped += delegate(object sender, EventArgs e)
        {
            Console.WriteLine("\nDisconnected...");
        };
        shell.Start();
        Thread.Sleep(1000 * 1000);
        shell.Stop();

错误就在这一行：

var shell = ssh.CreateShell(input, Console.Out, Console.Out, "xterm", 80, 24, 800, 600, "");

错误：错误3参数2：无法从'System.IO.TextWriter'转换为'System.IO.Stream'D：\ applications area \ test \ ConsoleApplication1 \ ConsoleApplication1 \ Program.cs 19 48 ConsoleApplication1

错误4参数3：无法从'System.IO.TextWriter'转换为'System.IO.Stream'D：\ applications area \ test \ ConsoleApplication1 \ ConsoleApplication1 \ Program.cs 19 61 ConsoleApplication1

任何解决方案？

Answer 1

Console.Out是TextWriter，而非流。

尝试这样的事情（警告：编译成头）

Stream stdout = Console.OpenStandardOutput();
var ssh = new SshClient("ip", "user", "pass");
var input = new MemoryStream(Encoding.ASCII.GetBytes("exit\r\n"));

var shell = ssh.CreateShell(input, stdout, stdout, "xterm", 80, 24, 800, 600, "");
shell.Stopped += delegate(object sender, EventArgs e)
{
    Console.WriteLine("\nDisconnected...");
};
shell.Start();
Thread.Sleep(1000 * 1000);
shell.Stop();

虽然注意两次传递似乎是错误的。您可能希望将第二个或第三个参数传递给ssh.CreateShell。可能Console.OpenStandardInput()，就此而言。

无法从'System.IO.TextWriter'转换为'System.IO.Stream'

1 个答案: