C#处理标准输入

时间:2013-10-23 20:50:39

标签: c# stdin

我目前正在尝试通过命令行断开网络文件夹,并使用以下代码:

System.Diagnostics.Process process2 = new System.Diagnostics.Process();
System.Diagnostics.ProcessStartInfo startInfo = new System.Diagnostics.ProcessStartInfo();
startInfo.WindowStyle = System.Diagnostics.ProcessWindowStyle.Hidden;
startInfo.FileName = "cmd.exe";
startInfo.Arguments = "/C NET USE F: /delete";
startInfo.RedirectStandardError = true;
startInfo.RedirectStandardInput = true;
startInfo.RedirectStandardOutput = true;
startInfo.UseShellExecute = false;
startInfo.CreateNoWindow = true;
process2.StartInfo = startInfo;
process2.Start();

StreamWriter sw = process2.StandardInput;
sw.WriteLine("Y");
sw.Close();

process2.WaitForExit();
process2.Close();

偶尔,我收到消息“继续断开并强行关闭它们是否可以?(是/否)[N]”,我想回答“Y”,但我似乎遇到了问题工作。

有谁知道为什么我的代码没有输入“Y”到标准输入?

1 个答案:

答案 0 :(得分:1)

使用 代码下面的消息“是否可以继续断开连接并强行关闭?(是/否)[N]”,回复“ Y“

static void Main(string[] args)
{
    System.Diagnostics.Process process2 = new System.Diagnostics.Process();
    System.Diagnostics.ProcessStartInfo startInfo = new System.Diagnostics.ProcessStartInfo();
    startInfo.WindowStyle = System.Diagnostics.ProcessWindowStyle.Hidden;
    startInfo.FileName = "cmd.exe";
    startInfo.Arguments = "/C NET USE F: /delete";
    startInfo.RedirectStandardError = true;
    startInfo.RedirectStandardInput = true;
    startInfo.RedirectStandardOutput = true;
    startInfo.UseShellExecute = false;
    startInfo.CreateNoWindow = true;
    process2.StartInfo = startInfo;
    process2.Start();

    Read(process2.StandardOutput);
    Read(process2.StandardError);

    while (true)
        process2.StandardInput.WriteLine("Y");

}

private static void Read(StreamReader reader)
{
    new Thread(() =>
    {
        while (true)
        {
            int current;
            while ((current = reader.Read()) >= 0)
                Console.Write((char)current);
        }
    }).Start();
}

我认为这可能对你有帮助..