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GooliveR
Asked:2020-08-15 05:38:27 +0000 UTC2020-08-15 05:38:27 +0000 UTC

如何将控制台的输出从另一个类传递到 TextBox

  • 772

我NewClass.cs在其中收集数据 如何显示来自此类（通过参数）的数据，这些数据ipconfig位于表单上？！cmdTextBox

public static class NewClass
{
     public static void Test()
     {
         ProcessStartInfo Info = new ProcessStartInfo
         {
            FileName = "cmd.exe",
            Arguments = "ipconfig /all",
            CreateNoWindow = false,
            UseShellExecute = false,
            WindowStyle = ProcessWindowStyle.Normal
         };
         Process result = Process.Start(Info);
         result.WaitForExit();
    }
}

在表单TextBox上，​​您需要Test(с передачей аргументов)从类进行方法调用，NewClass.cs以便命令行中的所有内容都(cmd.exe)输出到此，TextBox而不会出现控制台。
如何实施？\

c#

1 个回答

  1. Best Answer

    IProgress回调在这里非常方便。

    private void button1_click(object sender, EventArgs e)
{
    IProgress<string> progress = new Progress<string>(s =>
    {
        textBox1.SelectionStart = textBox1.Text.Length;
        textBox1.SelectedText = s + Environment.NewLine;
        textBox1.ScrollToCaret();
    });
    NewClass.Test(progress);
}

    public static class NewClass
{
    public static void Test(IProgress<string> progress)
    {
        Process process = new Process();
        process.StartInfo.FileName = "cmd.exe";
        process.StartInfo.Arguments = "/K ipconfig /all";
        process.StartInfo.CreateNoWindow = true;
        process.StartInfo.UseShellExecute = false;
        process.StartInfo.RedirectStandardOutput = true;
        process.StartInfo.RedirectStandardError = true;
        process.StartInfo.StandardOutputEncoding = Encoding.GetEncoding(866);
        process.StartInfo.StandardErrorEncoding = Encoding.GetEncoding(866);
        var handler = new DataReceivedEventHandler((s, e) => progress?.Report(e.Data));
        process.OutputDataReceived += handler;
        process.ErrorDataReceived += handler;
        process.Start();
        process.BeginOutputReadLine();
        process.BeginErrorReadLine();
    }
}

    在此处输入图像描述

    • 1

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