Question

我有一个程序WFA也有和命令窗口。我用AllocConsole（）打开窗口;当我关闭控制台窗口时，我使用FreeConsole（）;但是当我用AllocConsole（）再次打开它时;我想写它并从中读取它会抛出异常。代码：

using System;
using System.Collections.Generic;
using System.ComponentModel;
using System.Data;
using System.Drawing;
using System.Linq;
using System.Text;
using System.Threading.Tasks;
using System.Windows.Forms;
using System.Management;
using System.Management.Instrumentation;
using System.Collections.Specialized;
using System.Threading;
using System.Runtime.InteropServices;


    namespace WindowsFormsApplication2
{

class classx
{

    [DllImport("kernel32.dll")]
    public static extern Int32 AllocConsole();
    [DllImport("kernel32.dll")]
    public static extern bool FreeConsole();
    [DllImport("kernel32")]
    public static extern bool AttachConsole();
    [DllImport("kernel32")]
    public static extern bool GetConsoleWindow();
    public static bool z = false;
    [DllImport("kernel32")]
    public static extern bool SetConsoleCtrlHandler(HandlerRoutine HandlerRoutine, bool Add);
    public delegate bool HandlerRoutine(uint dwControlType);
}



public partial class Form1 : Form
{
    NotifyIcon icontask;
    Icon iconone_active;
    Icon iconone_inactive;
    /*Icon icontwo;
    Icon iconthree;
    Icon iconfour;
    Icon iconfive;*/
    Thread Threadworkermy;

    public Form1()
    {
        InitializeComponent();
        this.WindowState = FormWindowState.Minimized;
        this.ShowInTaskbar = false;
        iconone_active = new Icon(".../iconone_active.ico");
        iconone_inactive = new Icon(".../iconone_inactive.ico");
        icontask = new NotifyIcon();
        icontask.Icon = iconone_active;
        icontask.Visible = true;
        Threadworkermy = new Thread(new ThreadStart(checkActivityThread));
        Threadworkermy.Start();

        MenuItem Nameapp = new MenuItem("xr");
        MenuItem quitappitem = new MenuItem("quit program");
        MenuItem OpenGUI = new MenuItem("Open GUI");
        MenuItem Advancedmodewindow = new MenuItem("x");
        ContextMenu contextmenu = new ContextMenu();

        quitappitem.Click += quitappitem_click;
        OpenGUI.Click += OpenGUI_click;
        Advancedmodewindow.Click += Advancedmodewindow_click;
        contextmenu.MenuItems.Add(Nameapp);
        contextmenu.MenuItems[0].Enabled = false;
        contextmenu.MenuItems.Add("-");
        contextmenu.MenuItems.Add(OpenGUI);
        contextmenu.MenuItems.Add(Advancedmodewindow);
        contextmenu.MenuItems.Add("-");
        contextmenu.MenuItems.Add(quitappitem);
        icontask.ContextMenu = contextmenu;

        icontask.Icon = iconone_active;
        icontask.Visible = true;
    }

    private void Advancedmodewindow_click(object sender, EventArgs e)
    {
        classx.AllocConsole();
        Console.WriteLine("X");
        classx.FreeConsole();
    }

    private void OpenGUI_click(object sender, EventArgs e)
    {
        this.ShowInTaskbar = true;
        this.WindowState = FormWindowState.Normal;  
    }

    private void quitappitem_click(object sender, EventArgs e)
    {
        Threadworkermy.Abort();
        icontask.Dispose();
        this.Close();
    }

    public void checkActivityThread()
    {
        try
        {
            while(true)
            {
                Thread.Sleep(100);   
            }
        } catch(ThreadAbortException tbe)
        {

        }

    }

    private void button1_Click(object sender, EventArgs e)
    {
        this.WindowState = FormWindowState.Minimized;
        this.ShowInTaskbar = false;
    }


}
}

它抛出的异常＆＃39; System.IO.IOException＆＃39;在mscorlib.dll中附加信息：句柄无效。对于那些会说改变类型的人，我不能。（需要WFA申请）

为什么没有

CONSOLE.WRITELINE

和类似的代码工作？

Answer 1

异常不是来自打开/关闭控制台。当你想用控制台写/读东西时，它就会到来。您必须重置控制台的标准IO。每次你必须在AllocConsole（）之后执行此操作。请检查下面的代码，它已经过测试。

CODE：

//routes
var indexRoute = require('./app/routes/index');
app.use('/', indexRoute);

var apiRoute = require('./API/routes/index.js')
app.use('/api', apiRoute);

// uncomment after placing your favicon in /public
//app.use(favicon(path.join(__dirname, 'public', 'favicon.ico')));
app.use(logger('dev'));
app.use(bodyParser.json());
app.use(bodyParser.urlencoded({ extended: false }));

在表单应用程序

1 个答案: