Question

我正在尝试让应用程序获取用户单击的窗口的鼠标单击位置和标题（名称）。对于Ex。如果我点击任务栏中打开的outlook，那么我需要将应用程序名称存储为数据库中的“Microsoft Outlook”。

提前致谢....

Answer 1

检查此代码：How to get active window handle and title

命名空间：

using System;
using System.Runtime.InteropServices;
using System.Text;
using System.Threading;

方法：

[DllImport("user32.dll")]
private static extern IntPtr GetForegroundWindow();

[DllImport("user32.dll", CharSet = CharSet.Auto, SetLastError = true)]
private static extern int GetWindowText(IntPtr hWnd, StringBuilder lpString, int nMaxCount);

电话：

// get handle
IntPtr handle = GetForegroundWindow();

// get title
const int count = 512;
var text = new StringBuilder(count);

if (GetWindowText(handle, text, count) > 0)
{
    MessageBox.Show(text.ToString());
}

我几次使用这段代码。真的很容易使用。你可以设置一个每10ms启动一次的计时器。节省2个变量。一个是活动窗口，另一个是最后一个窗口。在伪代码中说：如果是newWindow！= oldWindow - ＆gt; listView.Add（窗口）。

看起来像这样：

public partial class Form1 : Form
    {
        // DECLARE GLOBALS //
        [DllImport("user32.dll")]
        private static extern IntPtr GetForegroundWindow();

        [DllImport("user32.dll", CharSet = CharSet.Auto, SetLastError = true)]
        private static extern int GetWindowText(IntPtr hWnd, StringBuilder lpString, int nMaxCount);

        public static string oldWindow = "";
        public static string currentWindow = "";

        // TIMER EVENT //
        private void timerCheckFocus_Tick(object sender, EventArgs e)
        {
            // get handle
            IntPtr handle = GetForegroundWindow();

            // get title
            const int count = 512;
            var currentWindow = new StringBuilder(count);

            if (currentWindow.ToString() != oldWindow)
            {
                // add your window to a listView //
                oldWindow = currentWindow.ToString();
            }
        }

Answer 2

你可以导入user32.dll

[DllImport("user32.dll", ExactSpelling = true, SetLastError = true)]
internal static extern IntPtr GetForegroundWindow();

然后例如使用计时器查找活动的应用程序 process.MainWindowTitle为您提供当前窗口/应用程序的名称。

private void dispatcherTimer_Tick(object sender, EventArgs e)
{
    IntPtr activeWindow = GetForegroundWindow();
    List<String> strListProcesses = new List<string>();
    foreach (Process process in Process.GetProcesses())
    {
        try
        {
            if (activeWindow == process.MainWindowHandle)
            {
                newApplication = process.MainWindowTitle;
            }
        }
        catch (System.ComponentModel.Win32Exception ex)
        {

        }
    }
}

如何根据c＃中的鼠标点击获取活动窗口名称

2 个答案: