Question

是否可以确定是否可以看到控件的至少一个像素（通过属性或可能使用事件通知）。

注意：即使其他窗口隐藏了控件，我也不会寻找可以返回true的Visible属性

Answer 1

实用的解决方案是使用表单的GetChildAtPoint（）方法，传递控件的4个角。如果其中一个返回true，则控件肯定可见。它不是100％可靠，所有4个角都可以被另一个控件重叠，但仍然可以看到内部的一部分。我不担心，太奇怪了。

public bool ChildReallyVisible(Control child) {
    var pos = this.PointToClient(child.PointToScreen(Point.Empty));

    //Test the top left
    if (this.GetChildAtPoint(pos) == child) return true;

    //Test the top right
    if (this.GetChildAtPoint(new Point(pos.X + child.Width - 1, pos.Y)) == child) return true;

    //Test the bottom left
    if (this.GetChildAtPoint(new Point(pos.X, pos.Y + child.Height -1)) == child) return true;

    //Test the bottom right
    if (this.GetChildAtPoint(new Point(pos.X + child.Width - 1, pos.Y + child.Height -1)) == child) return true;

    return false;
}

Answer 2

您可以使控件无效，然后调用GetUpdateRect（Win32 api函数）来查找它。但它确实会产生重画的副作用。

Answer 3

为了方便您之前answer的问题。

以下是使用GetUpdateRect函数处理jdv-Jan de Vaan时所需的源代码。

[System.Runtime.InteropServices.StructLayout(System.Runtime.InteropServices.LayoutKind.Sequential)]
internal struct RECT
{
    public int Left;
    public int Top;
    public int Right;
    public int Bottom;
    public int Width { get { return this.Right - this.Left; } }
    public int Height { get { return this.Bottom - this.Top; } }
}
[System.Runtime.InteropServices.DllImport("user32.dll")]
internal static extern bool GetUpdateRect(IntPtr hWnd, ref RECT rect, bool bErase);
public static bool IsControlVisibleToUser(Control control)
{
    control.Invalidate();
    Rectangle bounds = control.Bounds;
    RECT rect = new RECT { Left = bounds.Left, Right = bounds.Right, Top = bounds.Top, Bottom = bounds.Bottom };
    return GetUpdateRect(control.Handle, ref rect, false);
}

当您需要检查指定是否可见时，只需执行以下操作：

if (IsControlVisibleToUser(controlName) == true)
{
    // The Specified Control is visible.
    // ... do something 
}
else
{
    // Control is not visible.
    // ... do something else
}

祝你好运。

Answer 4

如果控件可见，将重复调用Paint事件。

通常对于不可见的控件，不会调用此事件。

Answer 5

受Hans的回答启发，我以这种方式实现了这种行为;

    [DllImport("user32.dll")]
    static extern IntPtr WindowFromPoint(POINT Point);

    [StructLayout(LayoutKind.Sequential)]
    public struct POINT
    {
        public int X;
        public int Y;

        public POINT(int x, int y)
        {
            this.X = x;
            this.Y = y;
        }

        public static implicit operator System.Drawing.Point(POINT p)
        {
            return new System.Drawing.Point(p.X, p.Y);
        }

        public static implicit operator POINT(System.Drawing.Point p)
        {
            return new POINT(p.X, p.Y);
        }
    }

    public static bool IsControlVisibleToUser(this Control control)
    {
        var pos = control.PointToScreen(control.Location);
        var pointsToCheck = new POINT[]
                                {
                                    pos,
                                    new Point(pos.X + control.Width - 1, pos.Y),
                                    new Point(pos.X, pos.Y + control.Height - 1),
                                    new Point(pos.X + control.Width - 1, pos.Y + control.Height - 1),
                                    new Point(pos.X + control.Width/2, pos.Y + control.Height/2)
                                };

        foreach (var p in pointsToCheck)
        {
            var hwnd = WindowFromPoint(p);
            var other = Control.FromChildHandle(hwnd);
            if (other == null)
                continue;

            if (control == other || control.Contains(other))
                return true;
        }

        return false;
    }

Answer 6

对上述内容进行了尝试，但是即使winform被另一个应用程序覆盖了，它也始终如一。

最后使用以下内容（在我的winform类内部）：

using System;
using System.Runtime.InteropServices;
using System.Windows.Forms;

namespace yourNameSpace
{
    public class Myform : Form
    {

        private void someFuncInvokedByTimerOnMainThread()
        {
            bool isVisible = isControlVisible(this);
            // do something.
        }

        [DllImport("user32.dll")]
        static extern IntPtr WindowFromPoint(System.Drawing.Point p);


        ///<summary><para>------------------------------------------------------------------------------------</para>
        ///
        ///<para>           Returns true if the control is visible on screen, false otherwise.                </para>
        ///
        ///<para>------------------------------------------------------------------------------------</para></summary>
        private bool isControlVisible(Control control)
        {
            bool result = false;
            if (control != null)
            {
                var pos = control.PointToScreen(System.Drawing.Point.Empty);
                var handle = WindowFromPoint(new System.Drawing.Point(pos.X + 10, pos.Y + 10)); // +10 to disregard padding   
                result = (control.Handle == handle); // should be equal if control is visible
            }
            return result;
        }
    }
}

Answer 7

您可以使用控件的布局事件。当控件进入屏幕并尝试布局其子控件时触发它。

例如，让我们说TabPage中有GroupBox。
单击相关选项卡后，布局事件将触发第一个标签页，然后触发GroupBox
您可以将它与可见性属性结合使用

Answer 8

您可以检查父控件的可见性。

    protected override void OnParentVisibleChanged(EventArgs e)
    {
        base.OnParentVisibleChanged(e);
        isVisible = true;
    }

Answer 9

我有点完成了Hans Passant的回答。下面的函数测试表单的所有四个角。

        /// <summary>
    /// determines if a form is on top and really visible.
    /// a problem you ran into is that form.invalidate returns true, even if another form is on top of it. 
    /// this function avoids that situation
    /// code and discussion:
    /// https://stackoverflow.com/questions/4747935/c-sharp-winform-check-if-control-is-physicaly-visible
    /// </summary>
    /// <param name="child"></param>
    /// <returns></returns>
    public bool ChildReallyVisible(Control child)
    {
        bool result = false;

        var pos = this.PointToClient(child.PointToScreen(Point.Empty));

        result = this.GetChildAtPoint(pos) == child;
        //this 'if's cause the condition only to be checked if the result is true, otherwise it will stay false to the end
        if(result)
        {
            result = (this.GetChildAtPoint(new Point(pos.X + child.Width - 1, pos.Y)) == child);
        }
        if(result)
        {
            result = (this.GetChildAtPoint(new Point(pos.X, pos.Y + child.Height - 1)) == child);
        }
        if(result)
        {
            result = (this.GetChildAtPoint(new Point(pos.X + child.Width - 1, pos.Y + child.Height - 1)) == child) ;
        }
        return result;
    }

C＃winform检查控件是否在物理上可见

9 个答案: