Control. Visible Свойство

Определение

Возвращает или задает значение, указывающее, отображаются ли элемент управления и все его дочерние элементы управления. Gets or sets a value indicating whether the control and all its child controls are displayed.

Значение свойства

Значение true , если элемент управления и все его дочерние элементы управления отображаются; в противном случае — значение false . true if the control and all its child controls are displayed; otherwise, false . Значение по умолчанию — true . The default is true .

Примеры

В следующем примере кода используются производные классы VScrollBar и HScrollBar устанавливаются Visible значения их свойств в зависимости от размера Image отображаемого PictureBox элемента управления. The following code example uses the derived classes VScrollBar and HScrollBar and sets their Visible property values, based on the size of an Image being displayed in a PictureBox control. В этом примере требуется, чтобы объект был PictureBox создан на форме, а HScrollBar VScrollBar элементы управления и были созданы в PictureBox . This example requires that a PictureBox has been created on a form and that HScrollBar and VScrollBar controls have been created on the PictureBox. Этот код должен вызываться, когда изображение загружается в графическое окно и Resize событие формы. This code should be called when the image is loaded into the picture box and by the Resize event of the form.

Комментарии

Обратите внимание, что даже если Visible параметр имеет значение true , элемент управления может быть невидимым для пользователя, если он скрыт позади других элементов управления. Note that even if Visible is set to true , the control might not be visible to the user if it is obscured behind other controls.

How can I check if a window has WS_VISIBLE to set? (or if is visible)

How can I do it? It’s an external window, not from my program. Thanks

2 Answers 2

Do you have an HWND to the window? If not, then you will need to obtain the window handle somehow, for example through FindWindow() (or FindWindowEx() ).

Once you have the HWND to the window, call IsWindowVisible() .

One nuance to be aware of. IsWindowVisible will return the true visibility state of the window, but that includes the visibility of all parent windows as well.

If you need to check the WS_VISIBLE flag for a specific window you can do GetWindowLong(hWnd, GWL_STYLE) and test for WS_VISIBLE.

. It sounds like you don’t need to do this for your case, but adding this for future reference in case others run across this question.

Not the answer you’re looking for? Browse other questions tagged c++ winapi api or ask your own question.

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How to check if my window gets hidden/visible?

If i press the «show desktop» button in Windows7, my program will still think its not minimized, and if i press WIN+D while my program is focused, only then my program will catch this minimize command. How can i check for 100% sure that my program is visible or not?

Here is my main loop:

Edit3: Is this good? looks like its working:

5 Answers 5

WM_ACTIVATE is sent when another window becomes active. When you say show desktop, no other window becomes active, so technically, your app is still active, even though it has been minimized.

You probably want to watch for WM_WINDOWPOSCHANGED. You can look at the flags to see what type of position event it was, or you can check IsIconic and IsWindowVisible whenever the window position changes.

There are a variety of potential functions that may get you the information you need depending on exactly what you want to do:

• GetForegroundWindow() : Gets the window the user is currently «working» with. You could use this if you only wanted to draw things when a user is using your application but not another one.
• GetActiveWindow() : Returns the active window within the calling thread which is probably not what you want. This might be useful if you wish to enable/disable drawing depending on which window was active within your own application.
• GetFocus() : Returns the window with the current keyboard focus in the calling thread. Probably not what you want and use GetForegorundWindow() instead.
• IsWindowVisible(): Returns whether the show/hide flag of the window is set to visible. This doesn’t actually tell you whether the window is actually visible on the screen.
• GetTopWindow(): Tells you the highest window in the z-order but not whether it actually has the focus/foreground. It may be possible for your window to be focused/activate/foreground but not have the highest z-order (I think anyways).

From your comments, however, you seem to actually want to see if there is at least one pixel of your window actually visible on the screen. For that I would probably use the technique mentioned in this SO question using the strangely named GetRandomRgn() although a simpler check may be to use GetClipBox() and check the return code for NULLREGION .

How to check if window is really visible in Windows Forms?

Normally you use Form.Visible to check if Window is visible at all. But sometimes on the screen window is below other windows so it’s really invisible.

So how to check in c# Windows Forms if window is really visible or not?

I would like to accomplish this: when I click CTRL+K on my keyboard and my window is visible on my screen it does nothing. But when it’s underneath other windows it pops to the top (Bring to front).

9 Answers 9

You can call the Activate method on the form to bring it to the front if it isn’t already.

However, note that if a different program is active, it will usually simply flash the desktop button (depending where you call it from). This is Windows’ standard protection against focus-stealing and you should not try to work around it.

I googled trough the web, but coudn’t find any straight answer to see if a part of a window is truly visible to the user. I actually needed a way to «hittest» the form, if the mouse is currently on top of the visible part of the window. I thought I’d share the code which took several days to accomplish:

You could use Windows API to enumerate all windows, retrieve their Z-Order and compare it with the Z-Order of your window. I think someone did this already here.

To answer the question as asked, you could try calling the WindowFromPoint API function to find the window at various points on your form, and check whether it returns the handle of whatever you expect to be at that point.

Hm. weird question. 😛

Perhaps you could ask the location of the forms, and if two forms interlap (figure out their coords, and make a simple method) check if one form has Focus(). If it has focus, then other must be «invisible» (in the sense that a user can’t see it because it’s underneath the other form).

Obviously this method is hacky at best, but it’s something you can start working with.

You also could.. 🙂 get the ClickablePoint property from the AutomationElement corresponding to the window. I am not 100%ly sure whether this is completely accurate though.. it has worked in 99% of the cases for me and I am still checking on the other 1%, where the problem lies (might be on my side or bad user handling, or.)

I aktually tried to implement SLaks suggestion. Although I wrote it in VB.NET, not C#

You should be able to find out if your window is visible by overriding the OnPaint method. You’ll want to pass control to the base class in order to do the actual painting, but you’ll be able to detect whether a paint message is received. Update: no, this doesn’t work, Sorry!

In principle, the Activate method should bring your window to the foreground, but in practice I’ve always found this problematic if other processes have the input focus. If you really want someone to see a window, set the topmost bit, but expect them to be annoyed! One surefire way to get some attention for a window is to close it and reopen it, if you can get away with that.

One way to achieve what you’re looking for is to use a notify icon, this will get the user’s attention in a way that is compliant with Windows UI guidelines.

Windows(ThisWorkbook.Name).Visible = True causes error on Excel’s auto-recovered files

Thanks to @YowE3K he referred me to Workbook_Open event as the problem was occurring when I try to recover(re-open) the file. Today I deeply focused on it and found that the reason I was getting:

I don’t know how it became False but, when I try to open auto-recover file, I realize that file is opening normally, my form is opening normally, but when form disappears, the error comes through, the name of the file changes to Microsoft Excel, everything disappear from screen, all of the Excel ribbon freeze, I can only go to Developer tab, nothing changes the situation.

So I think that,excel somehow can’t turn Windows(ThisWorkbook.Name).Visible to True after Form (my Excel Splash Screen) disappears.

How can I avoid this? I don’t want to delete my fancy splash screen but I sometimes need to use auto-recover files as well.

3 Answers 3

The code fails because when Excel recovers a file, it adds some text to the caption of the window, so that «FileName.xlsx» becomes something like «FileName.xlsx [Version last saved by user]». So use a routine like:

Windows(ThisWorkbook.Name) is a common, yet utterly wrong way to get the workbook’s window, which will give you Run-time error ‘9’ Subscript out of range sooner or later.

The right solution (I think, we’ll see how it goes) is to use the Workbook.Windows() collection.

Since Excel supports multiple windows («views» into the workbook, see an explanation here), doing it the right way requires thinking about which window or windows you need to operate on. In particular, ActiveSheet may be different in two different windows for the same workbook.

Given that most people are unaware of this functionality, I decided to always use the first window ( Workbook.Windows(1) ), like this:

To ensure this doesn’t lead to weird results, I wanted to do something when multiple windows are opened for my workbook.