Is there a way to manage my opened MDI Child windows in a toolbar, my project manage lot of windows and the user dont know if some window is open and then open a new one (i also need that the user can open more that one window with works fine) i just one something like a toolbar or something where user can see all open mdi child and when click, focus that window.
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I was wondering if there is a window property, for a wpf application, that disables any actions outside of the window. For example, when you open a message dialog and click outside of it, the message dialog background flashes. Kind of saying you can't do any actions outside of the message dialog untill you click the 'OK' button. I want that implemented in some windows that I open in a program i'm developing but can't seem to find any info on it. Looking for ways on how i might approach this.
Try ShowDialog() instead of Show()
Opens a window and returns only when the newly opened window is closed.
Window childWindow = new Window();
childWindow.ShowDialog();
Note that it will prevent clicks only on parent window(s). Clicks on other applications or Desktop can't be prevented by the Dialog.
I am currently working on AX form for automation . In one of the forms, I am trying to click a button , which opens another form and on this form I need to perform some action.
The window which I am opening on button click sometime opens in background and few times pop up in the foreground above the window where I have clicked the button.
I would want my window to open in the foreground so that I can perform operation , since it is opening in background the codedui playback is searching for the controls on the main window/parent window, which making test case fail
This is making my test case fails multiple times. I am using SetFocus property and SearchInMinimizedwindow options but none of them is working
Is there any solution to always get the window on foreground in codedui or in c#
I would suspect the popup window has some of the same automation properties as the parent window. If not able to fix this code side by using a different AutomationId, you may be able to workaround this by specifiying this is a different window instance.
SecondWindow.SearchProperties["Instance"] = 2;
SecondWindow.SetFocus();
The window that opens when you click the button, is that the child window of the Main window or does it exist on its own? Based on that, you will have to search on the parent (Main Window or Desktop).
I'm working on a program to simultaneously control a handful of other programs of the same type.
So far I use EnumWindows() and collect the handles of the appropriate windows(programs) that I'm going to be controlling. Upon recognizing some keyboard keystroke, I iterate through all of the programs(there could be any number of them open) and I post click messages to all of these programs simultaneously by using EnumChildWindows() to find the appropriate windows(buttons) based on which keystroke was pressed and then using PostMessage() to actually send the click.
Each of these programs that I'm controlling has a button named "Load Settings" which opens an Open file dialog when clicked. What I would like to do is first use PostMessage() to click "Load Settings"(I have successfully gotten this far, what follows is what I'm trying to do). Then I want to get the handle of that Open file dialog window, fill in the text field with the path to the appropriate settings file that I've already determined, then click the "Open" button to finish the task and close the Open window.
My main problem is finding that Open window. Now, I can use EnumWindows once again to find that window but the problem with this is upon clicking "Load Settings" the Open window doesn't open instantaneously. I could sleep after clicking the button, then enum through the windows to find it, but that's not practical as the window could take any amount of time to open.
I've used spy++ to gather some information about this window, but as far as I know, none of it helps enough. Anyways, here is what I have found:
The window "Open" is the Previous Window to the program which it was opened from.
The window "Open"'s parent is the program which it was opened from.
The window "Open"'s owner is the program which it was opened from.
But even though it's listed as a child of the main program under its window properties, it's listed on the same level as the main program (it's not nested like other child windows).
I've tried FindWindowEx() and EnumChildWindows() but neither could find the Open window.
I would think there should be some way to, based on some handle, get the handle of the previous window.
Once that's working, then I need to fill the Edit(text field) window with the appropriate file path.
UIA is what you want, if your goal is to automate UI actions (e.g. for testing). UIA exposes the entire UIA tree which contains all applications active on the machine.
So what you'll see is a root node which has one child node for each application. Each of those nodes should represent the top level window of that application. Any further children depend on the application itself.
See more info at UIA Overview
I'm opening some ie-windows from my c#-client
Is it possible to assign it to an exisiting tabgroup or at least open it as a tab in the already opened ie-window.
Maybe this would work with javascript in the onload of the window?
It has no parent, so is there a possibility to set this value before the window opens?
I use the IE-tab-settings to open new windows always in a tab, but this doesn't work for windows from my client...
I have a window that opens another window. I want that when this window it's opened i can't do anything on the parent window. (I'm not allowed to click buttons for example)
How I can do that?
Thanks.
You need to call the ShowDialog method instead of Show to show the second window as a modal dialog.
You want a modal window instead of a modeless window. A modal window means that the parent window is not usable while the child window is open.
You can open a modal window with ShowDialog.
You can open a modeless window with just Show.
I Recommend you to use MDI scenario behaving as child and parent windows but for instance according to the point you can use code below.
NewWindow.ShowDialog() method instead of NewWindow.Show() as it will disable background window
but that's not good practice we should implement MDI.
I've never seen a desktop application opening on multiple windows and disabling older ones.