I am new to Blend & Sketchflow. I hope someone can help me.
I have been doing a sketchflow project and have set up several buttons that navigate to specific screens (nice and easy).
Now the catch...
I have made a generic menu at the top, and have put a button on it, and what I want to achieve with it is that if someone clicks on that button, instead of navigating to a specific screen, it simply navigates to the previous screen that was viewed in sketchflow.
Does anyone know if this is possible? And if so, how I would achieve this?
Use the "back" behavior. There are 2 easy ways to apply this behavior to your button:
Right click the button in the artboard, select "Navigate To" -> "Back"
or
Open the assets panel, SketchFlow->Behaviors->NavigateBackAction, drag this behavior onto your button.
The xaml should look similar to this:
<UserControl
xmlns="http://schemas.microsoft.com/winfx/2006/xaml/presentation"
xmlns:x="http://schemas.microsoft.com/winfx/2006/xaml"
xmlns:d="http://schemas.microsoft.com/expression/blend/2008"
xmlns:mc="http://schemas.openxmlformats.org/markup-compatibility/2006"
mc:Ignorable="d"
xmlns:i="clr-namespace:System.Windows.Interactivity;assembly=System.Windows.Interactivity" xmlns:pb="clr-namespace:Microsoft.Expression.Prototyping.Behavior;assembly=Microsoft.Expression.Prototyping.Interactivity"
x:Class="SilverlightPrototype12Screens.Screen_1"
Width="640" Height="480">
<Grid x:Name="LayoutRoot" Background="White">
<Button Height="66" Margin="241,68,275,0" VerticalAlignment="Top" Content="Button">
<i:Interaction.Triggers>
<i:EventTrigger EventName="Click">
<pb:NavigateBackAction/>
</i:EventTrigger>
</i:Interaction.Triggers>
</Button>
</Grid>
I was looking for the same question, but I want to navigate from c#.net.
I found the following solution:
private void Navigate(object sender, RoutedEventArgs e)
{
Microsoft.Expression.Prototyping.Services.PlayerContext.Instance.ActiveNavigationViewModel.NavigateToScreen("WpfPrototype2Screens.Screen_2", true);
}
in this forum.
Related
I am working on an mobile app with Ionic2. When I navigate using navController.push(SomePage) method, I get back button. I want to show left menu button always and disable back button. How can I achieve that?
Thanks and Regards
You can use hideBackButton attribute to hide the back button when push is used.
Ref - https://ionicframework.com/docs/api/components/navbar/Navbar/
For example:
<ion-header>
<ion-navbar hideBackButton>
...
</ion-navbar>
</ion-header>
Ok so I figured out that rather to use
navController.setRoot(somePage);
with parameter
navController.setRoot(somePage, {"id" : 123});
I've got a ListControl box and I want to enable a textbox if the user clicks on a button there.
I've allready tried to update the dialog via an "OnLeftButtonUp" function which works quite well if I click outside the ListControl but when I click INSIDE the box nothing happens...
Any ideas why?
Cheers
This is the code I'm using to create a button:
<button type="button">Click Me!</button>
And the button being created is the same as the default button that Foundation creates for you.
Is it possible to reset Foundation's style and make the button looks like a browser default one?
I have a design like below:
So basically, I want to embed three dialogs in the application main dialog and switch between them, for each button click i.e., button 1 will show dialog one , button 2 will hide dialog 1 and show dialog 2 .. and so on.
Each dialog will be having a different design and functions.
I tried using CPropertySheet class to Add pages but its GUI is different. It has either option for navigating the dialogs using next / back button , or from a tab control.
None of which is as per my requirement.
So I want to know is it possible to have a design like this in MFC ? If yes how? Which Class/ control should I use.
Any help will be appreciated.
What you can do is use a normal CDialog class, add your buttons to it and also create a frame/rect as a placeholder for where your embedded dialogs are to appear. The following piece of code will create and position your embedded dialog.
CRect rect;
CWnd *pHost = GetDlgItem(ID_OF_YOUR_FRAME_RECT);
pHost->GetWindowRect(&rect);
ScreenToClient(&rect);
pDialog->Create(ID_OF_YOUR_DIALOG, this);
pDialog->MoveWindow(&rect);
pDialog->ShowWindow(SW_SHOW);
On button clicks, you hide the previously shown dialog (SW_HIDE) and show your selected dialog(SW_SHOW) with ShowWindow(...).
If you create your embedded dialogs with IDD_FORMVIEW style in the add resource editor it'll have the proper styles for embedding.
Another option is probably to use an embedded PropertySheet and hide the tab row and programatically change the tabs on the button clicks. I just find it to be too much fuzz with borders, positioning, validation and such for my liking.
If you have the MFC Feature Pack, that first came with VS2008 SP1 and is in all later versions, you might like to consider CMFCPropertySheet. There are a number of examples on the linked page, that are very similar to your design.
For example, this:
What worked for me just using dialog based application is SetParent() method. Dont know why nobody mentioned it. It seems to work fine.
I am doing like below:
VERIFY(pDlg1.Create(PanelDlg::IDD, this));
VERIFY(pDlg2.Create(PanelDlg2::IDD, this));
VERIFY(pDlg3.Create(PanelDlg2::IDD, this));
::SetParent(pDlg1.GetSafeHwnd(), this->m_hWnd);
::SetParent(pDlg2.GetSafeHwnd(), this->m_hWnd);
::SetParent(pDlg3.GetSafeHwnd(), this->m_hWnd);
Now I can show or hide a child dialog at will (button clicks) as below:
pDlg1.ShowWindow(SW_SHOW);
pDlg2.ShowWindow(SW_HIDE);
pDlg3.ShowWindow(SW_HIDE);
I have been searching online to no avail. Does anyone know how to access a button in a button box (created using the "Dialog with Buttons Right" template)?
In Designer, select the OK or Cancel button. Then open the property editor and scroll down to the QDialogButtonBox section. You can then expand the standardButtons item to see the various buttons that are available. Other properties, such as the centerButtons property, are also available.
However, designer gives you very little control over the button box.
In code, you can do many other things, such as change the text that appears on the "standard buttons." From the documentation:
findButton = new QPushButton(tr("&Find"));
findButton->setDefault(true);
moreButton = new QPushButton(tr("&More"));
moreButton->setCheckable(true);
moreButton->setAutoDefault(false);
buttonBox = new QDialogButtonBox(Qt::Vertical);
buttonBox->addButton(findButton, QDialogButtonBox::ActionRole);
buttonBox->addButton(moreButton, QDialogButtonBox::ActionRole);
As long as you give the button box a name in designer, you can set these properties in code.
I am writing this answer for the Python community. I am using PySide and faced a similar problem. I have a QDialogButtonBox and I would like to have my own buttons instead of the default ones.
I am using PySide which is more or less the exact replica of the c++ code, so I believe other c++ developers can also get something from it.
Here how I would do that:
my_ok_button = QtGui.QPushButton("My Ok Button")
my_cancel_button = QtGui.QPushButton("My Cancel Button")
ok_cancel_button = QtGui.QDialogButtonBox(QtCore.Qt.Horizontal)
ok_cancel_button.addButton(my_ok_button, QtGui.QDialogButtonBox.ButtonRole.AcceptRole)
ok_cancel_button.addButton(my_cancel_button, QtGui.QDialogButtonBox.ButtonRole.RejectRole)
I would then insert my button box to my layout like ususal:
layout.addWidget(ok_cancel_button, 1, 1)
Now later in my code I can do anything with my button. Lets change its name:
my_ok_button.setText("Some Other Name")
So then things to note here is that:
you must set the role of the buttons in the addButton() method if you
want to use functionalities given by standard buttons. E.g. if you
wish to do something like below, you need to have the button roles
set.
ok_cancel_button.accepted.connect(self.ok_method_handler)
ok_cancel_button.rejected.connect(self.close)
More information can be found here.