I've got a panel, which contains a bunch of (multiselect) ListBoxes each inside their own panel, and I need to figure out what the selected values in the ListBoxes are. Going through the API, the only way I can see of doing this is (pseudocode):
for (Wigdet w : outerPanel)
Panel innerPanel = (Panel) w;
for (Widget s : innerPanel) // only has the ListBox in it
ListBox box = (ListBox) s;
// do stuff with the ListBox to populate the list of selected options
The trouble is with the casting - eclipse doesn't complain and it compiles fine, but when run it produces a ClassCastException (on the first cast - I assume it would also have the same problem on the second, but since I can't get to it I can't say for sure)
What is the correct way to do this?
Got it working. Apparently GWT inserts its own widgets into your panel, besides the ones you add yourself (so creating a Panel then adding a ListBox to in will result in more than just the one widget on iteration).
So to solve the problem, throw in an instanceof check before the casting
Why can't you just declare them as instance variables?
Related
The tests I am developing access the DataGrid cells the following way:
window.Get<ListView>(gridName).Rows.First().Cells.First()
Then clicking a cell with UIItem.Click() method works fine, unless the cell is bound to a null value. In such a case, I get an exception:
Failed to click on ListViewCell. AutomationId:, Name:, ControlType:text, FrameworkId:WPF, bounds empty
Which makes sense, as apparently, an AutomationElement related to that cell is a TextBlock of 0 boundaries.
Is there some possible workaround to click such a cell so the tests work?
What seems to be a good enough workaround is the following:
this.window.Mouse.Location = Point.Subtract(cellToClick.ClickablePoint, new Vector(12, 6));
this.window.Mouse.Click(this.window.Mouse.Location);
So I just set the mouse location using property TestStack.White.UIItems.Mouse so it is situated right above the cell, then click.
In my (programmatic) Matlab GUI, I have a listbox uicontrol.
What I want is to display checkboxes in front of each option. When a user clicks the checkbox, it's marked (and the element will be considered during the calculations later). While if the user clicks the label, a description of the selected option will be displayed in a text uicontrol to inform the user what the option means.
Basically, I want functionality similar to installation programs where you can select components to install and can get information about said components by clicking them (which does not necessarily mark them as selected).
Is there a way to do this with checkboxes or something similar?
There are actually 2 built-in controls that you could use within Matlab:
com.jidesoft.swing.CheckboxList
com.mathworks.mwswing.checkboxlist.CheckBoxList
Usage example (more details in my Matlab-Java book):
jList = java.util.ArrayList; % any java.util.List will be ok
jList.add(0,'First');
jList.add(1,'Second');
jList.add(2,'Third');
jList.add(3,'and last');
jCBList = com.mathworks.mwswing.checkboxlist.CheckBoxList(jList);
jScrollPane = com.mathworks.mwswing.MJScrollPane(jCBList);
[jhCBList,hContainer] = javacomponent(jScrollPane,[10,10,80,65],gcf);
set(jCBList, 'ValueChangedCallback', #myMatlabCallbackFcn);
jCBModel = jCBList.getCheckModel;
jCBModel.checkAll;
jCBModel.uncheckIndex(1);
jCBModel.uncheckIndex(3);
There's no "ready" way for doing that - as listboxes take only plain strings as entries.
You could "manually" draw checkbox fitted into the area of the listbox, but that might mean quite a lot of work to get everything working...
Another alternative is to go for a java-componenent - e.g. using the jide components available in matlab. See e.g.
http://undocumentedmatlab.com/blog/using-jide-combo-boxes/
for a few examples.
Lets say I have a layout panel having multiple textboxes.
Now I want to get a particular textbox in that panel by its name. How can I do that?
Thanks in advance.
One way to do it is to keep references of the textboxes in a hashMap while you create them.
You need to be careful about memory leaks though.
Another way is to use GWtQuery. You could access then by name any element in the DOM.
To get it without without GWTQuery you can do that:
Element element = Document.get().getElementById("myTextboxID");
TextBox box = new TextBox();
box.wrap(element);
Since gwt has not a css-selector implementation, I would use gwtquery aka gquery whose selector implementation performs really well.
Apart from many other features, it is able to return the widget asociated with any element:
import static com.google.gwt.query.client.GQuery.*;
[...]
// if the textbox is already a widget
TextBox b = $("input[name=first_name]").widget();
//Or it its an element
TextBox b = TextBox.wrap($("input[name=search]").get(0));
b.setValue("Foo");
I'm building a Qt plugin with multiple forms. I have a main form which has a tree widget placed on the left of the form.
I want to add items to this tree, such that clicking on these items would load the corresponding form on the same form. But I want the tree widget to be active so that I can select any other form also.
I was able to display a form on the main form using the following code:
Form1 *myform;
myform=new Form1(this);
myform->show();
where Form1 is the class of the form i intend to display. However this, covers up the tree widget also. And I have to do a string comparison of the item in tree being clicked to display the appropriate form.
Can someone please help me with this as I'm very new to Qt programming.
Thanks
ixM has a good suggestion. The first step should definitely be to use layouts in your main window - separating the tree from the rest of the window - where you are going to put your form. I would suggest using a splitter, because then the user can resize the two halves. You can set the splitter as the main widget of your CentralWidget in your main window.
QSplitter splitter = new QSplitter(CentralWidget);
splitter->setOrientation(Qt::Horizontal);
splitter->setHandleWidth(3);
splitter->setChildrenCollapsible(false);
MyTree= new QTreeWidget(splitter);
splitter->addWidget(MyTree);
Then add your tree widget to the splitter, which will be on the left side.
The next step is to add a placeholder widget on the right side of your splitter. We are also going to add a layout inside that widget. This layout is very important we are going to use it later.
QWidget WidgetRightSide = new QWidget(splitter);
QVBoxLayout setupLayout= new QVBoxLayout(WidgetRightSide);
setupLayout->setSpacing(0);
setupLayout->setContentsMargins(0, 0, 0, 0);
Now, at this point, this is where my answer really differs from the previous answer. You could use a QStackedWidget. That is certainly an option. The problem with that is that you have to create and load all your forms at the beginning. That uses way more memory, and will take longer to start up. That's not so bad if you have 2-5 forms, but when we are talking about 20, 30 or more forms that's really ugly.
So what I would suggest instead, is that when the user selects something in the tree, we will remove the old form, and add the newly selected form at that point.
When the selected item in the tree changes this is now what we have to do.
First, remove all the stuff from the previously selection form.
QLayoutItem *_Item;
while ((_Item = setupLayout->takeAt(0)))
delete _Item;
Next, figure out what form to show next, and create it.
QWidget *ActiveSetupForm = NULL;
if ( I need to load form 1)
{
ActiveSetupForm = new YourNewForm( WidgetRightSide);
}
else ...
And lastly, add your new form to our layout.
if(ActiveSetupForm)
{
setupLayout->addWidget(pActiveSetupForm);
}
Just as a side note. Layouts are tricky to do by hand. I would strongly suggest that you look into using the QtDesigner when you are creating your forms. It makes life soooo much easier. If you would like to know more about it check out this link.
I don't exactly understand what you are trying to achieve but the bit of code you are showing suggests that you do not use the layouts provided by Qt.
If your goal is to be able to dynamically load a form depending on the item that was clicked in the tree, you could achieve that by having a layout (let's say QHBoxLayout) where you would insert your tree and a QStackedWidget in which you could "store" each form (by using addWidget()) and choose which one you want to display by calling setCurrentIndex().
I am using GWT 2.4's new DataGrid in a project. I configured the DataGrid with a pagesize of 50.
The available screen is not big enough to display all items and thus a vertical scrollbar is shown (this is actually the main purpose for using a DataGrid in the first place).
I attached a SingleSelectionModel to the DataGrid in order to be able to select items.
This works fine so far.
However I also have another widget with which the user can interact. Based on that user action a item from the DataGrid should be selected.
Sometimes the selected item is not in the visible screen region and the user has to scroll down in the DataGrid to see it.
Is there any way to automatically or manually scroll down, so that the selected item is visible?
I checked the JavaDocs of the DataGrid and found no appropriate method or function for doing that.
Don't know if this works, but you could try to get the row element for the selection and use the scrollIntoView Method.
Example Code:
dataGrid.getRowElement(INDEX_OF_SELECTED_ITEM).scrollIntoView();
The answer above works pretty well, though if the grid is wider than your window and has a horizontal scroll bar, it also scrolls all the way to the right which is pretty annoying. I was able to get it to scroll down and stay scrolled left by getting the first cell in the selected row and then having it scroll that into view.
dataGrid.getRowElement(dataGrid.getVisibleItems().indexOf(object)).getCells().getItem(0).scrollIntoView();
Don't have time to try it out, but DataGrid implements the interface HasRows, and HasRows has, among other things, a method called setVisibleRange. You just need to figure out the row number of the item that you want to focus on, and then set the visible range from that number n to n+50. That way the DataGrid will reset to put that item at the top (or near the top if it is in the last 50 elements of the list backing the DataGrid). Don't forget to redraw your DataGrid.
Have you already looked at this? If so, I'd be surprised that it didn't work.
Oh, and since this is one widget talking to another, you probably have some messaging set up and some message handlers so that when the user interacts with that second widget and "selects" the item, the message fires on the EventBus and a handler for that message fixes up the DataGrid along the lines I've described. I think you'll have to do this wiring yourself.
My solution, a little better:
dataGrid.getRow(model).scrollIntoView();
I got a Out of bounds exception doing the above.
I solved it getting the ScrollPanel in the DataGrid and used .scrollToTop() and so on on the ScrollPanel. However, to access the ScrollPanel in the DataGrid I had to use this comment:
http://code.google.com/p/google-web-toolkit/issues/detail?id=6865
As Kem pointed out, it's annoying the "scrollToRight" effect after the scrollIntoView. After me, Kem's solution gives a better behaviour than the base one as usually the first columns in a table are the more meaningful.
I improved a bit his approach, which scrolls horizontally to the first column of the row we want to be visible, by calculating the first visible column on the left before applying the scroll and then scrolling to it.
A final note: Columns absolute left is tested against "51". This is a value I found "experimentally" by looking the JS values in the browser's developer tool, I think it depends on the table's style, you may need to change/calculate it.
Below the code:
public void scrollIntoView(T next) {
int index = datagrid.getVisibleItems().indexOf(next);
NodeList<TableCellElement> cells = datagrid.getRowElement(index).getCells();
int firstVisibleIndex = -1;
for(int i=0; i<cells.getLength() && firstVisibleIndex<0;i++)
if(UIObject.isVisible(cells.getItem(i)) && (cells.getItem(i).getAbsoluteLeft() > 51) && (cells.getItem(i).getAbsoluteTop() > 0))
firstVisibleIndex = i;
cells.getItem(firstVisibleIndex>=0? firstVisibleIndex : 0).scrollIntoView();
}