I'm using Javascript to dynamically create a QML component via:
var component = Qt.createComponent(qmlURL)
var item = component.createObject(parentLayout, properties)
In this case, the parentLayout is a ColumnLayout and it always creates the object at the end of the children list (which means it is displayed at the bottom of the column).
How can I specify the index position within the children of where the object is placed, or how can I move it in the children list post instantiation?
(I tried assigning parentLayout.children[i] to no avail)
Use case: I have a large existing multi-level (nested) Menu system that is static, but I'd like insert and remove custom submenu items in the hierarchy in response to external plugins registering/unregistering via a service API. While we could completely re-implement the menu system as a dynamic treeview, that seems overkill just to insert a couple menu subitems dynamically.
I'm trying to select an item from a list that's sorted the same way as a ToggleGroup I have besides it. However, I found that toggleGroup.getToggles().indexOf(toggleGroup.getSelectedToggle()) always returns -1 (visible in the IndexOutOfBoundsException thrown as I pass it). Is there another way of figuring out the index, or am I at a loss with my approach and need to figure out something completely different?
UPDATE: Apparently, for the first time an item is selected (I have this code attached to changes of selectedToggleProperty()), it works fine (I just get no notice of it because the elements I make visible have no proper layout). However, when an item is selected while another item already is selected, getselectedToggle() becomes null, causing aforementioned behavior.
All of the JavaFX toggle controls have a property called UserData. You should use that to create the links between the toggles and data list. Relying on the index of the toggles in the toggle group is probably a bad idea.
(I use Qt 4.7, Windows 7, 64bit).
I created a custom table. Each row is a horizontal layout with widgets.
The rows are kept in a QList for easy access, and the children too. The rows are also added inside the parent widget.
If I resize the parent widget, I calculate the new sizes, delete everything, and recreate it again.
My problem is that I don't want to delete any widget. Only when I clear the table, I do it.
Since I have the widgets inside a QList and inside the parent layouts, How can I remove all widgets in each row, delete all layouts, and then add those to new layouts?
If I do: takeAt(0) for every element inside each layout I have a QLayoutItem with a widget inside... How can I delete the layoutItem without deleting the widget?.... How do I remove the widget without killing it, no matter if it's in the parent or the child? Because there are many methods for deleting: removeItem, removeWidget... in a layout, but not takeWidget... just takeAt() and it gives a Qlayoutitem.
I tried several ways, but I still see the widgets no matter what happened to them.
Questions about this:
When does a widget get deleted? If I takeWidget(index) from a layout, is it deleted some time by itself? does it happen if I have a pointer to it in another list?
removeAt(index) does execute the delete method of a widget?
Ok. I got it working.
Let me explain how this Removing, keeping widgets works.
A widget is known by its parent layout. And you remove it through the layout. By doing:
layout()->removeAt(widget);
delete widget;
If you use takeAt(index) in a QLayout (or its children), it gives you a QLayoutItem. To access the widget inside, just use widget(). But there's no way to remove the widget without deleting it. So this approach is non valid.
In the Docs it tells a way to delete the elements:
QLayoutItem *child;
while ((child = layout->takeAt(0)) != 0) {
...
delete child;
}
A special thing to note in Qt is the following:
If you have a hierarchy tree of layouts, added with addLayout() inside layouts, no matter how deep your widget is inserted, you can remove it from the child layouts or any of the parent layouts, if the tree path from the layout and this item is built from child layouts.
The easiest thing is to keep a list of pointers to all the items, in a custom table. When clearing the table to reconstruct it, just do this inside your widget:
CustomTableItem* item;
while ( !items_.isEmpty() && ( (item = items_.takeFirst()) != 0 ) ){
layout()->removeWidget(item);
delete item; // It works no matter where the item is
}
items_.clear(); // clear the list afterwards.
And it works perfectly, updates the layout too by itself.
If you want to keep the elements, just skip the "delete item;" and use them afterwards.
An important thing to note is that different "remove" functions work differently (as i understand on Qt Docs) in QList or similar widgets, and in a QLayout.
In the QList, removeAt actually removes the object.
(Qt 4.7 QList Docs)"Removes the item at index position i. i must be a valid index position in the list (i.e., 0 <= i < size())."
In a QLayout, removeWidget or removeItem don't remove the item/widget, you have the responsability to delete it, as I did before.
(Qt 4.7 QLayout Docs) "Removes the widget widget from the layout. After this call, it is the
caller's responsibility to give the widget a reasonable geometry or to
put the widget back into a layout."
Hope it helps. If you see any error, you could tell me and I will edit the answer!
More on deleting here:
Other stackoverflow post
A widget in Qt is a regular C++ object and can be deleted with the C++ delete operator as any other object:
delete myWidget;
In Qt there is always a parent-child relation between widgets. When the parent widget is destroyed, it will delete all its children. Usually, you do not need to think about explicitly deleting any widgets but the top level widgets, i.e., windows and dialogs. Qt will take care of deleting any child widgets.
QList::removeAt(int) does not delete the object that is removed, it only removes the object from the list. If you also want to delete the object you would have to do something like:
delete myList.takeAt(0);
This applies to all functions such as removeAt(int), takeAt(int), takeFirst(), etc. They never delete objects, they only remove them from the container (list, layout, scrollarea, etc). In most cases the ownership of the widget is then transferred to the caller, (the caller becomes responsible for deleting the widget as the parent-child relation breaks), but do not assume this is always the case, always read the documentation of the function.
Suppose I fill some cells of a QTableWidget with buttons:
ui->table->setCellWidget (i, 2, new QPushButton ("Details"));
Just a little bit later I scratch them all and do it again:
ui->table->clear();
Will this buttons get deleted automagically? Or do I need to store a list of them and delete manually? Also, Do I need to specify parent for every button like this: QPushButton ("Details", ui->table)?
The table widget takes ownership of the widget. The documentation states this:
Sets the given widget to be displayed in the cell in the given row and column, passing the ownership of the widget to the table.
You also don't need to specify a parent on creation, that will be set by the tablewidget accordingly.
i created one QWidget(Parent). in the inside of the parent widget i created another one QWidget(Child). In the run time i need to remove the child widget. how to do that?
i am not using any layout. i am directly putting in the Parent Widget.
Please Help me to fix this.
If you add the widget with e.g.:
QWidget *w = new QWidget(parent);
...then you can remove it with:
delete w;
Another approach would be to just hide it:
w->hide();
This answer is for those arriving from search engines and want an answer to the question as stated in the title.
If you want to remove a child from a parent without deleting it or hiding it (which does NOT remove it from its parent), set the child's parent to NULL.
QWidget::setParent(NULL)
Note that explicitly reparenting a widget like this carries several implications (e.g visibility automatically set to hidden). See QWidgets documentation for more information.