QStackedLayout¶

The QStackedLayout class provides a stack of widgets where only one widget is visible at a time. More…

Synopsis¶

Detailed Description¶

QStackedLayout can be used to create a user interface similar to the one provided by QTabWidget . There is also a convenience QStackedWidget class built on top of QStackedLayout .

A QStackedLayout can be populated with a number of child widgets (“pages”). For example:
QWidget *firstPageWidget = new QWidget;
QWidget *secondPageWidget = new QWidget;
QWidget *thirdPageWidget = new QWidget;

QStackedLayout *stackedLayout = new QStackedLayout;
stackedLayout->addWidget(firstPageWidget);
stackedLayout->addWidget(secondPageWidget);
stackedLayout->addWidget(thirdPageWidget);


QVBoxLayout *mainLayout = new QVBoxLayout;
mainLayout->addLayout(stackedLayout);
setLayout(mainLayout);
QStackedLayout provides no intrinsic means for the user to switch page. This is typically done through a QComboBox or a QListWidget that stores the titles of the QStackedLayout ‘s pages. For example:
QComboBox *pageComboBox = new QComboBox;
pageComboBox->addItem(tr("Page 1"));
pageComboBox->addItem(tr("Page 2"));
pageComboBox->addItem(tr("Page 3"));
connect(pageComboBox, SIGNAL(activated(int)),
        stackedLayout, SLOT(setCurrentIndex(int)));
When populating a layout, the widgets are added to an internal list. The indexOf() function returns the index of a widget in that list. The widgets can either be added to the end of the list using the addWidget() function, or inserted at a given index using the insertWidget() function. The removeWidget() function removes the widget at the given index from the layout. The number of widgets contained in the layout, can be obtained using the count() function.

The widget() function returns the widget at a given index position. The index of the widget that is shown on screen is given by currentIndex() and can be changed using setCurrentIndex() . In a similar manner, the currently shown widget can be retrieved using the currentWidget() function, and altered using the setCurrentWidget() function.

Whenever the current widget in the layout changes or a widget is removed from the layout, the currentChanged() and widgetRemoved() signals are emitted respectively.

See also

QStackedWidget QTabWidget

class QStackedLayout¶

QStackedLayout(parentLayout)

QStackedLayout(parent)

param parent

QWidget

param parentLayout

QLayout

Constructs a QStackedLayout with no parent.

This QStackedLayout must be installed on a widget later on to become effective.

See also

addWidget() insertWidget()

Constructs a new QStackedLayout and inserts it into the given parentLayout .

Constructs a new QStackedLayout with the given parent .

This layout will install itself on the parent widget and manage the geometry of its children.

PySide2.QtWidgets.QStackedLayout.StackingMode¶

This enum specifies how the layout handles its child widgets regarding their visibility.

Constant	Description
QStackedLayout.StackOne	Only the current widget is visible. This is the default.
QStackedLayout.StackAll	All widgets are visible. The current widget is merely raised.

PySide2.QtWidgets.QStackedLayout.currentChanged(index)¶

Parameters: index – int

PySide2.QtWidgets.QStackedLayout.currentIndex()¶

Return type: int

See also

setCurrentIndex()

PySide2.QtWidgets.QStackedLayout.currentWidget()¶

Return type: QWidget

Returns the current widget, or 0 if there are no widgets in this layout.

See also

currentIndex() setCurrentWidget()

PySide2.QtWidgets.QStackedLayout.insertWidget(index, w)¶

Parameters

index – int
w – QWidget

Return type

int

Inserts the given widget at the given index in this QStackedLayout . If index is out of range, the widget is appended (in which case it is the actual index of the widget that is returned).

If the QStackedLayout is empty before this function is called, the given widget becomes the current widget.

Inserting a new widget at an index less than or equal to the current index will increment the current index, but keep the current widget.

PySide2.QtWidgets.QStackedLayout.setCurrentIndex(index)¶