Hi Raymond, For the window title, you can try designer.window().windowTitle().
To get the QgsPrintLayout object, you can try designer.masterLayout() which has the .name() method. Cheers, John On Fri, 29 Jul 2022 at 12:04, Raymond Nijssen via QGIS-Developer < qgis-developer@lists.osgeo.org> wrote: > Hi developers, > > I'm trying to get the name for a layout designer window, which is in a > python variable (designer) of type QgsLayoutDesignerInterface. > > designer.layout() returns a QgsLayout object but that class does not > have a .name() function. > > The QgsPrintLayout class does have a .name() function but I don't know > how to get a QgsPrintLayout object from my designer. > > Anyone? > > Kind regards, > Raymond > > > > _______________________________________________ > QGIS-Developer mailing list > QGIS-Developer@lists.osgeo.org > List info: https://lists.osgeo.org/mailman/listinfo/qgis-developer > Unsubscribe: https://lists.osgeo.org/mailman/listinfo/qgis-developer > -- John Gitau
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