[QGIS-Developer] How to get the name for a QgsLayoutDesignerInterface object?

[Raymond Nijssen]

Wow, thank you so much, masterLayout() it is!

Raymond

On 29-07-2022 11:42, John Gitau wrote:
> 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 at lists.osgeo.org <mailto:qgis-developer at 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
> 
> 
> 
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> -- 
> John Gitau