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

[Raymond Nijssen via QGIS-Developer]

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 
mailto: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



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John Gitau

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

[John Gitau via QGIS-Developer]

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
>
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John Gitau
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