[QGIS-Developer] Probably a stupid python question regarding QgsVectorlayer

[Alessandro Pasotti]

Hi Bo,

iface.activeLayer().source() should give you what you want.




On Thu, Apr 29, 2021 at 2:21 PM Bo Victor Thomsen <
bo.victor.thomsen at gmail.com> wrote:

> Hi list -
>
> I have a (probably totally obvious, stupid) Pyhton question:
>
> I need the datasource URI string of a maplayer chosen by the user.
>
> I have found  the QgsMapLayer chosen by the user. And the QgsVectorLayer
> associated with the Maplayer.
>
> How do I find the datasource string associated with the QgsVectorLayer?
>
> There is oodles of QGIS information about generating a vector layer from
> a specific datasource. I need the opposite to generate a virtual layer
> in Python using a join between the chosen layer with a another
> predefined layer (which I have the datasource string for).
>
> --
>
> Med venlig hilsen / Kind regards
>
> Bo Victor Thomsen
>
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-- 
Alessandro Pasotti
QCooperative:  www.qcooperative.net
ItOpen:   www.itopen.it
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