[QGIS-Developer] Probably a stupid python question regarding QgsVectorlayer
Bo Victor Thomsen
bo.victor.thomsen at gmail.com
Thu Apr 29 06:30:50 PDT 2021
Thanks Alessandro :-)
(I need some new reading glasses, or maybe a new brain. When you read
the /Properties/ -> /Information/ tab for the layer, the needed
information is shown in the tab with the prefix /*Source*: /! It
shouldn't be that hard for me to guess!!)
Med venlig hilsen / Kind regards
Bo Victor Thomsen
Den 29-04-2021 kl. 14:32 skrev 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 <mailto: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 <https://www.qcooperative.net>
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