Indeed

On 19/09/2019 17:56, DelazJ wrote:
You mean, it's the selected field in the primary key drop-down menu in DB manager when you want to add the layer?

Le jeu. 19 sept. 2019 à 16:42, Janneke van Dijk <[email protected] <mailto:[email protected]>> a écrit :

    Hi Harrissou,
    Thanks for the quick response - indeed, I indicated the assetid
    which is a unique value. That did not solve the problem, alas...
    Janneke

    On 19/09/2019 17:31, DelazJ wrote:
    Hi Janneke,

    Maybe you need a field that would uniquely identify the 170 asset
    rows (their primary key). You can use an ID field of your asset
    layer or construct one with row_number function?


    Regards,
    Harrissou

    Le jeu. 19 sept. 2019 à 16:10, Janneke van Dijk
    <[email protected] <mailto:[email protected]>> a écrit :

        Dear list,

        I have a parent - child relationship set up between a table
        location (1,
        the parent) and a table assets (many, the child). The
        location table
        contains geometry, the asset table does not. I would like to
        be able to
        display each record in the asset table with the corresponding
        geometry
        of the location table - so if there are 3 assets at a certain
        location,
        I would like to display 3 points (each with identical geometry).

        What I can do with the DB manager is:

        select location_point.geom, asset_point.assetid,
        location_point.fid
        from asset_point cross join location_point on
        asset_point.locationid =
        location_point.fid

        this query yields 170 records, corresponding with the number
        of records
        in the assets table. When I load as a new layer, the result
        is a table
        with 170 records, but 100 filtered (the total number of
        locations). I
        chose assetid to be the unique id (which it is).

        Is that expected behaviour? How else can I display all the
        records in
        the assets table (instead of only the 100 that are filtered
        now)? I know
        I can do it in Postgres but for my purpose now I don't want
        to go there.

        Any insights deeply appreciated!

        Janneke


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