QGIS virtual layer: Query preparation error on PRAGMA table_info(_tview): no such column: t.poly

Question

I am trying to load a virtual layer from two existing PostGIS tables, using a simple LEFT JOIN statement:

SELECT t.id, t.poly
FROM table_t t
LEFT JOIN table_f f ON f.fk_poly_id = t.id 
WHERE fk_i_id = 10;

The table_f is a join table containing an id, and two foreign keys: fk_i_id and fk_poly_id.
The table_t is a standard table containing an id and the poly geometry that I'd like to see in my virtual layer for a feature i (referenced by its foreign key fk_i_id in the join table).

This query is perfectly working in psql or pgAdmin4 and returns me some geometric features (ST_Polygon - POLYGON Z) with their ids.

But when I add the two PostGIS tables that I need into the "Embedded layers" of the "Datasource Manager | Virtual Layer" window and when I paste the query in the dedicated space under, I keep getting this error when clicking either "Add" or the "Test" button:

Query preparation error on PRAGMA table_info(_tview): no such column: t.poly

I don't know what I may be doing wrong?
Is this a know bug?
I already tried to rewrite the query in plenty of way, including giving aliases or not, joining the other way, etc etc. All tests were working on pgAdmin but not in QGIS.

Here is the version information:

Name	Version
QGIS version	3.24.2-Tisler
QGIS code revision	13c1a02865
Qt version	5.15.3
Python version	3.10.4
GDAL/OGR version	3.4.1
PROJ version	8.2.1
EPSG Registry database version	v10.041 (2021-12-03)
GEOS version	3.10.2-CAPI-1.16.0
SQLite version	3.37.2
PDAL version	2.3.0
PostgreSQL client version	unknown
SpatiaLite version	5.0.1
QWT version	6.1.4
QScintilla2 version	2.11.6
OS version	Ubuntu 22.04 LTS

Follow up there: Adding a Virtual Layer makes QGIS unresponsive but the test of the query says "No error"

Answer 1

As described in the doc, the geometry column is always called geometry. When you add a layer from PostGres, you choose which is the geometry column and after that the name geometry refers to this selected column.

The same is true when you defined a layer in the embedded layers section.

SELECT t.id, t.geometry
FROM table_t t
LEFT JOIN table_f f ON f.fk_poly_id = t.id 
WHERE fk_i_id = 10;