QGIS: Use of a model output in the same model as a layer in the field calculator formula

Question

I am trying to use field caluclator within a model. In this formula I want to reference the output (temporary layer) of another step of the same model (output objektfilter_strassen_wege, which is named strassenpolygone_flurstuecke). Since this layer is not loaded in QGIS, I get no result. If I save the correct layer after running through the model and have the model repeated, I get a result. But that's not my goal! I want to get it the first time. I enter the following into the field calculator:

aggregate(
 layer:= 'strassenpolygone_flurstuecke',
 aggregate:='concatenate',
 expression:=oid,
 concatenator:=', ',
 filter:=intersects($geometry, geometry(@parent))
 )

strassenpolygone_flurstuecke should be called differently as input, I suspect. So not layer:= but something else that suggests that it is a generated output. The second possibility that went through my head was to choose to save the layer strassenpolygone_flurstuecke instead of the temporary layer. But even if I do this using the model, I don't get any result.

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Answer 1

This is currently (QGIS 3.16.1) not possible, see the following comment from https://github.com/qgis/QGIS/issues/30397#issuecomment-505699624 :

The issue here is that expression functions which operate on map layers [...] don't currently have access to any temporary output layers created a step in a model. This applies to a lot of functions, including aggregates.

Here is another ticket describing your use case which is marked as a duplicate of the ticket linked above: https://github.com/qgis/QGIS/issues/37347

Answer 2

I came across this post while looking for a solution to a similar problem. You just had to replace the layer name with @layer.

aggregate(
 layer:= @layer,
 aggregate:='concatenate',
 expression:=oid,
 concatenator:=', ',
 filter:=intersects($geometry, geometry(@parent))
 )