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Combobox filter from two tables

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Posted on by PierreCA

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Hello,

I'm trying to filter data for a combobox from two tables. The back end is a SharePoint list and the column in the Choices is a Lookup pointing on another table.

The first filter works fine but not the second. I've try to combine with the help of Ungroup and ForAll, but looks I missing a clue to end it well. What is it?

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v-mengmli-msft Microsoft Employee on at

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Hi @PierreCA ,

Could you please show the error details?
Or you can try this first:
Filter(Choices(Communiquer.client);Id in Filter(collListePersonnesPourListeDeroulante;profilActif=true).ID;Text(Id) in Filter(EntitePersonne;Value(idEntite)=15).idPersonne))

Best regards,
Rimmon
PierreCA 364 Moderator on at

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Hello @v-mengmli-msft ,

Those are the errors message I get. I've made the translation.

Also I've put a drawing of the relations between the tables, if it may help.

I try the code you proposed but it fails.

I wonder if I shouldn't try to create a temporary table with Personne and EntitiePersonne( with conditions) and then test it with Choices (Communiquer.client).
PierreCA 364 Moderator on at

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Still struggling with the code. I guess that's the right logic, but it expects a table with a single column. But how and where is the question.
v-mengmli-msft Microsoft Employee on at

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Hi @PierreCA ,

The Filter() returns a table, please try to add ".ColumnName" at last.
Just like this: Filter().ColumnName
PierreCA 364 Moderator on at

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Hello @v-mengmli-msft ,

Adding a reference to a column was something I tried, but I stay with two problem.

1) Id in Filter: is related to a delegation issue.

2) idEntite.Value = 15: Translation = Incompatible type to compare. Impossible to compare those types : Text,Number.

3) If I use idEntite.ID = 15: It return a delegation issue and no result when I expect two choices.

I've try different approaches to solve the second issue. Can't find the solution.
v-mengmli-msft Microsoft Employee on at

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Hi @PierreCA ,

1.Delegation error:

in isn't a delegable function.

2.data type error:

The error shows idEntive.Value is text type.

Two ways for your reference:

1)Value(idEntive.Value)

2)Replace 15 with "15".

Best regards,

Rimmon
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PierreCA 364 Moderator on at

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Hello @v-mengmli-msft , 🙂

Unfortunately, the solutoin did't work and decided to try something else.

Thanks.