Distinct Values from an InfoTable in Mashup (without extra service)?

Forum|Forum|7 months ago
September 10, 2025
1 reply
877 views

Hi Community,

I have an InfoTable with several fields, and I would like to use one of those fields as the source for a Dropdown widget. The issue is that this field can contain duplicate values, but in the dropdown I only want to show distinct values.

Right now, the only way I see is to create a separate service that takes the InfoTable and returns only the distinct entries.

My question: Is there any way to achieve this directly in the mashup (e.g., via Data Filter, Query, or some built-in option) without having to write an additional service?

Thanks in advance for your ideas!

Best answer by MA8731174

Yes There is no way to have distinct values in the dropdown. One can only do this with a service. Secondly There is also no way to use infotable in validators or expressions. if a developer wants to do small operations directly in the mashup then he cannot do that untill he has a small extension which can convert infotable to JSON and then one can use it validator and expressions. Anyways i have my answer.

P

PEHOWE

17-Peridot

Hello @MA8731174 ,

I do not believe that there are any options to filter the items which appear in the dropdown in a grid.
Is there a reason why you are trying to avoid using a service to define the list?

Regards,
Pehowe

M

MA8731174

AuthorAnswer

16-Pearl

Yes There is no way to have distinct values in the dropdown. One can only do this with a service. Secondly There is also no way to use infotable in validators or expressions. if a developer wants to do small operations directly in the mashup then he cannot do that untill he has a small extension which can convert infotable to JSON and then one can use it validator and expressions. Anyways i have my answer.

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