# Filter on a Collection

**URL:** <https://community.fibery.io/t/filter-on-a-collection/3411>\
**Category:** Misc\
**Created:** [October 18, 2022, 1:07pm UTC](https://community.fibery.io/t/filter-on-a-collection/3411 "2022-10-18T13:07:18Z")\
**Posts on this page:** 1\
**Showing post:** 9

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**Author:** ![antoniokov](https://sea2.discourse-cdn.com/flex020/user_avatar/community.fibery.io/antoniokov/32/201_2.png) [@antoniokov](https://community.fibery.io/u/antoniokov)\
**Post date:** [January 3, 2023, 9:32am UTC](https://community.fibery.io/t/filter-on-a-collection/3411/9 "2023-01-03T09:32:23Z")

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> [@thumDer](#):
>
> When using `.Filter()` on a Relation Field in a Formula is there a way, to validate against another Field of my current database?

Now the answer is “yes”: [December 29, 2022 / SOC 2 Type II compliance, [This ...] in Formulas](https://community.fibery.io/t/december-29-2022-soc-2-type-ii-compliance-this-in-formulas/3751).

I think you can achieve the desired result, perhaps with the use of an extra auxiliary Formula Field (we still can’t do aggregations within the `Filter(...)` function).

Please ping us via Intercom if you need assistance — @Chr1sG or I would be happy to jump on a screen-sharing call.

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_[View the full topic](https://community.fibery.io/t/filter-on-a-collection/3411)._