# \[DONE\] Siblings as Children of Parents: filter self

**URL:** <https://community.fibery.io/t/done-siblings-as-children-of-parents-filter-self/981>\
**Category:** Get Help\
**Created:** [September 16, 2020, 1:35am UTC](https://community.fibery.io/t/done-siblings-as-children-of-parents-filter-self/981 "2020-09-16T01:35:05Z")\
**Posts on this page:** 1\
**Showing post:** 6

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**Author:** ![Chr1sG](https://sea2.discourse-cdn.com/flex020/user_avatar/community.fibery.io/chr1sg/32/3941_2.png) [@Chr1sG](https://community.fibery.io/u/Chr1sG)\
**Post date:** [July 2, 2021, 8:26am UTC](https://community.fibery.io/t/done-siblings-as-children-of-parents-filter-self/981/6 "2021-07-02T08:26:32Z")

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If I’ve understood your problem, then I think there is a way to achieve it:

- Use a max formula to find the latest start date (as you’ve described)
- Use a formula in the position (or place name) type to check if its start date = person’s max start date (returns a boolean with the name ‘Current’)
- Use a formula in the person type which is `positions.filter(Current = true)`  
This should return a single position (assuming there is always only one with the highest start date)

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_[View the full topic](https://community.fibery.io/t/done-siblings-as-children-of-parents-filter-self/981)._