How can I achieve this result?

How do I make the “Connected in any way” column automatically use any circle that is in any way connected to some other circle.

For example:
If :green_circle:’s child (:orange_circle:) has a parent (:red_circle:), then :green_circle: and :red_circle: are connected.

Another example, if circle’s child has a parent whose parent has a child, and that child has a child, and a child of that child’s child is then connect to the initial circle. Those two circles on the far ends and any other circles in-between are connected.

The “Connected in any way” column on the screenshot is just what the final result SHOULD look like, it’s not an active formula. Is there a way to actually achive this result though?
Doc link

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Hey @Fran_Vidicek!

Requested edit access to the doc. This looks similar to the task I was solving here:

I had a table of items with child items (or actually the other way around, items and their parent items, because each item could have many children but only one parent) and I wanted to prevent circular dependencies. For that I had to know which items were already some descendants of the current item so that I could filter them out of the selection list.


If anyone from coda sees this, I’m assuming Paul is employed in Coda or is in some way compensated for helping others in the community, BUT if he’s not, he definitely should be!

Coda has the best costumer support I have seen, and it’s this community.
Paul’s currently working on my solution as I type this…


Thanks for the kind words, @Fran_Vidicek!

I’m compensated in various ways, namely being a Community Champion I enjoy the benefits of it: a free Team workspace and a direct line to Codans from whom I can learn how the internal stuff works (if I cannot reverse-engineer it myself, lol). Also I’ve been Maker Funded for the courses and getting a lot of support for my initiatives otherwise. So I wouldn’t complain :slight_smile:


The strength of coda is indeed its community, and all those who help other without any direct compensation :wink: Congrat’s Paul (again!) !


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