Help with workflow to delete item inside of nested RG

Need help App Organization
1

Hello friends,

I have nested RG so that I can display the list of items of the first RG as separate lines. My goal is to be able to then delete an individual line from the nested RG.
My use case is: I have a page for meeting notes and that meeting has tags. The nested RG displays each tag as it’s own line but i want to be able to delete a specific tag within the list.

I put a group inside of the child RG to then place the text of the parent group which is working fine - I can still see the name of the tag. I also added a group with an icon inside for the delete button so that it is clickable to start the workflow but when I try to build the workflow I can’t figure how to target just the specific line of the child RG.

My structure is:
Parent RG (type of content: Meeting data source: Search for Meetings) - working
Child RG (type of content: text data source: current page’s meeting’s tag name) - working
Inside of the child RG is a group that contains text cell of parent’s cells text - working
and inside of that group is a another group holding the icon for the delete button.

How can i connect that delete button to that specific tag?

Thank you!!!

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2

Should be that when you are creating the workflow for when that group that holds the icon is clicked you delete a thing, and the dynamic expression for what thing is 'this groups name.

The problem though is your datasource on the Child RG is type text and not a data type itself, so the above won’t work.

Instead you should be confident that the items in the Child RG will be displayed in the same order they are in the database…so the first cell of the Child RG is the first tag in the list of tags for tag field on the meetings data type…because of that you could have the workflow modify a thing, and the modification would be for the current page’s meeting and change the tag field by removing an item from the list and isolate the item using the :item # operator and use the current cell’s index as the number value

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