How to use references/variables in FINDROW()

Simple question here,
Im a doctor and Im creating a system to register emergency calls(kinda like 911), but im having HUGE problems with findrow()...
I NEED to use valiables and table references on the conditions and parameters and I only get #VALUE! ...

as simple as =FINDROW(table1, "table1[col1]=%", $[variable]) or =FINDROW(table1, "table1[col1]=%", table2!A2) wont work.... I REALLY NEED THIS....
Anyone can help?

Im also trying to figure out some other function that returns me the row/cell info not the link so I can work with it... nothing work

Hi @Dr_Leandro, I'm happy to help you out :slight_smile: :honeybee:

Could you provide a screenshot of your screen with the variable you're attempting to call, and another screenshot where you've set up your formula? Also, if you're getting any error with your formula, please provide that as well (that can help identify syntax errors).

Hi Alyssa, it was the user pick list the problem.... Ive changed it for a simple inputbox and It worked! thx
Its a shame tho cuz I didnt want the user to manualy input its name... but it is what it is

Hi @Dr_Leandro, findrow should also work with picklists. You may need to specifically call the column name though along with your picklist variable. Using your example, maybe try something like this:

=FINDROW(table1, "table1[col1]=%", $[picklist_variable][columnname])

Here's a link on how to retrieve columns from picklists/rowlinks: Dynamically Retrieve a Rowlink's Column

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