Hello,
I am trying to create a widget that counts Unique Items in a board. As an example, a board has 500 rows, but many of the rows contain different logged information on the same item. I want to know the number of Unique items amongst the 500. How do I do this?
Josh
Hey @JT1978! Can you elaborate further as to how you’re differentiating items from unique items? Are you using a specific column? This will help me better understand how we can get this set up for you
Each row in Monday represents a call for us. The rows have numerous columns, two of which are unique, the item number and phone number. So in a board with 500 rows (calls), Hundreds maybe from the same person(ie same item number or phone number). I need to know out of the 500 how to count the number of rows that are unique phone number. The functionality I am talking about is found in excel, see this link to understand better:
https://www.howtogeek.com/792407/count-unique-values-in-excel/amp/
I’m very curious about the solution for this too. I have the same issue but with agent ID numbers. I’d like to have a unique count of the IDs even if they appear in multiple groups within the board.
Hello everyone
You can actually use the Duplicates and Uniques app for that.
Using the app, you can label all unique values with status column and then use widget to count all items with that status column.
Here is a demo of how to find duplicates but the process is the same for finding unique values (by selecting Unique Values in Step 3 of the app)
Any way to do this that won’t cost me $50 a month just to count one column?
I am looking for this same feature. I have multiple items that are ticket requests. I have a text column called “Requested By” and I want to count how many unique people are using this ticket service.
I have configured the widgets a million different ways and can’t find a solution.
Right now, my solution is to pull information into a Google Sheet via Zapier and then run a UNIQUE formula and then pull the information back into Monday, but it seems like there should be a much simpler solution.