Forum Discussion
Running Total returns wrong value
Hi all,
I'm trying to figure out the running total quick measures. I've added in the base as the sum of couriers and the field as the start date.
As you can see below, the running totals aren't correct. Could you let me know what I've done wrong please? I'm trying to calculcate how many couriers there are per month in each quarter & year.
Thanks!
Anonymous
You need to create a calendar table with distinct dates, and connect that date column to your current created at date. Then you would use the new calendar date in the filter expression. For example, if you data is from 2020 to 2021, try1. Create a calendar table =Calendar(date(2020,11), date(2021,12,31))
2. Connect the new date table with Created_At.3. create the measure:
Measure = calculate(sum(sheet1[value]), Filter(allselected(calendar table), isonorafter(calendar table[date], max(calendar table[date]), desc))
Paul Zheng _ Community Support Team
If this post helps, please Accept it as the solution to help the other members find it more quickly.
7 Replies
Are you using Is_created_at in the table? Or different columns? The same formula works like a charm on my end:
You'd need to create a relationship between that table and your Sheet1 on Created_at column. And When you create a quick measure, don't place created_at column but instead, place date column that's on the related table.
Hope that makes sense
Anonymous
You need to create a calendar table with distinct dates, and connect that date column to your current created at date. Then you would use the new calendar date in the filter expression. For example, if you data is from 2020 to 2021, try1. Create a calendar table =Calendar(date(2020,11), date(2021,12,31))
2. Connect the new date table with Created_At.3. create the measure:
Measure = calculate(sum(sheet1[value]), Filter(allselected(calendar table), isonorafter(calendar table[date], max(calendar table[date]), desc))
Paul Zheng _ Community Support Team
If this post helps, please Accept it as the solution to help the other members find it more quickly.
