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INVOICE NUMBER | CUSTOMER | AMOUNT | DATE | year |
4564 | TEST | 100 | 01/01/2018 | 2018 |
123 | TEST2 | 0 | 23/05/2019 | 2019 |
4544 | TEST3 | 213 | 15/06/2019 | 2019 |
4554 | TEST | 45 | 01/01/2018 | 2018 |
THIS IS MY DATA FORMAT
i want calculate how many new customer compared to previous year(column)for period selected month/year based on slicer on Date column
and compare with previous year new customer
p.s.i dont have customer id but just customer name,
Who can help me?
Solved! Go to Solution.
Can you be a little more specific? Are you going to filter on date or month? Is your definition of new customer based on a date or month/year? Do you want to compare how many new customers you have (based on the filter selection) and compare the output with how many new customers you had in that same period (date? month?) in the previous year?
In any case, these links might be helpful:
In depth customer churn, new customers, lost customers
Calculating new and existing customers
Calculating new customers each month
Proud to be a Super User!
Paul on Linkedin.
Can you be a little more specific? Are you going to filter on date or month? Is your definition of new customer based on a date or month/year? Do you want to compare how many new customers you have (based on the filter selection) and compare the output with how many new customers you had in that same period (date? month?) in the previous year?
In any case, these links might be helpful:
In depth customer churn, new customers, lost customers
Calculating new and existing customers
Calculating new customers each month
Proud to be a Super User!
Paul on Linkedin.
Hi @snifer ,
Based on your description, you can try to get the customer of the previous year, and then use the lookupvalue function to get the result. But unfortunately, the customer name may have the same name. Only according to the customer name to judge whether it is a new customer, the result is not necessarily accurate.
Best Regards,
Liang
If this post helps, then please consider Accept it as the solution to help the other members find it more quickly.
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