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mbsptr's avatar
mbsptr
Frequent Visitor
4 years ago
Solved

Subscriptions Revenue Totals

Hello,   I am currently trying to calculate total subscription revenue for subscriptions based on the months filtered, but having trouble getting the totals to add up correctly (specifically the re...
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  • v-yalanwu-msft's avatar
    v-yalanwu-msft
    4 years ago

    Hi, mbsptr ;

    You should change the measure of Number of Subscriptions. if we select the order from 1 to 6.(A small amount of data makes it easier to see.）

    we could only order 2 Meet the conditions.

    so we should change the Number of Subscriptions to this.

    Number of Subscriptions = 
var CurrentDate = MAX(DateDim[Date])
var CurrentDate2 = MIN(DateDim[Date])
var ActiveSubscriptions = 
CALCULATE(DISTINCTCOUNT(SubscriptionBillingValue[Clip_Factory_Purchase_Order__c]), 
DateDim,
SubscriptionBillingValue[Subscription Start Date]<=CurrentDate, 
SubscriptionBillingValue[Subscription End Date]>=CurrentDate2
)
return
ActiveSubscriptions

    The final show:


    Best Regards,
    Community Support Team _ Yalan Wu
    If this post helps, then please consider Accept it as the solution to help the other members find it more quickly.

v-yalanwu-msft's avatar
v-yalanwu-msft
Community Support
4 years ago

Hi, mbsptr ;

You should change the measure of Number of Subscriptions. if we select the order from 1 to 6.(A small amount of data makes it easier to see.）

we could only order 2 Meet the conditions.

so we should change the Number of Subscriptions to this.

Number of Subscriptions = 
var CurrentDate = MAX(DateDim[Date])
var CurrentDate2 = MIN(DateDim[Date])
var ActiveSubscriptions = 
CALCULATE(DISTINCTCOUNT(SubscriptionBillingValue[Clip_Factory_Purchase_Order__c]), 
DateDim,
SubscriptionBillingValue[Subscription Start Date]<=CurrentDate, 
SubscriptionBillingValue[Subscription End Date]>=CurrentDate2
)
return
ActiveSubscriptions

The final show:


Best Regards,
Community Support Team _ Yalan Wu
If this post helps, then please consider Accept it as the solution to help the other members find it more quickly.