Sum of MinDate for Each ID with status each month

[Pbiuserr]

Hello,

I am trying to get sum of broken items per month for each ID within status for the earliest date available in the dataset (as they continously are in dataset for next months with the same status, but I want to count it only once in the month it has broke)

Status has to be = "B". I also need to use relationship because I have already established active relation between Active items and calendar 

So I was thinking of 

Calculate ( SUM ( ItemID ),
Table[ItemStatus] = "B",
MIN( Table[InActiveItemsDat] ),
USERELATIONSHIP ( Table[InActiveItemsDat], Calendar[Date] )
 )
however it doesnt work

[amitchandak]

@Pbiuserr , You need to try a measure like

Calculate ( SUM ( ItemID ),
Filter( Table, Table[ItemStatus] = "B" && Table[InActiveItemsDat] = minx(filter(allselected(Table), Table[ID] = Max(Table[ID])), Table[InActiveItemsDat]))
USERELATIONSHIP ( Table[InActiveItemsDat], Calendar[Date] )
)

refer

https://amitchandak.medium.com/power-bi-get-the-last-latest-value-of-a-category-d0cf2fcf92d0

[Pbiuserr]

Hi,

unfortunately it gives me blank value, so it doesnt count 

[v-kkf-msft · Accepted solution]

Hi @Pbiuserr ,

Please try these measures to see if they return correct result.

FirstDate = 
CALCULATE (
    MIN ( 'Calendar'[Date] ),
    FILTER (
        ALL ( 'Table' ),
        'Table'[ItemStatus] = "B"
            && 'Table'[ItemID] = MAX ( 'Table'[ItemID] )
    ),
    ALL (),
    USERELATIONSHIP ( 'Calendar'[Date], 'Table'[InActiveItemsDat] )
)

IDCount = 
SUMX (
    'Table',
    CALCULATE (
        COUNTROWS ( 'Calendar' ),
        FILTER ( 'Calendar', 'Calendar'[Date] = [FirstDate] ),
        USERELATIONSHIP ( 'Calendar'[Date], 'Table'[InActiveItemsDat] )
    )
)

If the problem is still not resolved, please provide detailed error information or the expected result you expect. Let me know immediately, looking forward to your reply.
Best Regards,
Winniz
If this post helps, then please consider Accept it as the solution to help the other members find it more quickly.