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Hi everybody, i hope to find a help for this question :
I have this :
I want count of the FNL (dimension) only if by each part number, the total of quantity is greater > 1000 and the status is (4,5,45)
example for the part number M114-102-136*T4 i have 1152 pcs so greater than 100, i count the number of FNL (4) in this case .
if i add the dimension 'Part N°' to the graph on the left it's working , but if i retry the dimension 'Part N°' --> no
if i don't select a dimension (Part N°) it's not working
if i select a specific part N° , it's working
my formula :
if (aggr (sum({<STATUS ={"4","5","45"}>} P1_ACC_QTY),PART_NO) < 1000, null (),
aggr (count({<STATUS ={"4","5","45"}>} FINISHING_LOT),[DATEQVD.autoCalendar.Date],PART_NO))
i think it's not working because qliksense lost the relationship with the part N° but it's inside my formula
thanks for all helps
Ph
@mahieuphilippe Yes then you can replace count(Part) with count (FNL)
sum(aggr(if(sum({<STATUS={"4","5","45"}>} P1_ACC_QTY)>1000,count(distinct {<STATUS={"4","5","45"}>} FINISHING_LOT)),[DATEQVD.autoCalendar.Date],PART_NO))
@mahieuphilippe try below
sum(aggr(if(sum({<STATUS={"4","5","45"}>} P1_ACC_QTY)>1000,count(distinct PART_NO)),[DATEQVD.autoCalendar.Date],PART_NO))
@mahieuphilippe If you have primary key in your data try below
count(distinct {<Primary_Key={"=sum({<STATUS={"4","5","45"}>} P1_ACC_QTY>100"}>} PART_NO)
Hi , first thank you very much to respond to my question 🙂 !!!
this is the result of your formula :
this not correct, with your formula you count how many part n° has more 1000 PCS and it's correct !! (22)
but for my i want to have the sum of count FNL so 596 FNL the colums '2'
thanks
@mahieuphilippe Yes then you can replace count(Part) with count (FNL)
sum(aggr(if(sum({<STATUS={"4","5","45"}>} P1_ACC_QTY)>1000,count(distinct {<STATUS={"4","5","45"}>} FINISHING_LOT)),[DATEQVD.autoCalendar.Date],PART_NO))
i do that and it's work
sum(aggr(if(sum({<STATUS={"4","5","45"}>} P1_ACC_QTY)>1000,count({<STATUS ={"4","5","45"}>} FINISHING_LOT)),[DATEQVD.autoCalendar.Date],PART_NO)) answer 596 great !!!
with your last formula i have 1207 ?
@mahieuphilippe Yeah. It's just addition of set inside inner aggregation as well.