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luisccmm
Contributor

Count distinct Stores Openend per Day - Regardless of Time Dimension

Hi,

I have a dataset of sales from different Stores and those related to the customers ID, product Id and so on... and I need to be able to SUM the number shops that have made any sales per day. 

I have 100 Stores (Maximum)

If I use the function COUNT (Distinct "Store Name) the result will depend on the dimension, so:

For this Example supose that The month only have "3 Days"

Day    Month  Count (Disntinct "Store Names")

1          1         100

2          1         95

3          1         80

If I take out Dimension Day from the above table, then  Count (Distinct "Store Names") will take account the month as a whole and the result will be 100.

Month  Count (Disntinct "Store Names")

1         100

1         95

1         80

Expected Result:

- I need to get the sume of Number Stores Open Every Day:  100 +95 +80    Regardless of the dimension I use, I need this function to be limitd to count per day dimension.

Any help???

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1 Solution

Accepted Solutions
MVP
MVP

Re: Count distinct Stores Openend per Day - Regardless of Time Dimension

Maybe like

=Sum( Aggr( Count(DISTINCT "Store Names"), Month, Day))

edit: and if you Need to ignore your Chart dimensions:

=Sum(TOTAL Aggr( Count(DISTINCT "Store Names"), Month, Day))

2 Replies
MVP
MVP

Re: Count distinct Stores Openend per Day - Regardless of Time Dimension

Maybe like

=Sum( Aggr( Count(DISTINCT "Store Names"), Month, Day))

edit: and if you Need to ignore your Chart dimensions:

=Sum(TOTAL Aggr( Count(DISTINCT "Store Names"), Month, Day))

luisccmm
Contributor

Re: Count distinct Stores Openend per Day - Regardless of Time Dimension

Thanks very much Stefan for you help!!!

It was quite helpful!!!