Unlock a world of possibilities! Login now and discover the exclusive benefits awaiting you.
Dear All,
My Data Set
Year | Product | Qty | Value |
2017 | Product A | 10 | 200 |
2017 | Product B | 20 | 200 |
2017 | Product C | 5 | 100 |
2017 | Product D | 25 | 500 |
2018 | Product A | 40 | 400 |
2018 | Product B | 30 | 250 |
My Concept is Product Loss (i.e)
If I select 2018 in a filter then i want to take 2017 sold product list and compare with 2018 sold product if doesn't exist then i want to shown that item
In this example Product C and Product D have to been shown if i select 2018 .
Need help
You can use an expression with a pattern like this:
Concat({<Product = P({<Year = {2017}>} Product) * E({<Year = {2017}>} Product)>} DISTINCT Product, ',')
(All the possible products in 2017 intersecting with all the products not in 2018)
Adapt to your exact requirements.
You need to generate products for all relevant cases some things like below
Product_Year: Load distinct Product from DataSet; join Load distinct Year from DataSet;
Final:
Load Year, Product resident Product_Year; Drop Table Product_Year ;
Left Join
Load Year, Product, Qty, Value from DataSet;
Now it will work. Only thing is your data volume will be increased.
Thanks for your reply agni
In My transaction table I have Date level transaction not in year we maintaining master calendar in separate table and also we are using fiscal year(APR-MAR)
instead of year what i should insert in load statement??
You can use an expression with a pattern like this:
Concat({<Product = P({<Year = {2017}>} Product) * E({<Year = {2017}>} Product)>} DISTINCT Product, ',')
(All the possible products in 2017 intersecting with all the products not in 2018)
Adapt to your exact requirements.
That should read
Concat({<Product = P({<Year = {2017}>} Product) * E({<Year = {2018}>} Product)>} DISTINCT Product, ',')
If i select customer name in filter the expression not working correctly
Please post a new question as this topic has already been answered. You can add a link back to this topic in the new question.