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kunkumnaveen
Specialist
Specialist

number of parts with multiple prices with same supplier

Hi All,

        the KIP i need to show is

                      # parts with multiple prices with same supplier 

what i understand is , i need to show number of materials  which is supplied by same supplier having more then one distinct  price and validto has to be greater than today()


Dimension :Supplier

Expression:

aggr(Count({<SCOPE={'Yes'},VALIDTO={">$(=Date(today(),'DD-MM-YYYY'))"},supplier={"=Count(Distinct Price)>1"}>}distinct part),supplier)


i am getting materials , suppliers which has only one  price ,which should not ,so i think  the greater than syntax in my expression

has problem,

any  suggestion plz..

Thanks






1 Solution

Accepted Solutions
sunny_talwar

1) I don't think you need Aggr() in your expression

2) May be try this

Count({<SCOPE = {'Yes'}, VALIDTO = {">$(=Date(today(),'DD-MM-YYYY'))"}, supplier = {"=Count({<SCOPE={'Yes'}, VALIDTO ={"">$(=Date(today(),'DD-MM-YYYY'))""}>} DISTINCT Price)>1"}>} DISTINCT part)

View solution in original post

15 Replies
Chanty4u
MVP
MVP

try below put "= "


aggr(Count({<SCOPE={'Yes'},VALIDTO={">=$(=Date(today(),'DD-MM-YYYY'))"},supplier={"=Count(Distinct Price)>1"}>}distinct part),supplier)


sunny_talwar

1) I don't think you need Aggr() in your expression

2) May be try this

Count({<SCOPE = {'Yes'}, VALIDTO = {">$(=Date(today(),'DD-MM-YYYY'))"}, supplier = {"=Count({<SCOPE={'Yes'}, VALIDTO ={"">$(=Date(today(),'DD-MM-YYYY'))""}>} DISTINCT Price)>1"}>} DISTINCT part)

kunkumnaveen
Specialist
Specialist
Author

thanks for your reply ,

i have used aggr because in different KIP i need to group it with plant wise ,so without using aggr how can i able to group with plant

Count({<SCOPE = {'Yes'}, VALIDTO = {">$(=Date(today(),'DD-MM-YYYY'))"}, supplier = {"=Count({<SCOPE={'Yes'}, VALIDTO ={"">$(=Date(today(),'DD-MM-YYYY'))""}>} DISTINCT Price)>1"}>} DISTINCT<Plant> part)

is above expression a valid one

sunny_talwar

Group by plant? Not sure I understand? can you elaborate?

kunkumnaveen
Specialist
Specialist
Author

i mean for a  particular plant ,supplier combination how many materials has distinct multiple prices supplied by that supplier .

sunny_talwar

Then I believe you will have to use Aggr() function instead of set analysis

kunkumnaveen
Specialist
Specialist
Author

Hello , i have used the below expression which suggested by you

Count({<SCOPE = {'Yes'},MATL_TYPE={'HALB','ROH'}, VALIDTO = {">$(=Date(today(),'DD-MM-YYYY'))"}, [MATERIAL CODE]= {"=Count({<SCOPE={'Yes'},MATL_TYPE={'HALB','ROH'}, VALIDTO ={"">$(=Date(today(),'DD-MM-YYYY'))""}>} DISTINCT PRICE_MAT)>1"}>} DISTINCt [MATERIAL CODE])


before selecting vendor i am getting wrong outputimage2.PNG

after selecting that vendor i am getting correct o/p

image3.PNG

can u suggest me y this is happening

sunny_talwar

Looking at screenshots and expressions doesn't give me much details Naveen. Would you be able to share a sample to dig deeper. Also, it would be helpful to know the expected number you are looking to get with and without selections

kunkumnaveen
Specialist
Specialist
Author

i wish to do so but it is client data which i am not allowed to share ,to keep it simple the number i am getting after selecting vendor (1,2)should appear even if i don't select vendor because they are correct num