Set analysis; use Field as Filter

[#1]

hi,

iam a rookie. in set Analysis, i often use them but useally copy and past or with help of the Toolbox.

i'm looking for for the Syntax or possibility to use a field as filter in a set Analysis Expression.

iam trying like this

Only({$<"VendorNo"="b-VendorNo">}"Price")

the 2 VendorNo. are 2 different fields in 2 different Tables. i want the "Price" where the 2 fields are matching.

there is already a Key between a the VendotNo and a "a-VendorNo"

now i want to list the Price of the B-Vendor in the same list.

hope my bad english dont scare you away...

[giakoum]

Only({$<VendorNo=[b-VendorNo]>} Price)

[MK_QSL]

Only({$<VendorNo={'b-VendorNo'}>}Price)

[PrashantSangle]

Hi,

Try this

Sum({<VendorNo={B-VendorNo}>}Price)

[jvitantonio]

Hello Hien,

Put this as  a Dimension like:

=If([VendorNo]=[b-VendorNo], [b-VendorNo])

And in your expression use:

=sum(Price)

JV

[alexandros17]

Unfortunately you cannot compare 2 fields in set analysys but you may compare a field and a value or a rage of values;

so you may write field1 =  {'$(=max(field2))'} but you cannot field1 = {field2}

[jerem1234 · Accepted solution]

You may be able to get away with the p() function.

Only({$<VendorNo=p([b-VendorNo])>}Price)

Hope this helps!

[#8]

thank for replies.

i tried all. but didnt work.

now i trying to use the max-Expression like allesandro wrote. the VendorNo r all 5-digits Numbers so ...should work?

[giakoum]

you better post a sample application, otherwise this may go on for ever

[#10]

hope this help..

first list got the Artikel.no. the main-vendor and the alternativ-vendor

second list got the Price of all vendors.

the are already linked with the main-vendor. and Artikel.no.

iam looking for a way to Show them in one list:

like

Artikelnr.     Kreditornr.     Ek-Preis     A-Kreditornr.     Ek-Preis

1062407     730189          62,14€        970279              109,09€