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table1 {A, B} table2 {A, ids} table3 {A, ids}
table1 -> table2 Active Relation
table1 -> table3 Active Relation
How can I get the ids (show the list in the report) that are in table3 but not in table2? (filtered by A)
Table1 (1:*) relationship with Table2
Table 1 (1:*) relationship with Table3
table1 table 2 table 3
A | B A | C | ids A | ids
______ _________ _______
p | a p | x | 1 P | 1
q | b p | x | 2 P | 2
r | c p | y | 3 p | 3
s | d p | y | 4 p | 4
t | e q | x | 2 p | 5
So in my report filter if I select C = 'x' I want to see one Table View with id's ( 1, 2 ) and another table view with id's ( 3, 4, 5 )
and if I select C = 'y' I want the first Table to show id's ( 3, 4 ) and the second Table to show id's ( 1, 2, 5 )
In case table one has all the ids. The last measure you can show in visual with if of Table 1
cnt3 = count(Table3[id])
cnt2 =count(table2[id])
id in 3 not in 2 =countx(Table1,if(not(isblank(cnt3)) && isblank(cnt2),1,blank()))
If table 1 do not have id and only table 3 has
this will give you a table for that
id in 3 not in 2 = except(all(Table3[id]),all(Table2[id]))
thanks, @amitchandak for the reply, but the solution you provided shows the Excluded ID across of the whole table but I want to get something like below.
Table1 (1:*) relationship with Table2
Table 1 (1:*) relationship with Table3
table1 table 2 table 3
A | B A | C | ids A | ids
______ _________ _______
p | a p | x | 1 P | 1
q | b p | x | 2 P | 2
r | c p | y | 3 p | 3
s | d p | y | 4 p | 4
t | e q | x | 2 p | 5
So in my report filter if I select C = 'x' I want to see one Table View with id's ( 1, 2 ) and another table view with id's ( 3, 4, 5 )
and if I select C = 'y' I want the first Table to show id's ( 3, 4 ) and the second Table to show id's ( 1, 2, 5 )
@SrinathLM10
Try create a measure:
Not equal = CALCULATE(MAX('Table (2)'[C]),FILTER(ALLSELECTED('Table (2)'),[C]<>SELECTEDVALUE('Table (2)'[C])))
Check the attached pbix file for detail.
Paul Zheng _ Community Support Team
If this post helps, then please consider Accept it as the solution to help the other members find it more quickly.
@Anonymous Thanks for the reply, Looks like the solution you provided has an active connection between the table(2) ids and table(3) ids. but I am looking for a solution without the relation, and also the ids in table 3 are not unique, we can also have entries like (q, 1) (q, 2) (q,3)
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