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Good day team,
SOLICITUD_ID | IDENTIFICATION | SURNAME |
1195416 | 790503 | NEPHEW |
1196011 | 52407932 | BROWN |
1200443 | 72152864 | FUENMAYOR |
1195416 | 1153463113 | MONTEJO |
I have a table with the following columns, as you can see there are requests that have more than one '1195416' ID associated with them
I want to create a measure that, regardless of whether I have filtered by ID '790503' shows me all the IDs associated with the SOLICITUD_ID that is filtered.
I already have a measurement that, regardless of the filter by ID '790503', brings me the number of different IDs.
If anyone knows how to find out what those IDs are that are leaking, I appreciate it
Solved! Go to Solution.
Hi @Syndicate_Admin ,
According to your description, here are my steps you can follow as a solution.
(1) My test data is the same as yours.
(2) We can create a slicer table and a measure [flag].
Slicer = VALUES('BASE_FINAL'[IDENTIFICATION])
Flag =
var _a=SELECTEDVALUE('Slicer'[IDENTIFICATION])
var _b=CALCULATE(MAX('BASE_FINAL'[SOLICITUD_ID]),FILTER(ALLSELECTED('BASE_FINAL'),'BASE_FINAL'[IDENTIFICATION]=_a))
return IF(MAX('BASE_FINAL'[SOLICITUD_ID]) = _b,1,0)
(3) Place [Flag=1] on visual object screening and then the result is as follows.
Best Regards,
Neeko Tang
If this post helps, then please consider Accept it as the solution to help the other members find it more quickly.
Hi @Syndicate_Admin ,
According to your description, here are my steps you can follow as a solution.
(1) My test data is the same as yours.
(2) We can create a slicer table and a measure [flag].
Slicer = VALUES('BASE_FINAL'[IDENTIFICATION])
Flag =
var _a=SELECTEDVALUE('Slicer'[IDENTIFICATION])
var _b=CALCULATE(MAX('BASE_FINAL'[SOLICITUD_ID]),FILTER(ALLSELECTED('BASE_FINAL'),'BASE_FINAL'[IDENTIFICATION]=_a))
return IF(MAX('BASE_FINAL'[SOLICITUD_ID]) = _b,1,0)
(3) Place [Flag=1] on visual object screening and then the result is as follows.
Best Regards,
Neeko Tang
If this post helps, then please consider Accept it as the solution to help the other members find it more quickly.
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