Forum Discussion
How to create a string column out of 3 other string columns
- Anonymous2 years ago
Hi Yacine_VC ,
First of all thanks to Tahreem24 for the quick reply and solution. He did a great job solving your initial problem. I have some suggestions for your follow-up questions. If I have misunderstood you, please create a new case, we suggest a case to solve only one problem, because it can get good help and give a better reference for other users!
The recommendations are as follows:
(1)We can create a table.
Table 2 = FILTER(DISTINCT(UNION( SELECTCOLUMNS('Table',"Tool", 'Table'[Tool 1]), SELECTCOLUMNS('Table',"Tool", 'Table'[Tool 2]), SELECTCOLUMNS('Table',"Tool", 'Table'[Tool 3])) ),[Tool]<>BLANK())(2)We can create a measure.
Measure = COUNTROWS(FILTER(ALLSELECTED('Table'),[Tool 1] in VALUES('Table 2'[Tool]) || [Tool 2]in VALUES('Table 2'[Tool]) || [Tool 3] in VALUES('Table 2'[Tool])))(3) 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 Yacine_VC ,
First of all thanks to Tahreem24 for the quick reply and solution. He did a great job solving your initial problem. I have some suggestions for your follow-up questions. If I have misunderstood you, please create a new case, we suggest a case to solve only one problem, because it can get good help and give a better reference for other users!
The recommendations are as follows:
(1)We can create a table.
Table 2 = FILTER(DISTINCT(UNION(
SELECTCOLUMNS('Table',"Tool", 'Table'[Tool 1]),
SELECTCOLUMNS('Table',"Tool", 'Table'[Tool 2]),
SELECTCOLUMNS('Table',"Tool", 'Table'[Tool 3]))
),[Tool]<>BLANK())
(2)We can create a measure.
Measure = COUNTROWS(FILTER(ALLSELECTED('Table'),[Tool 1] in VALUES('Table 2'[Tool]) || [Tool 2]in VALUES('Table 2'[Tool]) || [Tool 3] in VALUES('Table 2'[Tool])))
(3) 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.