Forum Discussion
List.Transform list of columns with return value with function which needs the actual columnName
Community AdminHey!
Is there any way that you can share a functional example M code that contains that list and what you're trying to achieve? wondering how your list actually looks like
I have one step with Table.ColumnNames which contains f.e. 3 columnsNames resourcetype, msdyn_startlocation,usertype.
This step is called OptionsetColumns and used in the List.Transform.
the each _ gives the value of the column f.e.669010000.
which I need as well the provide as the 1st parameter to the function but need the columns name like resourcetype as well.
I will post some more details later. Now on my phone....
DennesTorresPower Participant
Hi,
So, once again, I may be able to provide some guesses and hope they help:
If OptionsetColumns is already an array of columnNames, such as {"CustomerID", "Name", "Phone"},
the "_" on the first each, the 2n parameter of the List.Transform, is exactly the column name you want to retrieve.
For example, just to have a better view of the code, let me show a silly example:
List.Transform({"CustomerID", "Name", "Phone"}, each _ + "column")
May result in {"CustomerIDcolumn", "Namecolumn", "Phonecolumn"},
But instead of the array you have OptionsetColumns, but it's the same. The "_" on the first each has the column name and in this case I don't fully understand how your 2nd each is working...
Kind Regards,
DennesSo for this first each I had the idea already this is the column name.
within my function I need the value of the record within that column, "static entity name", "column name"
I beleave the second each gives me the value correct. But how do I get the column name of the first each again?
- DennesTorres
Hi,
Sorry, I didn't get your 2nd each.
The first each is inside the List.Transform. The List.Transform is applied over a single dimension array. The "_" is a string, the column name from the single dimension array.
I didn't get what iteration the 2nd each is trying to do. What the "_" on the 2nd each is ? Does the 2nd each works, if you make a more simple transformation, without the function call ?Kind Regards,
Dennes