Hi,
I'm trying to unselect/remove a column based on a condition. I have an email column where it could all be Empty or have values. If that column is empty it should be deselected. I tried this using dynamic select but for some reason I keep getting this error "unknown variable". Just wondering if I'm doing anything wrong or is there a better way to do this? Thank you in advance!
Solved! Go to Solution.
@martinding Thank you for the response. I already tried that and it didn't work.
Hi @aparna0208
The only way of doing this with the dynamic select tool that i know is by passing the columns that you want to select to a dynamic select tool that is inside a macro. But the easiest way is using the data cleansing tool.
You can find a example of how to do it using both methods on the attached file.
@martinding @Felipe_Ribeir0 @binuacs Ah I don't know how did I forget this simple data cleansing tool that did the magic. Thank you so much for all your inputs and reminding me about this tool:) It worked:)
Having said this I need help with one last thing. Now that I'm using data cleansing tool to remove the email field if it's all null and keep it if it has values. I have one more step where I have to see if that email column exists in the output then need to apply a filter to remove all personal emails. Is there a way to do this to see if that column exists in the output and if so apply this logic if not ignore? @martinding @Felipe_Ribeir0 @binuacs
Hi @aparna0208
I believe that you can filter out the fields that contain "@gmail.com", "@yahoo.com" etc
or filter in the fields that contain "@company1.com", "@company2.com". I would take the second way.
Before of the data cleansing tool, ofc.
Hi @aparna0208
You can use a filter or a formula tool depending on what you are trying to do.
In the filter tool, you can filter by contains @gmail.com, @mail.com etc
In the formula tool, you can do If Contains([ColumnName], "@gmal.com") then "Remove" else "Keep" endif... etc
Something like this would help you accomplish that.
Hope this helps.