How to enter a derived column in a SQL Server query being built using Access that provides if/then/else functionality
Posted on 2004-10-05
I need to enter an expression for a derived field in a view. The view is being defined using Access 2002 connected to a SQL Server database. I want to test a numeric value and, if greater than 0, set the column value to 'Yes', otherwise set it to 'No'.
I have tried various verions of 'if' and 'iif' and I keep getting errors.