I load a .txt file, in every line there is a string with day, month, year and other words; I used "(mid (@2,9,2)) AS Date" to separate day, month and year. Now I have 3 columns day,month,year(with month converted from 'Jan' to 01, 'Feb' to 02 etc.) but I need a new column with complete date (DD/MM/YYYY).
Date(Date#(Date&'/'&Month&'/'&Year, 'DD/MM/YYYY')) as Full_Date
Best Anil, When applicable please mark the correct/appropriate replies as "solution" (you can mark up to 3 "solutions". Please LIKE threads if the provided solution is helpful
Date(Date#(Date&'/'&Month&'/'&Year, 'DD/MM/YYYY')) as Full_Date
Best Anil, When applicable please mark the correct/appropriate replies as "solution" (you can mark up to 3 "solutions". Please LIKE threads if the provided solution is helpful
