Derived Column using Variable

Hi,
I have a requirement to derive month value from the variable.
The variable 'carInputMonth' holds the values 01 or 02 or 03 ...12 based on the input parameter.
My requirement is to load the target file with Month column and the value should be 'Jan', 'Feb',..'Dec'. So I have to derive the month name from month number.. I though to write a case statement in the derived column but it is not working for me. Please
advise.
Thanks.

Hi Amaya14,
Based on your description, you want to create a derived column that corresponding to the values of variable 'carInputMonth': 'Jan', 'Feb',…'Dec' for '01', '02',… '12'.
After testing the issue in my environment, we can use the expression as the derived column in Derived Column Transformation:
@[User::carInputMonth] == "01" ? "Jan" : @[User::carInputMonth] == "02" ? "Feb" : @[User::carInputMonth] == "03" ? "Mar" : @[User::carInputMonth] == "04" ? "Apr" : @[User::carInputMonth] == "05" ? "May" : @[User::carInputMonth] == "06" ? "Jun" : @[User::carInputMonth] == "07" ? "Jul" : @[User::carInputMonth] == "08" ? "Aug" : @[User::carInputMonth] == "09" ? "Sep" : @[User::carInputMonth] == "10" ? "Oct" : @[User::carInputMonth] == "11" ? "Nov" : @[User::carInputMonth] == "12" ? "Dec" : ""
Besides, please verify that the variable name is completely matched. We can drag the correct variable from ‘Variables and Parameters’ group to replace @[User::carInputMonth] in my expression above.
If there are any other questions, please feel free to ask.
Thanks,
Katherine Xiong
Katherine Xiong
TechNet Community Support

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    Message was edited by:
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