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SOLVED

Directory Tool -Extract Date

suby
11 - Bolide

Hi All,

 

I'm using a Directory tool to bring 2 files by using the  FileName Output column from the Directory tool I want to extract the date and put the date in new column and this I what I have in my formula tool

 

Formula

FileName (Output From Directory tool)

Right([Filename],10)

 

Formula

Date (New Column)
DateTimeFormat([FileName],'%d/%m/%Y')

 

I get NULL values Any thoughts around this

 

File Names looks as below

xxxxxxxx_xxx_xxxxx_2020-08-31.xlsx
xxxxxxxx_xxx_xxxxx_2020-09-20.xlsx

6 REPLIES 6
T_Willins
14 - Magnetar
14 - Magnetar

Hi @suby,

 

Based on your formulas your Right([Filename],10) is going to return "08-31.xlsx".  Try using Left(Right([Filename],15),10)  This should give you "2020-08-31" that you can apply the second formula to.  

suby
11 - Bolide

Hi,

 

Thanks for your reply and still brings up Null Values.

HomesickSurfer
12 - Quasar

Hi @suby 

 

Try:  DateTimeFormat(ToDate(Right(FileGetFileName([FileName]), 10)),'%d/%m/%Y')

 

Hope this works for you...

suby
11 - Bolide

Hello,

 

Great this works Thank you so Much.

 

Could you please explain me how the syntax works.

HomesickSurfer
12 - Quasar

Hi @suby 

 

I'm pleased.  Please like and accept as an acceptable solution.  Thx 🙂

 

DateTimeFormat(ToDate(Right(FileGetFileName([FileName]), 10)),'%d/%m/%Y')

 

FileGetFileName([FileName]) removes the file extension and '.'

(Right(FileGetFileName([FileName]), 10)) captures the last 10 characters

ToDate(Right(FileGetFileName([FileName]), 10)) converts to date format

DateTimeFormat(ToDate(Right(FileGetFileName([FileName]), 10)),'%d/%m/%Y') formats the date as specified dd/mm/yyyy

suby
11 - Bolide

Thank you so Much for the explanantion.

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