Forum Discussion
Helper IIColumn DATE from column (20310312)
Dear all,
I've a column in this format: 20210312 (YYYYMMDD).
I want to create a new colum date.
How can I do it?
Because with date(column) it doesn´t works.
Hi godtooro ,
You could try below code:-
new_date = VAR _year = MID ( 'Table'[Column1], 1, 4 ) VAR _month = MID ( 'Table'[Column1], 5, 2 ) VAR _day = MID ( 'Table'[Column1], 7, 2 ) RETURN DATEVALUE ( _year & "-" & _month & "-" & _day )Output:-
Thanks,
Samarth
4 Replies
- ichavarria
Solution Specialist
Hi godtooro,
You can use the "FORMAT" function to convert the date in the YYYYMMDD format to a date format that can be used for analysis. Here's an example formula:
Date = DATEVALUE(FORMAT(Table1[Column], "####/##/##"))
In this formula, "Table1" refers to the name of your table and "Column" refers to the name of the column containing the YYYYMMDD date format. The "FORMAT" function converts the date to the format "####/##/##", which is the format recognized by the "DATEVALUE" function.
Best regards,
Isaac Chavarria
If this post helps, then please consider Accepting it as the solution and giving Kudos to help the other members find it more quickly- godtooro
Thank you so much!
I've this error: number entry "20210329" is out of the interval YYYY/MM/DD.
Do you know why it is?
Samarth_18Community Champion
Hi godtooro ,
You could try below code:-
new_date = VAR _year = MID ( 'Table'[Column1], 1, 4 ) VAR _month = MID ( 'Table'[Column1], 5, 2 ) VAR _day = MID ( 'Table'[Column1], 7, 2 ) RETURN DATEVALUE ( _year & "-" & _month & "-" & _day )Output:-
Thanks,
Samarth
- godtooro
Hi!!
Superb!! Works great! Thank you, thank you so much.
