WebOct 20, 2024 · In Summary: Use the datetimeadd() function to subtract the weekday number (%w) from the date in your data to calculate the first day of the week. In Detail : You are … WebAug 31, 2016 · I didn't find anything on Alteryx's website about this, but I stumbled upon a possible solution. DateTimeFormat ( [DateField],"%u") will give you the day of the week where Monday = 1 and Sunday = 7. This NEEDS to be in the documentation! Thanks @danejensen4 ! Reply 0 3 Share wolnov 5 - Atom 02-17-2024 04:52 AM Thanks, …
convert todays date into the day name - Alteryx Community
WebJul 22, 2024 · One possible solution would be to calculate the day of the week, and then add and subtract days accordingly. I have done this in the attached workflow. However, I couldn't help but notice for completed dates 7/1/2024 and 7/9/2024 the week endings are a Wednesday - whereas for completed dates 7/12/2024 and 7/22/2024 the week endings … WebFeb 11, 2024 · You simply want to subtract the dow value from the current date. select dateadd (d, -datepart (dow, my_date), my_date) from (select date ('2024-02-14') as my_date) > 2024-02-11 00:00:00.0 For example, if dow is 3 for 2024-02-14 - a Wednesday - you can subtract 3 days to get back to "day 0". sibylle berg nacht text
Convert date to week number of that year - Alteryx Community
WebAug 31, 2016 · Date conversion to Day of Week Options BarleyCorn 8 - Asteroid 08-31-2016 08:19 AM Trying to convert a date to day of the week. So field of actual date to new variable that is Monday, Tuesday, Wednesday, etc... based on the field date. Have tried DateTimeParse and DateTimeFormat functions in Formula Tool - no luck! This is a basic … WebMar 24, 2024 · On March 24, 2024, invested in software company Alteryx from ICONIQ Capital. Pricing; Log In; Pro - Try 1 Week Free; Investment Summary Exits Alteryx. ... Try For Free 7-Day Free Trial. Seller(S) 1 . SELLER. ICONIQ Capital. website. Category: Growth Capital Firm ... DATE TARGET DEAL TYPE VALUE; 2024-03-15: WebNov 13, 2024 · Alteryx 11-13-2024 07:41 AM @mike_j1028 In which case the below should work If DateTimeFormat (DateTimeNow (),"%A") = "Monday" THEN [Date] >= DateTimeAdd ( [Date], -3,"days") AND [Date] <= DateTimeAdd ( [Date], -1,"days") ELSE [Date] = DateTimeAdd ( [Date], -1,"days") ENDIF Reply 0 0 Share OllieClarke 15 - Aurora 11-13 … the perfume center