Have a way to easily get date parts out of a date field

[Philip]

Functions such as Year([Date Field]), Month([Date Field]), and Day([Date Field]) would really help with date-based formulas and filter tests.

[MarqueeCrew]

Until such a time, how about?

DateTimeFormat([Date Field],"%Y") >= '2014'

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DateTimeFormat([Date Field],"%m") == '10'

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Just a thought....

[jdunkerley79]

I built some Date functions as a Formula AddIn to do exactly this.

Code is on GitHub: https://github.com/jdunkerley/AlteryxFormulaAddOns

Would be nice to see functions expanded in the next release.

[ARich]

Hi all,

Thanks for the idea (and code!). We're looking at re-vamping all of our DateTime functionality across the product, so we'll keep these ideas in mind when those initiatives kick off.

Best,

Alex

[gsaval]

Good news for anyone reading this post, these functions are now available in Alteryx Designer 11.5:

• DateTimeYear(dt)
• DateTimeMonth(dt)
• DateTimeDay(dt)

[FreeRangeDingo]

@MarqueeCrew suggested -- DateTimeFormat([Date Field],"%Y") >= '2014'

Thanks for the suggestion.  I did use this in a recent workflow.  However, for anyone from Alteryx reading this, this is non-intuitive.  '2014' implies that 2014 is a string and will be treated like a string, not an integer.  I didn't try this when I was working it out on my own for that reason.

[KylieF]

Thank you for your idea! This idea was implemented back in 11.5 with the functions: 

• DateTimeYear(dt)
• DateTimeMonth(dt)
• DateTimeDay(dt)

[Cndro_Consulting]

What function gives the date? for example DateTimeDay(2020-07-05) will give a numeric figure which is 5, what if i want a Sunday?

[Cndro_Consulting]

You can actually use DateTime to achieve this, convert date to string and use custom method

Type Month for just month and day for day. @FreeRangeDingo  @gsaval @Philip @KylieF @ARich @