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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.

Implemented

General

Comments

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