convert %10.0g to %td stata

[Michael Andersonnnn]

I have a long varialable in format %10.0g, which in an integer showing monthdayyear, ie it could be 61289, or 221189
I need to convert this to a %td ie daily date, so I can work out an age using two dates
I am struggling to work out the right code to do this, and any help would be really appreciated?

[Michael Andersonnnn]

Just to add to this, I have tried the following code:

tostring mydob, gen(datevar)
gen date2 = date(datevar, "MDY")

and unfortunately date2 just comes up with missing values

[Andrew Musau]

Stata dates are SIF values, so you have to find a function that does the conversion.

help datetime

clear
input float date
61289
221189
160223
end

gen wanted= daily(string(date, "%06.0f"), "DMY", 2023)
format wanted %td

Res.:

. l

     +--------------------+
     |   date      wanted |
     |--------------------|
  1. |  61289   06dec1989 |
  2. | 221189   22nov1989 |
  3. | 160223   16feb2023 |
     +--------------------+

[Michael Andersonnnn]

Thanks so much for your help, this is still coming up with missing values. Anything else you would recommend?

[Andrew Musau]

Showing an example using dataex.