On 19 Oct 2012 14:42, Stone, Joel wrote:
I have a 6 character field named "todaysDate" that I want to load
with today's date using SQL.
The stmt below works well, but one would think that there should be a
much simpler stmt to load a commonly used date format.
Is there a simpler SQL stmt?
set todaysDate = right(replace(char(current date,iso),'-',''),6)
Not sure about "simpler", but more succinct...
Try the following [sorry, I can not test\verify its efficacy]:
Links to the documentation for the two scalar functions utilized: