On Fri, 2011-09-09 at 10:42 -0400, Henry Drexler wrote:
> take any table and run
>
> Query
> ---------------------
> select
> 1/3
> from
> storage
> limit 1
>
>
> Result
> ---------------------
> ?column?
> integer
> 0
>
>
> Expected Result
> ---------------------
> ?column?
> double precision
> 0.33333...
>
>
>
> Question
> ---------------------
> Since there is no column type to begin with as this is a made-up column,
> shouldn't postgres know it is double precision due to the remainder?
>
You divide an integer with an integer, that should give you an integer.
And that's exactly what it does.
> I thought perhaps I could cast it as double precision as noted on
> http://www.postgresql.org/docs/8.3/interactive/sql-expressions.html
>
> though doing the following:
>
> select
> float8(1/3)
> from
> storage
> limit 1
>
> results in:
>
> float8
> double precision
> 0
>
You still divide an integer with an integer. 1/3 as integers has a
result of 0. You then cast it to float which gives you the value 0 in
double precision.
> any ideas on how to get this type of a manufactured column (not sure the
> right term for it) to show the double precision result?
Sure, do select 1./3 from... or select float8(1)/3...
--
Guillaume
http://blog.guillaume.lelarge.info
http://www.dalibo.com
