Обсуждение: operator @> does not work with box.

I do

select circle '((0,0),3)' @> point '(0,0)';

and I get true.

I do

select box '((0,0),(1,1))' @> point '(0,0)';

and I get this:

ERROR:  operator does not exist: box @> point
ROW 1: select box '((0,0),(1,1))' @> point '(0,0)';
                                  ^
HINT:  No operator matches the given name and argument type(s). You
might need to add explicit type casts.

I'm using 8.4.4 and I feel confused. What am I doing wrong here?

A B <gentosaker@gmail.com> writes:
> I do
> select box '((0,0),(1,1))' @> point '(0,0)';
> and I get this:
> ERROR:  operator does not exist: box @> point

> I'm using 8.4.4 and I feel confused. What am I doing wrong here?

Nothing --- there isn't any box @> point operator, although there
is the commutator case point <@ box.  (I see Teodor fixed this
omission for 9.0.)

            regards, tom lane

Tom Lane wrote:
> A B <gentosaker@gmail.com> writes:
>> I do
>> select box '((0,0),(1,1))' @> point '(0,0)';
>> and I get this:
>> ERROR:  operator does not exist: box @> point
>
>> I'm using 8.4.4 and I feel confused. What am I doing wrong here?
>
> Nothing --- there isn't any box @> point operator, although there
> is the commutator case point <@ box.  (I see Teodor fixed this
> omission for 9.0.)


I notice that @ is accepted as an operator (at least in 8.4.1)?

box @ point is equivalent to box @> point
point @ box is equivalent to point <@ box

It's not listed in 8.4 doc pages online.