Thread: 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.