sql - 运算符不是唯一的。无法选择最佳候选运算符。您可能需要添加显式类型转换

Question

我有这个查询，它只选择两个日期之间的差异:

select date_part('day', 'toDate' - 'fromDate') from "Reservation" where "reservationID" = 1;

但这给了我这个错误:

operator is not unique. Could not choose a best candidate operator. You might need to add explicit type casts.

我该如何解决这个问题？

附言这就是我创建“预订”表的方式:

CREATE TABLE "Reservation"
(
  "reservationID" serial NOT NULL,
  "fromDate" timestamp without time zone NOT NULL DEFAULT '-  infinity'::timestamp without time zone,
  "toDate" timestamp without time zone NOT NULL DEFAULT '-infinity'::timestamp without time zone,
  "staffID" integer NOT NULL DEFAULT 0,
  "customerID" integer NOT NULL DEFAULT 0,
  "roomID" integer NOT NULL DEFAULT 0,
  CONSTRAINT "PK_public.Reservation" PRIMARY KEY ("reservationID"),
  CONSTRAINT "FK_public.Reservation_public.Customer_customerID" FOREIGN KEY    ("customerID")
  REFERENCES "Customer" ("customerID") MATCH SIMPLE
  ON UPDATE NO ACTION ON DELETE CASCADE,
  CONSTRAINT "FK_public.Reservation_public.Room_roomID" FOREIGN KEY ("roomID")
  REFERENCES "Room" ("roomID") MATCH SIMPLE
  ON UPDATE NO ACTION ON DELETE CASCADE,
  CONSTRAINT "FK_public.Reservation_public.Staff_staffID" FOREIGN KEY ("staffID")
  REFERENCES "Staff" ("staffID") MATCH SIMPLE
  ON UPDATE NO ACTION ON DELETE CASCADE
)

Answer 1

你不能使用 ' 你需要 " 代替列名

select date_part('day', "toDate" - "fromDate") from "Reservation" where "reservationID" = 1;