0

这是我的桌子

CREATE TABLE pago
(
  codigo serial NOT NULL,
  codigocaja integer NOT NULL,
  codigotarjeta integer,
  codigocheque integer,
  formapago character varying(1) NOT NULL,
  tipocomp character varying(1) NOT NULL,
  importetotal money NOT NULL,
  fecha date NOT NULL,
  codigoalumno integer NOT NULL,
  codigodeposito integer,
  vigencia boolean NOT NULL DEFAULT true,
  descripcion character varying(200),
  CONSTRAINT "PK_CodigoPago" PRIMARY KEY (codigo),
  CONSTRAINT "FK_CodigoAlumno" FOREIGN KEY (codigoalumno)
      REFERENCES alumno (codigoalumno) MATCH SIMPLE
      ON UPDATE NO ACTION ON DELETE NO ACTION,
  CONSTRAINT "FK_CodigoCaja" FOREIGN KEY (codigocaja)
      REFERENCES caja (codigo) MATCH SIMPLE
      ON UPDATE NO ACTION ON DELETE NO ACTION,
  CONSTRAINT "FK_CodigoCheque" FOREIGN KEY (codigocheque)
      REFERENCES cheque (codigo) MATCH SIMPLE
      ON UPDATE NO ACTION ON DELETE NO ACTION,
  CONSTRAINT "FK_CodigoDeposito" FOREIGN KEY (codigodeposito)
      REFERENCES deposito (codigo) MATCH SIMPLE
      ON UPDATE NO ACTION ON DELETE NO ACTION,
  CONSTRAINT "FK_CodigoTarjeta" FOREIGN KEY (codigotarjeta)
      REFERENCES tarjeta (codigo) MATCH SIMPLE
      ON UPDATE NO ACTION ON DELETE NO ACTION,
  CONSTRAINT "UN_CodigoPago" UNIQUE (codigo)
)
WITH (
  OIDS=FALSE
);
ALTER TABLE pago
  OWNER TO postgres;

这是我添加新行的功能

CREATE OR REPLACE FUNCTION fn_inspagar(vcodcaja integer, vcodtarjeta integer, vcodcheque integer,  vformapago character varying, vtipocomp character varying,vimporte money ,vfecha date, vcodal integer, vcoddeposito integer,vdesc character varying)
  RETURNS void AS
$BODY$
begin

INSERT INTO pago(codigocaja,codigotarjeta,codigocheque,formapago,tipocomp,importetotal,fecha,codigoalumno,codigodeposito,descripcion)
VALUES (vcodcaja,vcodtarjeta,vcodcheque,vformapago,vtipocomp,vimporte,vfecha,vcodal,vcoddeposito,vdesc);

end;
$BODY$
  LANGUAGE plpgsql VOLATILE
  COST 100;

当我执行这样的功能时

select * from fn_inspagar(1,2,null,'T','B',200,'2013-01-01',196,0,'matricula')

结果我有以下

错误：不存在函数 fn_inspagar（整数，整数，整数，未知，未知，整数，未知，整数，整数，未知）

4

1 回答 1

2

因为货币类型不是“微不足道的转换”，所以money在调用函数时需要在示例中转换 200 值（否则 PostgreSQL 会优先考虑整数转换）。

工作示例：

select * from fn_inspagar(1,2, null, 'T', 'B',CAST ( 200 as money) , '2013-01-01' ,196,0, 'matricula')

查看此处以获取有关 PostgreSQL 如何处理函数签名的更多信息：

于 2013-05-29T22:52:24.850 回答