13

我的数据库中有这个功能：

CREATE OR REPLACE FUNCTION "insertarNuevoArticulo"(nombrearticulo character varying, descripcion text, idtipo integer, idfamilia bigint, artstock integer, minstock integer, maxstock integer, idmarca bigint, precio real, marcastock integer)
RETURNS boolean AS
$BODY$
DECLARE
    articulo "Articulo"%ROWTYPE;
BEGIN
    SELECT * INTO articulo FROM "Articulo" WHERE "Nombre" = $1 AND "idTipo"=$3 AND "idFamilia"=$4;
    IF NOT FOUND THEN
        INSERT INTO "Articulo" ("Nombre", "Descripcion", "idTipo", "idFamilia", "Stock", "MinStock", "MaxStock") Values ($1, $2, $3, $4, $5, $6, $7);
        SELECT last_value
        INTO articulo."idArticulo"
        FROM "public"."Articulo_idArticulo_seq";
    END IF;
    SELECT * FROM "ArticuloMarca" AS am WHERE am."idArticulo" = articulo."idArticulo" and am."idMarca" = $8;
    IF NOT FOUND THEN
        Insert into "ArticuloMarca"("idArticulo", "idMarca", "PrecioReferencial", "Stock") Values (articulo."idArticulo", $8, $9, $10);
        RETURN TRUE;
    END IF;
    RETURN FALSE;
END;
$BODY$
 LANGUAGE plpgsql VOLATILE
 COST 100;
 ALTER FUNCTION "insertarNuevoArticulo"(character varying, text, integer, bigint, integer, integer, integer, bigint, real, integer)
 OWNER TO postgres;

但是一旦我尝试使用它，它就说PERFORM如果我想丢弃结果，我需要使用它！这里的问题是我不想！我希望它们在articulo我声明的那一行！

我正在使用这个语句：

SELECT "insertarNuevoArticulo"('Acetaminofen', 'caro', '1' , '1', '8', '1', '10', '1', '150.7', '10');

我得到的错误是 42601，一个语法错误！如果我使用 IDE 来创建它会怎样？关于这个问题的任何想法？

4

2 回答 2

17

在 plpgsql 代码中，SELECT没有目标会触发错误。但你显然不想，你只想SELECT INTO设置状态FOUND。你会用PERFORM它。

更好的是，使用IF EXISTS .... 考虑一下你的函数的重写：

CREATE OR REPLACE FUNCTION "insertarNuevoArticulo"( nombrearticulo text, descripcion text, idtipo int, idfamilia bigint, artstock int, minstock int, maxstock int, idmarca bigint, precio real, marcastock int)
  RETURNS boolean
  LANGUAGE plpgsql AS
$func$
DECLARE
    _id_articulo "Articulo"."idArticulo"%TYPE;
BEGIN
    SELECT a."idArticulo" INTO _id_articulo
    FROM   "Articulo" a
    WHERE  a."Nombre" = $1 AND a."idTipo" = $3 AND a."idFamilia" = $4;

    IF NOT FOUND THEN
        INSERT INTO "Articulo"("Nombre", "Descripcion", "idTipo", "idFamilia", "Stock", "MinStock", "MaxStock")
        VALUES ($1, $2, $3, $4, $5, $6, $7)
        RETURNING "Articulo"."idArticulo" INTO _id_articulo;
    END IF;

   IF EXISTS (SELECT FROM "ArticuloMarca" a
              WHERE a."idArticulo" = _id_articulo AND a."idMarca" = $8) THEN
      RETURN false;
   ELSE
      INSERT INTO "ArticuloMarca"("idArticulo", "idMarca", "PrecioReferencial", "Stock")
      VALUES (_id_articulo, $8, $9, $10);
      RETURN true;
    END IF;
END
$func$;

关于EXISTS：

另一个重点：

Postgres 9.5+

在 Postgres 9.5 或更高版本中使用INSERT ... ON CONFLICT DO NOTHING（又名“UPSERT”）代替。
你会对和然后有UNIQUE限制："Articulo"("Nombre", "idTipo", "idFamilia")"ArticuloMarca"("idArticulo", "idMarca")

CREATE OR REPLACE FUNCTION insert_new_articulo( nombrearticulo text, descripcion text, idtipo int, idfamilia bigint, artstock int, minstock int, maxstock int, idmarca bigint, precio real, marcastock int)
  RETURNS boolean
  LANGUAGE plpgsql AS
$func$
DECLARE
    _id_articulo "Articulo"."idArticulo"%TYPE;
BEGIN
   LOOP
      SELECT "idArticulo" INTO _id_articulo
      FROM   "Articulo"
      WHERE  "Nombre" = $1 AND "idTipo" = $3 AND "idFamilia" = $4;

      EXIT WHEN FOUND;

      INSERT INTO "Articulo"("Nombre", "Descripcion", "idTipo", "idFamilia", "Stock", "MinStock", "MaxStock")
      VALUES ($1, $2, $3, $4, $5, $6, $7)
      ON     CONFLICT (tag) DO NOTHING
      RETURNING "idArticulo" INTO _id_articulo;

      EXIT WHEN FOUND;
   END LOOP;

   LOOP
      INSERT INTO "ArticuloMarca"("idArticulo", "idMarca", "PrecioReferencial", "Stock")
      VALUES (_id_articulo, $8, $9, $10)
      ON     CONFLICT ("idArticulo", "idMarca") DO NOTHING;

      IF FOUND THEN
         RETURN true;
      END IF;

      IF EXISTS (SELECT FROM "ArticuloMarca"
                 WHERE "idArticulo" = _id_articulo AND "idMarca" = $8) THEN
         RETURN false;
      END IF;
   END LOOP;
END
$func$;

这更快、更简单、更可靠。添加的循环排除了并发写入的任何剩余竞争条件（同时几乎不增加任何成本）。如果没有并发写入，您可以简化。详细解释：

另外：使用合法的小写标识符来避免所有嘈杂的双引号。

于 2012-04-12T03:31:42.607 回答
4

这条线在我看来很可疑，可能是导致你悲伤的原因：

SELECT * FROM "ArticuloMarca" AS am WHERE am."idArticulo" = articulo."idArticulo" and am."idMarca" = $8;

您正在函数中执行 SELECT，但没有对结果做任何事情。您需要像之前在函数中所做的那样执行 SELECT INTO。

于 2012-04-12T02:46:26.203 回答