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如何在删除后有条件地更新/插入找不到行?

更新时间:2023-02-14 19:09:13

如果DELETE未找到符合条件的行,则其RETURNING子句将返回无行

标题要求有条件地更新/插入删除后的,但如果没有要删除的行,则正文会报告&q;失败。如果条件不是存在要删除的行,则条件是什么?

冒险,这个可能就是您想要的:

CREATE FUNCTION updateoutfit(_id UUID, _title text DEFAULT NULL::text, _garments json)
  RETURNS TABLE (id UUID, title text, garments json)
  LANGUAGE sql AS
$func$
DELETE FROM outfit_garment WHERE outfit_id = _id;  -- DELETE if exists

INSERT INTO outfit (id, title)  -- UPSERT outfit
VALUES (_id, _title)
ON CONFLICT (id) DO UPDATE
SET    title = EXCLUDED.title;
   
WITH ins AS (  -- INSERT new rows in outfit_garment
   INSERT INTO outfit_garment (position_x, outfit_id)
   SELECT "positionX", _id
   FROM   json_to_recordset(_garments) AS x("positionX" float)  -- outfit_id UUID was unused!
   RETURNING json_build_object('positionX', position_x) AS garments
   )
SELECT _id, _title, json_agg(garments)
FROM   ins
GROUP  BY id, title;
$func$;
它删除表outfit_garment中给定UUID的所有行,然后在表outfit中插入或更新行,最后在表outfit_garment中添加新的详细信息行。将忽略传入的任何outfit_id
然后,它返回一行,所有服装都合并到一个JSON值中。