Question

我有以下数据库结构：

CREATE TABLE songs (_id SERIAL PRIMARY KEY, _artist TEXT, _name TEXT);
CREATE TABLE cafes (_id SERIAL, _ip TEXT PRIMARY KEY);
CREATE TABLE users (_id SERIAL UNIQUE, _imei TEXT PRIMARY KEY);
CREATE TABLE votes (_id SERIAL PRIMARY KEY, _cafe TEXT, _user INTEGER, _song INTEGER, _vote INTEGER);

我有一个存储过程来检查某些行是否存在，如果它们不存在则创建它们存储过程：

CREATE OR REPLACE FUNCTION testfunc(user_imei text, cafe_ip text, song_artist text, song_name text) RETURNS void AS $$
DECLARE 
id_cafe INTEGER;
id_user INTEGER;
id_song INTEGER;
temp_row RECORD;
BEGIN
IF NOT EXISTS (SELECT _id FROM users WHERE _imei = user_imei) THEN
    insert into users (_imei) values (user_imei);
END IF;

IF NOT EXISTS (SELECT _id FROM cafes WHERE _ip = cafe_ip) THEN
    insert into cafes (_ip) values (cafe_ip);
END IF;

IF NOT EXISTS (SELECT _artist FROM songs WHERE _artist = song_artist AND _name = song_name) THEN
    insert into songs (_artist, _name) values (song_artist, song_name);
END IF;    

END;
$$ LANGUAGE plpgsql;

毕竟我想在表vote中插入一个新行，其中：

_cafe = cafes._id
_song = songs._id
_user = users._id

有人可以帮帮我吗？

Answer 1

我做了一些调整。

user是reserved word，我使用usr作为列名。
假设列votes.cafe为integer，而不是text。
从列名中删除了前导下划线。这是一个特殊的命名惯例。我通常将它用于plpgsql函数中的变量，因此它们永远不会与列名冲突..

CREATE OR REPLACE FUNCTION testfunc(
    _user_imei text
   ,_cafe_ip text
   ,_song_artist text
   ,_song_name text) RETURNS void AS
$func$
DECLARE 
   id_cafe integer;
   id_user integer;
   id_song integer;
BEGIN

-- user
SELECT INTO id_user  id FROM users WHERE imei = _user_imei;

IF NOT FOUND THEN
   INSERT INTO users (imei) VALUES (user_imei)
   RETURNING id INTO id_user;
END IF;

-- cafe
SELECT INTO id_cafe  id FROM cafes WHERE ip = _cafe_ip;

IF NOT FOUND THEN
   INSERT INTO cafes (ip) VALUES (cafe_ip)
   RETURNING id INTO id_cafe;
END IF;

-- song
SELECT INTO id_song  id FROM songs
WHERE artist = _song_artist AND name = _song_name;

IF NOT FOUND THEN
   INSERT INTO songs (artist, name) VALUES (_song_artist, _song_name)
   RETURNING id INTO id_song;
END IF;

INSERT INTO votes (cafe, usr, song)  -- What about column vote??
VALUES (id_cafe, id_user, id_song);  -- Assuming cafe is type integer

END
$func$ LANGUAGE plpgsql;

存储过程：从表中获取3个值并将它们保存到另一个表中

1 个答案: