我收到错误:
在不能接受集合的上下文中调用的集合值函数
RETURN QUERY EXECUTE
在行上执行此功能时:
PLSQL $ cat lookup_email.pl
CREATE OR REPLACE FUNCTION app.lookup_email(ident_id bigint,sess bigint,company_id bigint,email varchar)
RETURNS SETOF RECORD as $$
DECLARE
rec RECORD;
comp_id bigint;
server_session bigint;
schema_name varchar;
query varchar;
BEGIN
schema_name:='comp' || company_id;
select app.session.session into server_session from app.session where app.session.identity_id=ident_id and app.session.session=sess;
IF FOUND
THEN
BEGIN
query:='SELECT i.email,u.user_id FROM app.identity as i,' || schema_name || '.uzer as u WHERE i.email like ''%' || email || '%'' and i.identity_id=u.identity_id';
RAISE NOTICE 'executing: %',query;
RETURN QUERY EXECUTE query;
RETURN;
EXCEPTION
WHEN OTHERS THEN
RAISE NOTICE ' query error (%)',SQLERRM;
END;
END IF;
END;
$$ LANGUAGE plpgsql;
这是来自psql的输出:
dev=> select app.lookup_email(4,730035455897450,6,'u');
NOTICE: executing: SELECT i.email,u.user_id FROM app.identity as i,comp6.uzer as u WHERE i.email like '%u%' and i.identity_id=u.identity_id
NOTICE: query error (set-valued function called in context that cannot accept a set)
lookup_email
--------------
(0 rows)
我知道查询不包含任何错误,因为它可以在另一个psql会话中使用:
dev=> SELECT i.email,u.user_id FROM app.identity as i,comp6.uzer as u WHERE i.email like '%u%' and i.identity_id=u.identity_id;
email | user_id
----------------+---------
[email protected] | 1
(1 row)
那么,如果我声明函数为as,为什么Postgres会抱怨RETURNS SETOF RECORD
呢?我的错误在哪里?
那么,如果我声明我的函数是RECORD SET,为什么Postgres会抱怨呢?我的错误在哪里?
它称为Set Returning Function,但是您要指定复合类型
这是完全有效的,
RETURNS SETOF RECORD $$
但是,您可能必须调用它,
SELECT email, user_id
FROM
app.lookup_email(4,730035455897450,6,'u')
AS t(email text, user_id integer)
您不能在其中调用无类型SRF的上下文是没有表定义的上下文。这种语法可能会令人讨厌,因此只需更改RETURNS SETOF RECORD
为
RETURNS TABLE(email text, user_id integer) AS $$
并使用没有列定义列表的函数
SELECT email, user_id
FROM app.lookup_email(4,730035455897450,6,'u')
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