【发布时间】:2016-02-17 16:10:34
【问题描述】:
我正在为 Postgres 中的表创建分区并具有以下功能:
CREATE OR REPLACE FUNCTION create_partition_and_insert() RETURNS trigger AS
$BODY$
DECLARE
partition_date TEXT;
partition TEXT;
BEGIN
partition_date := to_char(NEW.date,'YYYY_MM_DD');
partition := TG_RELNAME || '_' || partition_date;
IF NOT EXISTS(SELECT relname FROM pg_class WHERE relname=partition) THEN
RAISE NOTICE 'A partition has been created %',partition;
EXECUTE 'CREATE TABLE ' || partition || ' (check (date = ''' || NEW.date || ''')) INHERITS (' || TG_RELNAME || ');';
END IF;
EXECUTE 'INSERT INTO ' || partition || ' SELECT(' || TG_RELNAME || ' ' || quote_literal(NEW) || ').*;';
RETURN NULL;
END;
$BODY$
LANGUAGE plpgsql VOLATILE
COST 100;
我正在将此函数与触发器一起使用:
CREATE TRIGGER testing_partition_insert_trigger
BEFORE INSERT ON testing_partition
FOR EACH ROW EXECUTE PROCEDURE create_partition_and_insert();
测试用表:
CREATE TABLE testing_partition(patent_id BIGINT, date DATE) WITH ( OIDS=FALSE);
有没有办法从函数中返回专利 ID?现在它返回null。我试图通过以下方式扩展功能:
RETURNING patent_id INTO newid;
Postgres 似乎不支持以这种方式返回值:
ERROR: syntax error at or near "INTO"
LINE 1: ...rtition '(111,2018-01-11)').* RETURNING patent_id INTO newid...
^
QUERY: INSERT INTO testing_partition_2018_01_11 SELECT(testing_partition '(111,2018-01-11)').* RETURNING patent_id INTO newid;
CONTEXT: PL/pgSQL function create_partition_and_insert() line 12 at EXECUTE statement
【问题讨论】:
-
好友看看这个stackoverflow.com/questions/33212417/…。详细解释。
-
INTO必须是EXECUTE语句的一部分,而不是INSERT的一部分。将其移到查询字符串之外。 -
谢谢,尼克,你介意给我举个例子吗?我想我有它的一部分 INSERT
标签: postgresql plpgsql