1
我可以传递视图的名称作为函数的参数吗?例如:将视图的名称作为参数传递给函数
CREATE OR REPLACE FUNCTION example_test(test type_view) return void as $$
BEGIN
start_ts = CLOCK_TIMESTAMP();
REFRESH MATERIALIZED VIEW test;
GET DIAGNOSTICS total_rows = ROW_COUNT;
INSERT INTO control_dw_monitoring (name, start_time, end_time, total)
VALUES ('view points that never contacted', start_ts, CLOCK_TIMESTAMP(), total_rows);
END
$$ language plpgsql;
你需要为这个动态SQL:http://www.postgresql.org/docs/current/static/plpgsql-statements.html#PLPGSQL -STATEMENTS-EXECUTING-DYN – 2015-02-09 12:27:10
[那么你有答案吗?](http://meta.stackexchange.com/a/5235/169168) – 2015-07-24 03:21:25