Question

CREATE OR REPLACE FUNCTION analyzeTables()
  RETURNS INT AS $$
DECLARE
  v_rec information_schema.tables%ROWTYPE;
BEGIN
  FOR v_rec IN
  SELECT *
  FROM information_schema.tables mytable
  WHERE mytable.TABLE_SCHEMA = 'public'
        AND mytable.table_type = 'BASE TABLE'
  ORDER BY mytable.table_name
  LOOP
    ANALYZE v_rec.table_name;
  END LOOP;
  RETURN 0;
END;
$$
LANGUAGE plpgsql;

SELECT * FROM analyzeTables();

导致ERROR架构＆＃34; v_rec＆＃34;不存在SQL语句"ANALYZE v_rec.table_name"。正如您所看到的，我试图分析函数中的某些表并且它没有正确解释v_rec，当我尝试返回我要分析的表名时，它认为它是模式名称{{ 1}} fuel_station。尝试使用和不使用mytable别名并使用i.e ANALYZE和%ROWTYPE声明同样的错误。我做错了什么？

Answer 1

尝试动态sql：

CREATE OR REPLACE FUNCTION analyzeTables()
  RETURNS INT AS $$
DECLARE
  v_rec information_schema.tables%ROWTYPE;
BEGIN
  FOR v_rec IN
  SELECT *
  FROM information_schema.tables mytable
  WHERE mytable.TABLE_SCHEMA = 'public'
        AND mytable.table_type = 'BASE TABLE'
  ORDER BY mytable.table_name
  LOOP
    execute format('ANALYZE %I',v_rec.table_name);
  END LOOP;
  RETURN 0;
END;
$$
LANGUAGE plpgsql;

SELECT * FROM analyzeTables();

使用format

ERROR架构不存在

1 个答案: