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我想在Postgres函数中传递一个表名作为参数.我试过这段代码:

CREATE OR REPLACE FUNCTION some_f(param character varying) RETURNS integer 
AS $$
    BEGIN
    IF EXISTS (select * from quote_ident($1) where quote_ident($1).id=1) THEN
     return 1;
    END IF;
    return 0;
    END;
$$ LANGUAGE plpgsql;

select some_f('table_name');

我得到了这个:

ERROR:  syntax error at or near "."
LINE 4: ...elect * from quote_ident($1) where quote_ident($1).id=1)...
                                                             ^

********** Error **********

ERROR: syntax error at or near "."

以下是更改为此时出现的错误select * from quote_ident($1) tab where tab.id=1:

ERROR:  column tab.id does not exist
LINE 1: ...T EXISTS …

我正在尝试编写plpgsql以下形式的函数(注意这是一个简化版本):

CREATE FUNCTION check_valid(tablename regclass) RETURNS boolean AS $$

DECLARE valid_row tablename%ROWTYPE;

BEGIN

EXECUTE format('SELECT * FROM %s', tablename) into valid_row;

IF valid_row IS NULL THEN
      RETURN QUERY SELECT false;

ELSIF valid_row.is_valid = false;
      RETURN QUERY SELECT false;

ELSIF valid_row.hit_count > valid_row.hit_limit;
      RETURN QUERY SELECT false;

ELSE
      RETURN QUERY SELECT true;

END IF;

END

$$ LANGUAGE plpgsql;

失败的部分是DECLARE线.如何根据变量表名声明类型？或许我需要以某种方式施展它？

类似的DECLARE mytable%ROWTYPE;工作正常,但如果我使用变量名称,如tablename%ROWTYPE:

ERROR: relation "tablename" does not exist