相关疑难解决方法(0)

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 …

CREATE OR REPLACE FUNCTION getParentLtree(parent_id bigint, tbl_name varchar) 
  RETURNS ltree AS
$BODY$
DECLARE
   parent_ltree ltree;
BEGIN
-- This works fine:
-- select into parent_ltree l_tree from tbl1 where id = parent_id;

EXECUTE format('select into parent_ltree l_tree from %I
                where id = %I', tbl_name,parent_id);

RETURN parent_ltree;
END;
$BODY$ LANGUAGE plpgsql;

上述功能有2个问题:

1. parent_id是integer,但它会被替换引号？int变量的正确格式说明符是什么？
2. select into不起作用EXECUTE？如何使上面的注释查询使用表名传递？