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 …

我对公共模式中所有表的触发器有以下查询：

SELECT 'CREATE TRIGGER ' || tab_name|| '_if_modified_trg INSERT OR UPDATE OR DELETE ON  ' || tab_name|| ' FOR EACH ROW EXECUTE PROCEDURE audit.if_modified_func(); ' AS trigger_creation_query
FROM (
   SELECT quote_ident(table_schema) || '.' || quote_ident(table_name) as  tab_name
   FROM information_schema.tables
   WHERE table_schema='public'
   ) AS foo;

我知道如何检查触发器是否存在：

SELECT tgname
from pg_trigger
where not tgisinternal AND tgname='randomname'

但是，如何在第一个查询中检查同名触发器是否已存在 - 并跳过创建它并继续？这是我的解决方案，但它不起作用：

SELECT 'CREATE TRIGGER ' || tab_name|| '_if_modified_trg INSERT OR UPDATE OR DELETE ON  ' || tab_name|| ' FOR EACH ROW EXECUTE PROCEDURE audit.if_modified_func(); …