将参数传递给plpgsql函数中的EXECUTE

发布于 2024-10-13 18:37:48 字数 577 浏览 4 评论 0原文

我从下面的函数中得到以下错误:

ERROR:  column "_df" does not exist  

create or replace function lax()returns setof record as
$$
declare 
  rs record;
  _planunitcode int;
  _DF VARCHAR(25);
  str text;
begin
  _planunitcode:=1;
  _DF :='role.planner';
  str :='select role from userplanunit where role = _DF';
  --str :='select role from userplanunit where role = quote_literal('DF');';quote_ident
  for rs in execute str
  --for rs in select role from userplanunit where role = _DF
  loop
    return next rs;
  end loop;
  return;
end
$$ language 'plpgsql';

I get following error from the function below:

ERROR:  column "_df" does not exist  

create or replace function lax()returns setof record as
$
declare 
  rs record;
  _planunitcode int;
  _DF VARCHAR(25);
  str text;
begin
  _planunitcode:=1;
  _DF :='role.planner';
  str :='select role from userplanunit where role = _DF';
  --str :='select role from userplanunit where role = quote_literal('DF');';quote_ident
  for rs in execute str
  --for rs in select role from userplanunit where role = _DF
  loop
    return next rs;
  end loop;
  return;
end
$ language 'plpgsql';

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薄荷港 2024-10-20 18:37:48

我纠正了这是转义单引号的问题

CREATE OR replace FUNCTION lax ()
RETURNS setof record AS $

DECLARE rs record;

_planunitcode INT;

_DF VARCHAR(25);

str TEXT;

BEGIN
    _planunitcode: = 1;

    _DF : = 'role.planner';

    str : = 'select role from userplanunit where role ='''||_DF||'''';
    FOR

    rs IN

    EXECUTE str LOOP

    RETURN NEXT rs;
END

LOOP;

RETURN;END $

LANGUAGE 'plpgsql';

i rectified it is the problem of escaping the single quote

CREATE OR replace FUNCTION lax ()
RETURNS setof record AS $

DECLARE rs record;

_planunitcode INT;

_DF VARCHAR(25);

str TEXT;

BEGIN
    _planunitcode: = 1;

    _DF : = 'role.planner';

    str : = 'select role from userplanunit where role ='''||_DF||'''';
    FOR

    rs IN

    EXECUTE str LOOP

    RETURN NEXT rs;
END

LOOP;

RETURN;END $

LANGUAGE 'plpgsql';
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