我可以从Postgresql函数内部创建临时表吗？

这是Postgresql脚本。我想把它变成一个函数

CREATE TEMPORARY TABLE bad_survey (
survey_id int8 NOT NULL,
template_id int8 NOT NULL
);
analyze bad_survey;
insert into bad_survey(survey_id, template_id)
(select id as survey_id, template_id 
from survey_storage 
where status in ('Never Surveyed','Incomplete Configuration','Ready to Launch')
and id=original_row_id 
and tenant_id=owner_tenant_id
and tenant_id=5);
insert into bad_survey(survey_id, template_id)
(select pss.id, pss.template_id
 from survey_storage css
    inner join company_by_path cbp
        on css.company_by_path_id = cbp.id
        and css.tenant_id = cbp.tenant_id   
        and cbp.relationship_type = 'partner'
    inner join survey_storage pss
        on cbp.owner_tenant_id = pss.tenant_id
        and css.master_template_id = pss.master_template_id
        and css.tenant_id = pss.owner_tenant_id
        and css.source_id = pss.source_id
        and css.tenant_id != pss.tenant_id
        and css.template_id != pss.template_id
        and pss.id != pss.original_row_id
 where css.id in (select id as survey_id
                from survey_storage 
                where status in ('Never Surveyed','Incomplete Configuration','Ready to Launch')
                and id=original_row_id 
                and tenant_id=owner_tenant_id
                and tenant_id=5));
DELETE FROM survey_user su
    USING bad_survey bs
    WHERE su.survey_id = bs.survey_id;

DELETE FROM survey_library_users slu
    USING bad_survey bs
    WHERE slu.survey_library_id = bs.template_id;

DELETE FROM row_history rh
    USING bad_survey bs
    WHERE rh.row_id = bs.survey_id;

DELETE FROM survey_storage ss
    USING bad_survey bs
    WHERE ss.id = bs.survey_id;

DELETE FROM survey_library sl
    USING bad_survey bs
    WHERE sl.id = bs.template_id;

在这里，我看到我在两个地方硬编码了租户id。我想把它作为函数参数传递。 但当我试图将此代码包装到存储过程中时，它会在创建临时表时出现语法错误

更新 以下是失败的函数创建脚本：

CREATE OR REPLACE FUNCTION public.purge_bad_surveys(_tenant_id bigint)
RETURNS bool
LANGUAGE plpgsql
AS $function$ 
BEGIN
CREATE TEMPORARY TABLE bad_survey (
survey_id int8 NOT NULL,
template_id int8 NOT NULL
);
analyze bad_survey;
insert into bad_survey(survey_id, template_id)
(select id as survey_id, template_id 
from survey_storage 
where status in ('Never Surveyed','Incomplete Configuration','Ready to Launch')
and id=original_row_id 
and tenant_id=owner_tenant_id
and tenant_id=_tenant_id);
insert into bad_survey(survey_id, template_id)
(select pss.id, pss.template_id
 from survey_storage css
    inner join company_by_path cbp
        on css.company_by_path_id = cbp.id
        and css.tenant_id = cbp.tenant_id   
        and cbp.relationship_type = 'partner'
    inner join survey_storage pss
        on cbp.owner_tenant_id = pss.tenant_id
        and css.master_template_id = pss.master_template_id
        and css.tenant_id = pss.owner_tenant_id
        and css.source_id = pss.source_id
        and css.tenant_id != pss.tenant_id
        and css.template_id != pss.template_id
        and pss.id != pss.original_row_id
 where css.id in (select id as survey_id
                from survey_storage 
                where status in ('Never Surveyed','Incomplete Configuration','Ready to Launch')
                and id=original_row_id 
                and tenant_id=owner_tenant_id
                and tenant_id=_tenant_id));
DELETE FROM survey_user su
    USING bad_survey bs
    WHERE su.survey_id = bs.survey_id;
RAISE NOTICE 'Done with deleting survey users';
DELETE FROM survey_library_users slu
    USING bad_survey bs
    WHERE slu.survey_library_id = bs.template_id;
RAISE NOTICE 'Done with deleting survey_library_users';
DELETE FROM row_history rh
    USING bad_survey bs
    WHERE rh.row_id = bs.survey_id;
 
RAISE NOTICE 'Done with deleting row_history';

DELETE FROM survey_storage ss
    USING bad_survey bs
    WHERE ss.id = bs.survey_id

RAISE NOTICE 'Done with deleting survey_storage';

DELETE FROM survey_library sl
    USING bad_survey bs
    WHERE sl.id = bs.template_id;


RAISE NOTICE 'Done with deleting survey_library'; 

return true;
end;
 $function$

下面是我尝试持久化函数时的错误消息：

  SQL Error [42601]: ERROR: syntax error at or near "RAISE"
  Position: 1893
  ERROR: syntax error at or near "RAISE"
  Position: 1893
  ERROR: syntax error at or near "RAISE"
  Position: 1893

另一次更新

---

在注释掉所有的raisenotice之后，我能够继续使用该函数。现在，为什么“提高通知”不起作用？

您在报告行之前的语句中缺少分号

从测量存储中删除
使用bad_调查bs
其中ss.id=bs.survey_id；——向我们显示给您一个错误的代码。@Kamil，我已经更新了问题以显示给您一个错误的代码。很抱歉，没有用更正更新问题，我发现当我注释掉“提出通知”时“行&它仍然没有编译。但是在分号更正之后，它仍然在发出通知时失败。然而，我被告知我的代码是正确的，只有我执行代码的ide DBeaver不能处理raise通知或任何其他raise语句。我们将直接从shell在服务器中运行代码，看看是否有问题。