Question

假设您有以下表结构，您希望维基百科具有存储在不同表中的页面的标识和状态：

create table endUsers (
  uuid             UUID         primary key,
  created          timestamptz  default now()
);

create table endUserRevisions (
  id               bigserial    primary key,
  endUser          UUID         not null        references endUsers,
  modified         timestamptz  default now(),
  modifiedBy       UUID         not null        references portalUsers,
  name             text         not null,
  company          text         not null,
  email            text         not null
);

alter table endUsers add column
  latestRevision  bigint        not null        references endUserRevisions;

然后您想要将全新的用户插入此数据库，如：

with lastID as (
  insert into  endUserRevisions (endUser, name, company, email)
    values ('08e7882c-7596-43d1-b4cc-69f855210d72', 'a', 'b', 'c') returning id)
insert into endUsers (uuid, latestRevision)
  values ('08e7882c-7596-43d1-b4cc-69f855210d72', lastID);

-- or

with revision as (
  insert into  endUserRevisions (endUser, name, company, email)
    values ('08e7882c-7596-43d1-b4cc-69f855210d72', 'a', 'b', 'c') returning *)
insert into endUsers (uuid, latestRevision)
  values ('08e7882c-7596-43d1-b4cc-69f855210d72', revision.id);

这两种变体都失败了

专栏＆＃34; lastid＆＃34;不存在

或

缺少FROM-clause条目表＃34;最后＆＃34;

Answer 1

失败的原因是因为每个子查询都可以作为表访问周围的上下文，而不是普通值。换句话说，必须使用如下的select语句访问它：

with revision as (
  insert into  endUserRevisions (endUser, name, company, email)
    values ('08e7882c-7596-43d1-b4cc-79f855210d76', 'a', 'b', 'c') returning id)
insert into endUsers (uuid, latestRevision)
 values ('08e7882c-7596-43d1-b4cc-79f855210d76', (select id from revision));

-- or

with revision as (                                                                                                    
  insert into  endUserRevisions (endUser, name, company, email)                                                       
    values ('08e7882c-7596-43d1-b4cc-79f855210d74', 'a', 'b', 'c') returning id)                                      
insert into endUsers (uuid, latestRevision)                                                                           
  select '08e7882c-7596-43d1-b4cc-79f855210d74', revision.id from revision;

使用从postgresql中的with语句中的插入返回的id

1 个答案: