如何使用单个 PostgreSQL 查询进行深层复制？答案

【问题标题】：How do I do a deep copy with a single PostgreSQL query?如何使用单个 PostgreSQL 查询进行深层复制？
【发布时间】：2016-11-18 20:57:09
【问题描述】：

我有三张桌子：

CREATE TABLE offers
(
  id serial NOT NULL PRIMARY KEY,
  title character varying(1000) NOT NULL DEFAULT ''::character varying
);

CREATE TABLE items
(
  id serial NOT NULL PRIMARY KEY,
  offer_id integer NOT NULL,
  title character varying(1000) NOT NULL DEFAULT ''::character varying,
  CONSTRAINT items_offer_id_fkey FOREIGN KEY (offer_id)
      REFERENCES offers (id) 
);

CREATE TABLE sizes
(
  id serial NOT NULL PRIMARY KEY,
  item_id integer NOT NULL,
  title character varying(1000) NOT NULL DEFAULT ''::character varying,
  CONSTRAINT sizes_item_id_fkey FOREIGN KEY (item_id)
      REFERENCES items (id) 
);

我有 1 个包含 2 个项目的报价。每件商品有 2 种尺寸：

INSERT INTO offers (title) VALUES ('My Offer');
INSERT INTO items (offer_id, title) VALUES (1, 'First Item');
INSERT INTO items (offer_id, title) VALUES (1, 'Second Item');
INSERT INTO sizes (item_id, title) VALUES (1, 'First Size of Item #1');
INSERT INTO sizes (item_id, title) VALUES (1, 'Second Size of Item #1');
INSERT INTO sizes (item_id, title) VALUES (2, 'First Size of Item #2');
INSERT INTO sizes (item_id, title) VALUES (2, 'Second Size of Item #2');

有没有办法通过一次查询克隆所有商品和尺寸的商品？

我尝试用 CTE 解决它，这是我的 SQL：

WITH tmp_offers AS (
    INSERT INTO offers (title)
    SELECT title FROM offers WHERE id = 1
    RETURNING id
), tmp_items AS (
    INSERT INTO items (offer_id, title)
    (SELECT (SELECT id FROM tmp_offers), title FROM items WHERE offer_id = 1)
    RETURNING id
)
INSERT INTO sizes (item_id, title)
(SELECT (SELECT id FROM tmp_items), title FROM sizes WHERE id IN (
    SELECT sizes.id FROM sizes
    JOIN items ON items.id = sizes.item_id
    WHERE items.offer_id = 1
));

但是这个 SQL 导致了一个错误，我无法解决：

错误：用作表达式的子查询返回多行

非常感谢您的帮助。

附注我使用 PostgreSQL 9.5

【问题讨论】：

标签： sql postgresql common-table-expression bulkinsert

【解决方案1】：

这应该可行：

WITH tmp_offers AS (
    INSERT INTO offers (title)
    SELECT title 
    FROM offers 
    WHERE id = 1
    RETURNING id
), tmp_items AS (
    INSERT INTO items (offer_id, title)
    SELECT o.id, i.title 
    FROM items i
      cross join tmp_offers o
    WHERE i.offer_id = 1
    order by i.id
    RETURNING items.id
), numbered_new as (
  select ti.id, 
         row_number() over (order by ti.id) as rn
  from tmp_items ti
), numbered_old as (
  select i.id, 
         row_number() over (order by i.id) as rn
  from items i
  WHERE i.offer_id = 1
), item_mapper as (
  select n.id as new_item_id, 
         o.id as old_item_id
  from numbered_new n
     join numbered_old o on n.rn = o.rn
)
INSERT INTO sizes (item_id, title)
select im.new_item_id, s.title
from sizes s
  join item_mapper im on im.old_item_id = s.item_id;

在线示例：http://rextester.com/RYQUS11008

【讨论】：

谢谢您，先生。这简直太棒了，而且完全符合要求。
我想我在这里发现了一个问题。最后一个查询执行 where s.id in 并从 items 表中选择一个 id。
@IM_AG：你是对的。原来旧项目 ID 到尺寸表中新项目 ID 的映射并不像我想象的那么简单。我的新版本应该会解决这个问题。
天哪，这令人印象深刻。惊人的 SQL 技能。非常感谢。
我也很欣赏到 rextester 的链接及其与 postgresql“模拟”的功能。它还具有许多其他语言/功能。

【解决方案2】：

你很接近。这是需要工作的最终查询：

WITH tmp_offers AS (
      INSERT INTO offers (title)
          SELECT title FROM offers WHERE id = 1
          RETURNING id
     ),
    tmp_items AS (
     INSERT INTO items (offer_id, title)
        SELECT o.id, i.title
        FROM items i CROSS JOIN
             (SELECT id FROM tmp_offers) o
        WHERE i.offer_id = 1
        RETURNING id, title
)
INSERT INTO sizes (item_id, title)
    SELECT i.id, i.title
    FROM tmp_items i;

这里的主要区别在于tmp_items 现在有两列——它们似乎是您为此目的所需的列。

【讨论】：