将对象的插值数组转换为复合类型的postgresql数组的正确方法是什么?

fykwrbwg  于 2022-11-23  发布在  PostgreSQL
关注(0)|答案(1)|浏览(149)

我使用SQL函数来执行多重插入,但是因为它们不能接受记录集作为参数,所以我必须先将它们转换为数组。它对基元数组很有效,因为它们可以简单地用CAST (${value} as primitive_type[])来转换,然后用它来完成。
但是多插入查询需要复合类型数组,CAST()似乎不能处理这些数组,因为它需要一列输入。
所有查询都显示在这个小提琴上:https://dbfiddle.uk/w_Qbq-lw

表和类型

CREATE TABLE accounts (
  id bigint GENERATED ALWAYS AS IDENTITY PRIMARY KEY,
  created_at timestamptz NOT NULL DEFAULT CURRENT_TIMESTAMP,
  login text NOT NULL,
  password text NOT NULL,
  email text
);

CREATE TYPE account_init AS (
  login text,
  password text,
  email text
);

函数

CREATE FUNCTION get_accounts(
  pagination_limit bigint DEFAULT 25,
  pagination_offset bigint DEFAULT 0,
  account_ids bigint[] DEFAULT NULL
)
RETURNS TABLE (
  id bigint,
  created_at timestamptz,
  login text,
  password text,
  email text
)
LANGUAGE SQL
AS $BODY$
  WITH input_accounts AS (
    SELECT
      id,
      created_at,
      login,
      password,
      email
    FROM
      accounts
    WHERE
      account_ids IS NULL OR id = ANY (account_ids)
    ORDER BY
      id
    LIMIT pagination_limit
    OFFSET pagination_offset
  )
  SELECT
    id,
    created_at,
    login,
    password,
    email
  FROM
    input_accounts
  ORDER BY
    id
$BODY$;

CREATE FUNCTION create_accounts(
  account_inits account_init[]
)
RETURNS TABLE (
  id bigint,
  created_at timestamptz,
  login text,
  password text,
  email text
)
LANGUAGE SQL
AS $BODY$
  WITH new_accounts AS (
    INSERT INTO accounts ( 
      login, 
      password, 
      email 
    )
    SELECT 
      login, 
      password, 
      email
    FROM 
      unnest(account_inits)
    RETURNING
      id
  )
  SELECT
    id,
    created_at,
    login,
    password,
    email
  FROM
    get_accounts(
      NULL,
      NULL,
      ARRAY(
        SELECT
          id
        FROM
          new_accounts
      )
    )
  ORDER BY
    id
$BODY$;

初始化数据

const account_inits = [
  {
    login:"EC4A42323F", 
    password: "3DF1542F23A29B73281EEC5EBB55FFE18C253A7E800E7A541B"
  },
  {
    login:"1D771C1E52", 
    password: "2817029563CC722FBC3D53F9F29F0000898F9843518D882E4A", 
    email: "a@b"
  },
  {
    login:"FB66381D3A", 
    password: "C8F865AC1D54CFFA56DEBDEEB671C8EF110991BBB3B9EE57D2", 
    email: null
  }
]

用法

--- insert data
WITH input_inits AS (
  SELECT
    login,
    password,
    email
  FROM
    json_to_recordset(${account_inits:json}) AS input_init(
      login text,
      password text,
      email text
    )
),
input_data AS (
  SELECT
    array_agg(
      CAST (
        (
          login,
          password,
          email
        ) AS account_init
      )
    ) AS account_inits
  FROM
    input_inits
)
SELECT
  new_accounts.id,
  new_accounts.created_at,
  new_accounts.login,
  new_accounts.password,
  new_accounts.email
FROM
  input_data
  CROSS JOIN
  create_accounts(input_data.account_inits) AS new_accounts
ORDER BY
  new_accounts.id ASC
;

目前,我将其插值为:json,然后将其转换为CTE中的记录集,然后在第二个CTE中转换为复合类型数组,作为参数传递给函数。将对象数组传递给函数参数似乎需要做大量工作。我尝试过不使用:json转换,但遇到了与array[]相关的或malformed object literal语法错误。

k5ifujac

k5ifujac1#

“适当的”替代方案需要知道输入数组中对象的预期键(以及所有嵌套对象/对象数组),以便构造适当的元组数组。
这基本上需要某种运行时引用模式,即使实现得很好,仍然需要为每个新的查询文件编写样板代码。最好的情况是,它不会比helpers模块中的函数或查询中的json -> record set -> composite type array转换更好。

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