Postgresql：减号运算符不能与参数一起使用吗？只有硬编码值？

Question

以下查询使用索引删除条目：

 const deleteGameQuery = ` 
                update users 
                set games = games - 1
                where username = 
            `

如果我将索引作为参数传递，则不会删除任何内容：

const gameIndex = rowsCopy[0].games.findIndex(obj => obj.game == gameID).toString();


            const deleteGameQuery = ` 
                update users 
                set games = games - 
                where username = 
            `
    
          const { rows } = await query(deleteGameQuery, [gameIndex, username]);

          ctx.body = rows;

gameIndex参数只是一个字符串，和我输入的一样。那么为什么它似乎没有读取值呢？这是不允许的吗？

游戏栏是jsonb数据类型，数据如下：

[
    {
        "game": "cyberpunk-2077",
        "status": "Backlog",
        "platform": "Any"
    },
    {
        "game": "new-pokemon-snap",
        "status": "Backlog",
        "platform": "Any"
    }
]

Answer 1

问题是您传递的是文本而不是整数。您需要传递一个整数。我不确定您的数据库接口如何传递整数，请尝试删除 toString() 并确保 gameIndex 是 Number.

const gameIndex = rowsCopy[0].games.findIndex(obj => obj.game == gameID).

array - integer 和 array - text 是两个不同的意思。

array - 1 从数组中删除 second 元素。

select '[1,2,3]'::jsonb - 1;

 [1, 3]

array - '1' 搜索条目 '1' 并将其删除。

select '["1","2","3"]'::jsonb - '1';

 ["2", "3"]

-- Here, nothing is removed because 1 != '1'.
select '[1,2,3]'::jsonb - '1';

 [1, 2, 3]

当你传入一个参数时，它会根据它的类型被query翻译。如果您传递一个数字，它将被翻译为 1。如果您传递一个字符串，它将被翻译为 '1'。（或者至少它应该是这样工作的，我并不完全熟悉 Javascript 数据库。）

附带说明一下，这种数据最好用 join table.

来处理

create table games (
  id bigserial primary key,
  name text not null,
  status text not null,
  platform text not null
);

create table users (
  id bigserial primary key,
  username text not null
);

create table game_users (
  game_id bigint not null references games,
  user_id bigint not null references users,

  -- If a user can only have each game once.
  unique(game_id, user_id)
);

-- User 1 has games 1 and 2. User 2 has game 2.
insert into game_users (game_id, user_id) values (1, 1), (2, 1), (2,2);

-- User 1 no longer has game 1.
delete from game_users where game_id = 1 and user_id = 1;

您还会有一个平台 table 和一个 game_platforms 连接 table。

加入 table 有点麻烦，但它们是 SQL 存储关系的方式。 JSONB很有用，但不能代替关系

Answer 2

您可以尝试避免在 postgres 之外分解对象，并像这样在查询内部操作 jsonb 结构：

create table gameplayers as (select 1 as id, '[
{
    "game": "cyberpunk-2077",
    "status": "Backlog",
    "platform": "Any"
},
{
    "game": "new-pokemon-snap",
    "status": "Backlog",
    "platform": "Any"
},
{
    "game": "gameone",
    "status": "Backlog",
    "platform": "Any"
}
]'::jsonb games);



with
 ungroupped as (select * from gameplayers g, jsonb_to_recordset(g.games) 
  as (game text, status text, platform text)),
 filtered as (select id, 
     jsonb_agg(
      json_build_object('game', game, 
                        'status', status, 
                        'platfrom', platform
                       )
              ) games 
  from ungroupped where game not like 'cyberpunk-2077' group by id)     
UPDATE gameplayers as g set games=f.games 
       from filtered f where f.id=g.id;

Postgresql：减号运算符不能与参数一起使用吗？只有硬编码值？

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