获取表中第一个可用索引，包括0

Posted 2023-03-31

【中文标题】获取表中第一个可用索引，包括0

【英文标题】：Get first available index in the table, including 0

【发布时间】：2020-08-08 12:53:33

【问题描述】：

我有一个名为player_chest 的表，其中包含以下字段：

idx integer NOT NULL,
player_id integer NOT NULL,
chest_id integer NOT NULL,

我正在使用以下查询来获取表中的第一个可用索引：

SELECT c.idx + 1 FROM player_chest c WHERE c.player_id = 2 AND NOT EXISTS (
  SELECT 1 FROM player_chest c1 WHERE c1.player_id = 2 AND c1.idx = c.idx + 1
) ORDER BY c.idx;

查询工作的示例：

chest_id | idx | player_id
       0 |   0 |         2
       1 |   1 |         2
       2 |   2 |         2
       1 |   4 |         2

返回 3

查询不工作的示例： 查询工作的示例：

chest_id | idx | player_id
       1 |   1 |         2
       2 |   2 |         2
       1 |   4 |         2

返回 3，但我希望它返回 0

但它在可用时无法返回索引 0。我该如何解决？

【问题讨论】：

样本数据和期望的结果会有所帮助。

我添加了示例数据

【参考方案1】：

一种解决方案使用窗口函数：

select (case when min_idx > 0 then 0
             else 1 + min(idx) filter (where next_idx is distinct from idx + 1)
        end)
from (select pc.*,
             lead(idx) over (partition by player_id order by idx) as next_idx,
             min(idx) over (partition by player_id) as min_idx
      from player_chest pc
      where player_id = 2
     ) pc
group by player_id, min_idx;

如果你想要一个即使玩家不在桌子上也能返回0的版本，那么：

select (case when max(min_idx) > 0 or max(min_idx) is null then 0
             else 1 + min(idx) filter (where next_idx is distinct from idx + 1)
        end)
from (select pc.*,
             lead(idx) over (partition by player_id order by idx) as next_idx,
             min(idx) over (partition by player_id) as min_idx
      from player_chest pc
      where player_id = 2
     ) pc

【讨论】：

你的意思是如果玩家在 player_chest 表中没有条目，那么只有第二个查询会起作用？