Sql 获取表中第一个可用索引，包括0

我有一个名为player_cost的表，它有以下字段：

idx整数不为空， 玩家id整数不为空， 胸部id整数不为空， 我使用以下查询获取表中的第一个可用索引：

从播放器c中选择c.idx+1，其中c.player\u id=2且不存在 从player_胸部c1中选择1，其中c1.player_id=2和c1.idx=c.idx+1 c.idx订购； 查询工作的示例：

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

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

返回3

查询不起作用的示例： 查询工作的示例：

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

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

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但当索引0可用时，它无法返回索引0。如何修复它？

一个解决方案使用窗口功能：

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;

如果您想要一个即使播放机不在表中也返回0的版本，则：

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

---

一种解决方案使用窗口功能：

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;

如果您想要一个即使播放机不在表中也返回0的版本，则：

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

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示例数据和期望的结果会有帮助。我添加了示例数据示例数据和期望的结果会有帮助。我添加了示例数据u的意思是如果玩家在player\u chest表中没有条目，那么只有第二个查询可以工作？u的意思是如果玩家在player\u chest表中没有条目，那么只有第二个查询可以工作？