使用连接更新许多行非常慢

Question

我有一个包含五个相关字段的 table - id、source、iid、track_hash、alias。我想将所有条目分组到具有共同 track_hash 的组中，然后为每一行保存具有最低 source 的行的 id （打破联系以支持最高 iid) 从其组中进入 alias 字段。为此，我编写了以下查询：

with best as 
(SELECT id as bid, track_hash FROM
    (SELECT id, track_hash,
        RANK () OVER ( 
            PARTITION BY track_hash
            ORDER BY source asc, iid DESC
        ) rank
        from albums
    )
    where rank = 1
)
select bid, a.* from albums a inner join best
on a.track_hash = best.track_hash

这在 24k 行上花费了完全合理的 2 秒。现在，我不想简单地看到这个 id，而是想实际保存它。为此，我使用了以下非常相似的查询：

with best as 
(SELECT id as bid, track_hash FROM
    (SELECT id, track_hash,
        RANK () OVER ( 
            PARTITION BY track_hash
            ORDER BY source asc, iid DESC
        ) rank
        from albums
    )
    where rank = 1
)
update albums
set alias = bid FROM albums a inner join best
on a.track_hash = best.track_hash

然而，这个需要 1 到 10 分钟，我真的不明白为什么。无论如何，引擎不需要将每一行与其 best.id/alias 匹配，这正是我在更新时所做的吗？为什么会这样，我做错了什么？

查询计划如下所示：

MATERIALIZE 1
CO-ROUTINE 4
SCAN TABLE albums USING INDEX track_hash_idx
USE TEMP B-TREE FOR RIGHT PART OF ORDER BY
SCAN SUBQUERY 4
SCAN TABLE albums USING COVERING INDEX track_hash_idx
SEARCH SUBQUERY 1 USING AUTOMATIC PARTIAL COVERING INDEX (rank=?)
SEARCH TABLE albums AS a USING COVERING INDEX track_hash_idx (track_hash=?)

Answer 1

您不需要加入 albums（再次）。
UPDATE ... FROM 语法实际上提供了 albums 到 best:

的隐式连接

UPDATE albums AS a
SET alias = b.bid 
FROM best AS b
WHERE a.track_hash = b.track_hash

使用连接更新许多行非常慢

Update of many rows with join extremely slow

sql

sqlite

join

common-table-expression

sql-update