MySQL Lead函数返回NULL

我不是SQL专家,但这很有效:

With e AS (
    SELECT id, Name, Lead(Name) OVER (ORDER BY id) AS 'next' from MyTable 
) SELECT Id, Name, next FROM e WHERE Name='French'

可能有更有效的解决方案,但由于该表是关于语言的,我不希望它包含数百万条记录。因此,我认为性能几乎不会成为问题。

编辑 这段代码也做了同样的事情,我想它对大表来说确实更高效,因为它不会首先计算整个表的所有“next”字段:

SELECT d.*, e.Next
    FROM MyTable d OUTER APPLY
    (SELECT TOP(1) Name AS Next
    FROM Mytable e 
    WHERE e.Id > d.Id
    ORDER BY Id
    ) AS e 
    WHERE d.Name = 'French'