sql-server - 选择第 8 行后的前 3 行

Question

我有一张桌子：id, Group, Status, Subject。

我需要 group=2 AND status=2 的行，但我想跳过前 8 个结果（按 排序id）。这是我尝试过的，但它不会跳过前 8 行：

SELECT TOP 3 [id], LEFT ([subject],30) AS Title 
FROM [Katavot] 
    WHERE (([Status] = @Status) AND ([Group] = @Group)) 
ORDER BY [id] DESC

Answer 1

;WITH Katavod_Modified AS 
(
    SELECT id, LEFT ([subject],30) As Subject, [Group], [Status],  ROW_NUMBER() OVER (ORDER BY id) RN
    FROM [Katavot]
)
SELECT TOP 3 [id], LEFT ([subject],30) AS Title FROM [Katavod_Modified] WHERE 
(([Status] = @Status) AND ([Group] = @Group) AND (RN > 8)) ORDER BY [id] DESC 

看起来和这个很相似。

Answer 2

基于row_number

WITH MyResult AS
{
SELECT ROW_NUMBER() OVER(ORDER BY Id DESC) AS RowNumber, [id], LEFT ([subject],30) AS Title 
FROM [Katavot] 
    WHERE (([Status] = @Status) AND ([Group] = @Group)) 
ORDER BY [id] DESC
}
SELECT [id], Title 
FROM MyResult
WHERE RowNumber BETWEEN 9 AND 12

Answer 3

MySQL：

LIMIT x, y

微软SQL：

select top y *
from <table>
where <unique id> not in (select top x <unique id> from <table> order by <unique id>)
from <table> order by <unique id>