我一直在阅读有关 SQL SERVER LIMIT 子句的替代方案的大量资料.令人沮丧的是他们仍然拒绝适应它.无论如何,我真的无法解决这个问题.我要转换的查询是这样的...
I've been doing a lot of reading on alternatives to the LIMIT clause for SQL SERVER. It's so frustrating that they still refuse to adapt it. Anyway, I really havn't been able to get my head around this. The query I'm trying to convert is this...
SELECT ID, Name, Price, Image FROM Products ORDER BY ID ASC LIMIT $start_from, $items_on_page
如有任何帮助,我们将不胜感激,谢谢.
Any assistance would be much appreciated, thank you.
在 SQL Server 2012 中,支持 ANSI 标准 OFFSET
/FETCH
语法.我写了关于这个的博客 这里是 官方文档(这是ORDER BY
的扩展).您为 SQL Server 2012 转换的语法是:
In SQL Server 2012, there is support for the ANSI standard OFFSET
/ FETCH
syntax. I blogged about this and here is the official doc (this is an extension to ORDER BY
). Your syntax converted for SQL Server 2012 would be:
SELECT ID, Name, Price, Image
FROM Products
ORDER BY ID ASC
OFFSET (@start_from - 1) ROWS -- not sure if you need -1
-- because I don't know how you calculated @start_from
FETCH NEXT @items_on_page ROWS ONLY;
在此之前,您需要使用各种解决方法,包括 ROW_NUMBER()
方法.请参阅这篇文章和后续讨论.如果您不是在 SQL Server 2012 上,则不能使用标准语法或 MySQL 的非标准 LIMIT
,但可以使用更详细的解决方案,例如:
Prior to that, you need to use various workarounds, including the ROW_NUMBER()
method. See this article and the follow-on discussion. If you are not on SQL Server 2012, you can't use standard syntax or MySQL's non-standard LIMIT
but you can use a more verbose solution such as:
;WITH o AS
(
SELECT TOP ((@start_from - 1) + @items_on_page)
-- again, not sure if you need -1 because I
-- don't know how you calculated @start_from
RowNum = ROW_NUMBER() OVER (ORDER BY ID ASC)
/* , other columns */
FROM Products
)
SELECT
RowNum
/* , other columns */
FROM
o
WHERE
RowNum >= @start_from
ORDER BY
RowNum;
有很多其他方法可以给这只猫剥皮,这不太可能是最有效的,但语法方面可能是最简单的.我建议查看我发布的链接以及问题评论中提到的重复建议.
There are many other ways to skin this cat, this is unlikely to be the most efficient but syntax-wise is probably simplest. I suggest reviewing the links I posted as well as the duplicate suggestions noted in the comments to the question.
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