我想在一个查询中返回每个部分的前 10 条记录.任何人都可以帮助如何做到这一点?部分是表中的列之一.
I want to return top 10 records from each section in one query. Can anyone help with how to do it? Section is one of the columns in the table.
数据库是 SQL Server 2005.我想按输入的日期返回前 10 名.部分是业务、本地和功能.对于某个特定日期,我只需要前 (10) 个业务行(最新条目)、前 (10) 个本地行和前 (10) 个特征.
Database is SQL Server 2005. I want to return the top 10 by date entered. Sections are business, local, and feature. For one particular date I want only the top (10) business rows (most recent entry), the top (10) local rows, and the top (10) features.
如果您使用的是 SQL 2005,您可以执行以下操作...
If you are using SQL 2005 you can do something like this...
SELECT rs.Field1,rs.Field2
FROM (
SELECT Field1,Field2, Rank()
over (Partition BY Section
ORDER BY RankCriteria DESC ) AS Rank
FROM table
) rs WHERE Rank <= 10
如果您的 RankCriteria 有平局,那么您可能会返回超过 10 行,Matt 的解决方案可能更适合您.
If your RankCriteria has ties then you may return more than 10 rows and Matt's solution may be better for you.
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