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？节是表中的列之一。
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，我只想要前 (10) 个业务行（最近的条目）、前 (10) 个本地行和前 (10) 个特征。
jazzrai
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14 回答
255
如果您使用的是 SQL 2005，则可以执行以下操作...
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 有联系，那么您可能会返回 10 多行，而 Matt 的解决方案可能更适合您。
于 2008-10-07T02:13:52.423 回答
114
在 T-SQL 中，我会这样做：
WITH TOPTEN AS (
SELECT *, ROW_NUMBER()
over (
PARTITION BY [group_by_field]
order by [prioritise_field]
) AS RowNo
FROM [table_name]
)
SELECT * FROM TOPTEN WHERE RowNo <= 10
于 2012-06-25T10:16:59.370 回答
42
SELECT r.*
FROM
(
SELECT
r.*,
ROW_NUMBER() OVER(PARTITION BY r.[SectionID]
ORDER BY r.[DateEntered] DESC) rn
FROM [Records] r
) r
WHERE r.rn <= 10
ORDER BY r.[DateEntered] DESC
于 2012-06-15T14:26:32.557 回答
35
这适用于 SQL Server 2005（编辑以反映您的说明）：
select *
from Things t
where t.ThingID in (
select top 10 ThingID
from Things tt
where tt.Section = t.Section and tt.ThingDate = @Date
order by tt.DateEntered desc
)
and t.ThingDate = @Date
order by Section, DateEntered desc
于 2008-10-07T02:09:02.107 回答
20
我这样做：
SELECT a.* FROM articles AS a
LEFT JOIN articles AS a2
ON a.section = a2.section AND a.article_date <= a2.article_date
GROUP BY a.article_id
HAVING COUNT(*) <= 10;
更新： 此 GROUP BY 示例仅适用于 MySQL 和 SQLite，因为这些数据库比有关 GROUP BY 的标准 SQL 更宽松。大多数 SQL 实现要求选择列表中不属于聚合表达式的所有列也在 GROUP BY 中。
于 2008-10-07T04:07:06.453 回答
18
如果我们使用 SQL Server >= 2005，那么我们可以只用一个选择来解决任务：
declare @t table (
Id int ,
Section int,
Moment date
);
insert into @t values
( 1 , 1 , '2014-01-01'),
( 2 , 1 , '2014-01-02'),
( 3 , 1 , '2014-01-03'),
( 4 , 1 , '2014-01-04'),
( 5 , 1 , '2014-01-05'),
( 6 , 2 , '2014-02-06'),
( 7 , 2 , '2014-02-07'),
( 8 , 2 , '2014-02-08'),
( 9 , 2 , '2014-02-09'),
( 10 , 2 , '2014-02-10'),
( 11 , 3 , '2014-03-11'),
( 12 , 3 , '2014-03-12'),
( 13 , 3 , '2014-03-13'),
( 14 , 3 , '2014-03-14'),
( 15 , 3 , '2014-03-15');
-- TWO earliest records in each Section
select top 1 with ties
Id, Section, Moment
from
@t
order by
case
when row_number() over(partition by Section order by Moment) <= 2
then 0
else 1
end;
-- THREE earliest records in each Section
select top 1 with ties
Id, Section, Moment
from
@t
order by
case
when row_number() over(partition by Section order by Moment) <= 3
then 0
else 1
end;
-- three LATEST records in each Section
select top 1 with ties
Id, Section, Moment
from
@t
order by
case
when row_number() over(partition by Section order by Moment desc) <= 3
then 0
else 1
end;
于 2014-12-26T15:05:26.147 回答
10
如果您知道这些部分是什么，您可以执行以下操作：
select top 10 * from table where section=1
union
select top 10 * from table where section=2
union
select top 10 * from table where section=3
于 2008-10-07T02:10:05.113 回答
9
我知道这个帖子有点老了，但我刚刚遇到了一个类似的问题（从每个类别中选择最新的文章），这是我想出的解决方案：
WITH [TopCategoryArticles] AS (
SELECT
[ArticleID],
ROW_NUMBER() OVER (
PARTITION BY [ArticleCategoryID]
ORDER BY [ArticleDate] DESC
) AS [Order]
FROM [dbo].[Articles]
)
SELECT [Articles].*
FROM
[TopCategoryArticles] LEFT JOIN
[dbo].[Articles] ON
[TopCategoryArticles].[ArticleID] = [Articles].[ArticleID]
WHERE [TopCategoryArticles].[Order] = 1
这与 Darrel 的解决方案非常相似，但克服了可能返回比预期更多行的 RANK 问题。
于 2011-02-14T12:10:18.583 回答
7
尝试了以下方法，它也适用于领带。
SELECT rs.Field1,rs.Field2
FROM (
SELECT Field1,Field2, ROW_NUMBER()
OVER (Partition BY Section
ORDER BY RankCriteria DESC ) AS Rank
FROM table
) rs WHERE Rank <= 10
于 2017-11-12T06:55:10.417 回答
6
如果要生成按部分分组的输出，则仅显示每个部分的前n条记录，如下所示：
SECTION SUBSECTION
deer American Elk/Wapiti
deer Chinese Water Deer
dog Cocker Spaniel
dog German Shephard
horse Appaloosa
horse Morgan
...那么以下内容应该适用于所有 SQL 数据库。如果您想要前 10 名，只需在查询末尾将 2 更改为 10 即可。
select
x1.section
, x1.subsection
from example x1
where
(
select count(*)
from example x2
where x2.section = x1.section
and x2.subsection <= x1.subsection
) <= 2
order by section, subsection;
建立：
create table example ( id int, section varchar(25), subsection varchar(25) );
insert into example select 0, 'dog', 'Labrador Retriever';
insert into example select 1, 'deer', 'Whitetail';
insert into example select 2, 'horse', 'Morgan';
insert into example select 3, 'horse', 'Tarpan';
insert into example select 4, 'deer', 'Row';
insert into example select 5, 'horse', 'Appaloosa';
insert into example select 6, 'dog', 'German Shephard';
insert into example select 7, 'horse', 'Thoroughbred';
insert into example select 8, 'dog', 'Mutt';
insert into example select 9, 'horse', 'Welara Pony';
insert into example select 10, 'dog', 'Cocker Spaniel';
insert into example select 11, 'deer', 'American Elk/Wapiti';
insert into example select 12, 'horse', 'Shetland Pony';
insert into example select 13, 'deer', 'Chinese Water Deer';
insert into example select 14, 'deer', 'Fallow';
于 2013-02-01T05:10:52.287 回答
6
Q) Finding TOP X records from each group(Oracle)
SQL> select * from emp e
2 where e.empno in (select d.empno from emp d
3 where d.deptno=e.deptno and rownum<3)
4 order by deptno
5 ;
EMPNO ENAME JOB MGR HIREDATE SAL COMM DEPTNO
7782 CLARK MANAGER 7839 09-JUN-81 2450 10
7839 KING PRESIDENT 17-NOV-81 5000 10
7369 SMITH CLERK 7902 17-DEC-80 800 20
7566 JONES MANAGER 7839 02-APR-81 2975 20
7499 ALLEN SALESMAN 7698 20-FEB-81 1600 300 30
7521 WARD SALESMAN 7698 22-FEB-81 1250 500 30
6 rows selected.
于 2011-02-20T17:36:30.367 回答
4
UNION运算符可以为您工作吗？每个部分有一个 SELECT，然后将它们联合在一起。猜猜它只适用于固定数量的部分。
于 2008-10-07T02:05:28.633 回答
4
虽然问题是关于 SQL Server 2005，但大多数人已经继续前进，如果他们确实找到了这个问题，那么在其他情况下可能首选的答案是使用CROSS APPLY本博客文章中所示的答案。
SELECT *
FROM t
CROSS APPLY (
SELECT TOP 10 u.*
FROM u
WHERE u.t_id = t.t_id
ORDER BY u.something DESC
) u
此查询涉及 2 个表。OP 的查询仅涉及 1 个表，在这种情况下，基于窗口函数的解决方案可能更有效。
于 2018-05-14T09:07:40.903 回答
1
You can try this approach. This query returns 10 most populated cities for each country.
SELECT city, country, population
FROM
(SELECT city, country, population,
@country_rank := IF(@current_country = country, @country_rank + 1, 1) AS country_rank,
@current_country := country
FROM cities
ORDER BY country, population DESC
) ranked
WHERE country_rank <= 10;
于 2017-01-25T22:41:16.930 回答