为每个类别选择前10条记录-Java 学习之路

169

我想在一个查询中返回每个部分的前10条记录 . 任何人都可以帮忙解决这个问题吗？ Section是表中的一列 .

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 . 对于一个特定日期，我只需要前（10）个业务行（最近的条目），前（10）个本地行和前（10）个特征 .

14 回答

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

回复于 2024-05-04T01:38:22+08:00

你可以尝试这种方法 . 此查询返回每个国家/地区的10个人口最多的城市 .

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;

回复于 2024-05-04T01:38:22+08:00

16

UNION运算符可能适合您吗？每个部分都有一个SELECT，然后将它们组合在一起 . 猜猜它只适用于固定数量的部分 .

回复于 2024-05-04T01:38:22+08:00

如果您使用的是SQL 2005，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有关系，那么您可能会返回超过10行，而Matt的解决方案可能对您更好 .

回复于 2024-05-04T01:38:22+08:00

问）从每个组中查找TOP X记录（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

选择了6行 .

回复于 2024-05-04T01:38:22+08:00

10
虽然问题是关于SQL Server 2005的，但是大多数人都已经开始了，如果他们确实发现了这个问题，那么在其他情况下，首选答案是什么？using CROSS APPLY as illustrated in this blog post .
```
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 .
回复于 2024-05-04T01:38:22+08:00

如果您知道这些部分是什么，您可以：

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

回复于 2024-05-04T01:38:22+08:00

如果要生成按部分分组的输出，则只显示每个部分的前n个记录，如下所示：

SECTION     SUBSECTION

deer        American Elk/Wapiti
deer        Chinese Water Deer
dog         Cocker Spaniel
dog         German Shephard
horse       Appaloosa
horse       Morgan

...然后，以下内容应该与所有SQL数据库一起工作 . 如果您想要前10名，只需在查询结束时将2更改为10 .

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;

Build ：

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';

回复于 2024-05-04T01:38:22+08:00

22
我是这样做的：
```
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;
```
update: GROUP BY的这个示例仅适用于MySQL和SQLite，因为这些数据库比关于GROUP BY的标准SQL更宽松 . 大多数SQL实现要求select-list中不属于聚合表达式的所有列也在GROUP BY中 .
回复于 2024-05-04T01:38:22+08:00

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

回复于 2024-05-04T01:38:22+08:00

尝试了以下内容，它也与关系一起工作 .

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

回复于 2024-05-04T01:38:22+08:00

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

回复于 2024-05-04T01:38:22+08:00

我知道这个帖子有点旧，但我刚刚碰到了类似的问题（从每个类别中选择最新的文章），这就是我提出的解决方案：

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 .

回复于 2024-05-04T01:38:22+08:00

182

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;

回复于 2024-05-04T01:38:22+08:00

为每个类别选择前10条记录

14 回答

相关问题