Question

我有一个表，它是用户登录时的集合条目。

username, date,      value
--------------------------
brad,     1/2/2010,  1.1
fred,     1/3/2010,  1.0
bob,      8/4/2009,  1.5
brad,     2/2/2010,  1.2
fred,     12/2/2009, 1.3

etc..

如何创建一个能够为每个用户提供最新日期的查询？

更新：我忘记了我需要一个与最新日期一致的值。

Answer 1

select t.username, t.date, t.value
from MyTable t
inner join (
    select username, max(date) as MaxDate
    from MyTable
    group by username
) tm on t.username = tm.username and t.date = tm.MaxDate

Answer 2

使用窗口函数（适用于Oracle，Postgres 8.4，SQL Server 2005，DB2，Sybase，Firebird 3.0，MariaDB 10.3）

select * from (
    select
        username,
        date,
        value,
        row_number() over(partition by username order by date desc) as rn
    from
        yourtable
) t
where t.rn = 1

Answer 3

我看到大多数开发人员使用内联查询而不考虑其对大数据的影响。

简单地说，您可以通过以下方式实现这一目标：

SELECT a.username, a.date, a.value
FROM myTable a
LEFT OUTER JOIN myTable b
ON a.username = b.username 
AND a.date < b.date
WHERE b.username IS NULL
ORDER BY a.date desc;

Answer 4

要获得包含用户最长日期的整行：

select username, date, value
from tablename where (username, date) in (
    select username, max(date) as date
    from tablename
    group by username
)

Answer 5

SELECT *     
FROM MyTable T1    
WHERE date = (
   SELECT max(date)
   FROM MyTable T2
   WHERE T1.username=T2.username
)

Answer 6

这个应该为您编辑的问题提供正确的结果。

子查询确保只查找最新日期的行，而外部GROUP BY将处理关系。如果同一用户的同一日期有两个条目，则会返回value个最高的条目。

SELECT t.username, t.date, MAX( t.value ) value
FROM your_table t
JOIN (
       SELECT username, MAX( date ) date
       FROM your_table
       GROUP BY username
) x ON ( x.username = t.username AND x.date = t.date )
GROUP BY t.username, t.date

Answer 7

SELECT MAX(DATE) AS dates 
FROM assignment  
JOIN paper_submission_detail ON  assignment.PAPER_SUB_ID = 
     paper_submission_detail.PAPER_SUB_ID

Answer 8

根据我的经验，最快的方法是获取表中没有较新行的每一行。这是一个基准数据，其中包含我手头的一些数据。

另一个优点是所使用的语法非常简单，并且查询的含义非常容易掌握（采用所有行，以便考虑的用户名不存在更新的行）。

不存在

SELECT username, value
FROM t
WHERE NOT EXISTS (
  SELECT *
  FROM t AS witness
  WHERE witness.date > t.date
);

说明总费用：2.38136

ROW_NUMBER

SELECT username, value
FROM (
  SELECT username, value, row_number() OVER (PARTITION BY username ORDER BY date DESC) AS rn
  FROM t
) t2
WHERE rn = 1

总费用：61.5823

内部加入

SELECT t.username, t.value
FROM t
INNER JOIN (
  SELECT username, MAX(date) AS date
  FROM t
  GROUP BY username
) tm ON t.username = tm.username AND t.date = tm.date;

说明总费用：67.5439

左外联接

SELECT username, value
FROM t
LEFT OUTER JOIN t AS w ON t.username = w.username AND t.date < w.date
WHERE w.username IS NULL

说明总费用：62.964

解释计划来自一个大约有1万行的数据库，以XML格式存储。所使用的查询还包含谓词“ group_id ='1'”。

Answer 9

SELECT Username, date, value
 from MyTable mt
 inner join (select username, max(date) date
              from MyTable
              group by username) sub
  on sub.username = mt.username
   and sub.date = mt.date

将解决更新的问题。即使索引很好，它在大型表上也可能无法正常工作。

Answer 10

您还可以使用分析等级函数

    with temp as 
(
select username, date, RANK() over (partition by username order by date desc) as rnk from t
)
select username, rnk from t where rnk = 1

Answer 11

这与上面的答案之一相似，但是在我看来，它要简单得多。此外，它显示了交叉应用语句的良好用法。对于SQL Server 2005及更高版本...

select
    a.username,
    a.date,
    a.value,
from yourtable a
cross apply (select max(date) 'maxdate' from yourtable a1 where a.username=a1.username) b
where a.date=b.maxdate

Answer 12

我为自己的应用做了一些事情：

以下是查询：

select distinct i.userId,i.statusCheck, l.userName from internetstatus 
as i inner join login as l on i.userID=l.userID 
where nowtime in((select max(nowtime) from InternetStatus group by userID));

Answer 13

我的小编辑

自我join优于嵌套select
但group by并未向您primary key提供join
此密钥可由partition by与first_value（docs）

所以，这是一个查询：

select
 t.*
from 
 Table t inner join (
  select distinct first_value(ID) over(partition by GroupColumn order by DateColumn desc) as ID
  from Table
  where FilterColumn = 'value'
 ) j on t.ID = j.ID

优点：

使用任意列

where

select来自过滤行的所有列

缺点：

需要从2012年开始的MS SQL Server。

Answer 14

SELECT * FROM TABEL1 WHERE DATE= (SELECT MAX(CREATED_DATE) FROM TABEL1)

Answer 15

我用这种方式为我桌上的每个用户记录最后一条记录。这是根据PDA设备上检测到的最近时间获取销售员最后位置的查询。

CREATE FUNCTION dbo.UsersLocation()
RETURNS TABLE
AS
RETURN
Select GS.UserID, MAX(GS.UTCDateTime) 'LastDate'
From USERGPS GS
where year(GS.UTCDateTime) = YEAR(GETDATE()) 
Group By GS.UserID
GO
select  gs.UserID, sl.LastDate, gs.Latitude , gs.Longitude
        from USERGPS gs
        inner join USER s on gs.SalesManNo = s.SalesmanNo 
        inner join dbo.UsersLocation() sl on gs.UserID= sl.UserID and gs.UTCDateTime = sl.LastDate 
        order by LastDate desc

Answer 16

内部查询将返回当前用户的最新日期，外部查询将根据内部查询结果提取所有数据。

Answer 17

对于Oracle按降序对结果集进行排序并获取第一条记录，因此您将获得最新记录：

select * from mytable
where rownum = 1
order by date desc

Answer 18

SELECT t1.username, t1.date, value
FROM MyTable as t1
INNER JOIN (SELECT username, MAX(date)
            FROM MyTable
            GROUP BY username) as t2 ON  t2.username = t1.username AND t2.date = t1.date

Answer 19

SELECT *
FROM ReportStatus c
inner join ( SELECT 
  MAX(Date) AS MaxDate
  FROM ReportStatus ) m
on  c.date = m.maxdate

Answer 20

这也应该可以为用户获取所有最新条目。

SELECT username, MAX(date) as Date, value
FROM MyTable
GROUP BY username, value

Answer 21

SELECT DISTINCT Username, Dates,value 
FROM TableName
WHERE  Dates IN (SELECT  MAX(Dates) FROM TableName GROUP BY Username)


Username    Dates       value
bob         2010-02-02  1.2       
brad        2010-01-02  1.1       
fred        2010-01-03  1.0

Answer 22

您将使用聚合函数MAX和GROUP BY

SELECT username, MAX(date), value FROM tablename GROUP BY username, value

如何查询sql以获取每个用户的最新记录日期

22 个答案:

不存在

ROW_NUMBER

内部加入

左外联接