选择每位用户最近日期的行

我有一张用户签入和签出时间的表（“ lms_attendance”），如下所示：

id  user    time    io (enum)
1   9   1370931202  out
2   9   1370931664  out
3   6   1370932128  out
4   12  1370932128  out
5   12  1370933037  in

我正在尝试创建此表的视图，该视图将仅输出每个用户ID的最新记录，同时为我提供“ in”或“ out”值，如下所示：

id  user    time    io
2   9   1370931664  out
3   6   1370932128  out
5   12  1370933037  in

到目前为止，我已经很接近了，但是我意识到视图将不接受子查询，这使它变得更加困难。我得到的最接近的查询是：

select 
    `lms_attendance`.`id` AS `id`,
    `lms_attendance`.`user` AS `user`,
    max(`lms_attendance`.`time`) AS `time`,
    `lms_attendance`.`io` AS `io` 
from `lms_attendance` 
group by 
    `lms_attendance`.`user`, 
    `lms_attendance`.`io`

但是我得到的是：

id  user    time    io
3   6   1370932128  out
1   9   1370931664  out
5   12  1370933037  in
4   12  1370932128  out

这很接近，但并不完美。我知道最后一个分组依据不应存在，但是没有它，它将返回最近的时间，但不会返回相对IO值。

有任何想法吗？谢谢！

查询：

SQLFIDDLE示例

SELECT t1.*
FROM lms_attendance t1
WHERE t1.time = (SELECT MAX(t2.time)
                 FROM lms_attendance t2
                 WHERE t2.user = t1.user)

结果：

| ID | USER |       TIME |  IO |
--------------------------------
|  2 |    9 | 1370931664 | out |
|  3 |    6 | 1370932128 | out |
|  5 |   12 | 1370933037 |  in |

每次都会起作用的解决方案：

SQLFIDDLE示例

SELECT t1.*
FROM lms_attendance t1
WHERE t1.id = (SELECT t2.id
                 FROM lms_attendance t2
                 WHERE t2.user = t1.user            
                 ORDER BY t2.id DESC
                 LIMIT 1)

无需尝试重​​新发明轮子，因为这是常见的每组最多n个问题。提出了非常好的解决方案。

我更喜欢没有子查询（因此易于在视图中使用）的最简单的解决方案（请参阅SQLFiddle，已更新的贾斯汀的）：

SELECT t1.*
FROM lms_attendance AS t1
LEFT OUTER JOIN lms_attendance AS t2
  ON t1.user = t2.user 
        AND (t1.time < t2.time 
         OR (t1.time = t2.time AND t1.Id < t2.Id))
WHERE t2.user IS NULL

在同一组中有两个具有相同最大值的不同记录的情况下，这也可以使用-借助的技巧(t1.time = t2.time AND t1.Id < t2.Id)。我在这里所做的只是确保在同一用户的两条记录具有相同时间的情况下，仅选择一条。准则是否真的无关紧要Id-基本上任何保证唯一的准则都可以在这里完成工作。

基于@TMS答案，我喜欢它，因为不需要子查询，但是我认为省略该'OR'部分就足够了，而且更易于理解和阅读。

SELECT t1.*
FROM lms_attendance AS t1
LEFT JOIN lms_attendance AS t2
  ON t1.user = t2.user 
        AND t1.time < t2.time
WHERE t2.user IS NULL

如果您对空时间行不感兴趣，可以在WHERE子句中过滤它们：

SELECT t1.*
FROM lms_attendance AS t1
LEFT JOIN lms_attendance AS t2
  ON t1.user = t2.user 
        AND t1.time < t2.time
WHERE t2.user IS NULL and t1.time IS NOT NULL

已经解决，但仅作记录，另一种方法是创建两个视图...

CREATE TABLE lms_attendance
(id int, user int, time int, io varchar(3));

CREATE VIEW latest_all AS
SELECT la.user, max(la.time) time
FROM lms_attendance la 
GROUP BY la.user;

CREATE VIEW latest_io AS
SELECT la.* 
FROM lms_attendance la
JOIN latest_all lall 
    ON lall.user = la.user
    AND lall.time = la.time;

INSERT INTO lms_attendance 
VALUES
(1, 9, 1370931202, 'out'),
(2, 9, 1370931664, 'out'),
(3, 6, 1370932128, 'out'),
(4, 12, 1370932128, 'out'),
(5, 12, 1370933037, 'in');

SELECT * FROM latest_io;

单击此处查看它在SQL Fiddle中的运行情况

select b.* from 

    (select 
        `lms_attendance`.`user` AS `user`,
        max(`lms_attendance`.`time`) AS `time`
    from `lms_attendance` 
    group by 
        `lms_attendance`.`user`) a

join

    (select * 
    from `lms_attendance` ) b

on a.user = b.user
and a.time = b.time

 select result from (
     select vorsteuerid as result, count(*) as anzahl from kreditorenrechnung where kundeid = 7148
     group by vorsteuerid
 ) a order by anzahl desc limit 0,1

如果您使用的是MySQL 8.0或更高版本，则可以使用Window函数：

查询：

DBFiddleExample

SELECT DISTINCT
FIRST_VALUE(ID) OVER (PARTITION BY lms_attendance.USER ORDER BY lms_attendance.TIME DESC) AS ID,
FIRST_VALUE(USER) OVER (PARTITION BY lms_attendance.USER ORDER BY lms_attendance.TIME DESC) AS USER,
FIRST_VALUE(TIME) OVER (PARTITION BY lms_attendance.USER ORDER BY lms_attendance.TIME DESC) AS TIME,
FIRST_VALUE(IO) OVER (PARTITION BY lms_attendance.USER ORDER BY lms_attendance.TIME DESC) AS IO
FROM lms_attendance;

结果：

| ID | USER |       TIME |  IO |
--------------------------------
|  2 |    9 | 1370931664 | out |
|  3 |    6 | 1370932128 | out |
|  5 |   12 | 1370933037 |  in |

与使用Justin提出的解决方案相比，我看到的优势在于，它使您甚至可以从子查询中选择每个用户（或每个id或每个对象）具有最新数据的行，而无需中间视图或表。

如果您运行的是HANA，速度也会快7倍左右：D

好的，这可能是黑客入侵或容易出错，但无论如何它还是可以正常工作的-

SELECT id, MAX(user) as user, MAX(time) as time, MAX(io) as io FROM lms_attendance GROUP BY id;

试试这个查询：

  select id,user, max(time), io 
  FROM lms_attendance group by user;

可能您可以按用户分组，然后按时间排序。如下所示

  SELECT * FROM lms_attendance group by user order by time desc;

这为我工作：

SELECT user, time FROM 
(
    SELECT user, time FROM lms_attendance --where clause
) AS T 
WHERE (SELECT COUNT(0) FROM table WHERE user = T.user AND time > T.time) = 0
ORDER BY user ASC, time DESC