下面是另一种获取最后一条相关记录的方法，使用GROUP_CONCAT和SUBSTRING_INDEX从列表中选择一条记录

SELECT 
  `Id`,
  `Name`,
  SUBSTRING_INDEX(
    GROUP_CONCAT(
      `Other_Columns` 
      ORDER BY `Id` DESC 
      SEPARATOR '||'
    ),
    '||',
    1
  ) Other_Columns 
FROM
  messages 
GROUP BY `Name` 

上面的查询将组所有Other_Columns在同一名称组和使用ORDER BY id DESC将连接所有Other_Columns在一个特定的组降序与提供的分隔符在我的情况下，我已经使用||，使用SUBSTRING_INDEX在这个列表将选择第一个

小提琴演示

如果您真正关心的是性能，则可以在表上引入一个名为IsLastInGroup的类型为BIT的新列。

在最后的列上设置为true，并在每一行插入/更新/删除时保持该值。写的速度会变慢，但读的时候会受益。这取决于您的用例，我只建议在以读取为重点的情况下使用它。

因此，您的查询将如下所示:

SELECT * FROM Messages WHERE IsLastInGroup = 1

这是另一个没有子查询的选项。

本解决方案使用MySQL的LAST_VALUE窗口函数，利用窗口函数框架提供MySQL工具。

SELECT DISTINCT 
    LAST_VALUE(Id)            
        OVER(PARTITION BY Name 
             ORDER     BY Id 
             ROWS BETWEEN 0 PRECEDING 
                      AND UNBOUNDED FOLLOWING),
    Name,
    LAST_VALUE(Other_Columns)            
        OVER(PARTITION BY Name 
             ORDER     BY Id 
             ROWS BETWEEN 0 PRECEDING 
                      AND UNBOUNDED FOLLOWING)
FROM   
    tab

在这里试试。

显然，有很多不同的方法得到相同的结果，你的问题似乎是什么是一个有效的方法得到最后的结果在每组MySQL。如果你正在处理大量的数据，并且假设你正在使用InnoDB，即使是最新版本的MySQL(比如5.7.21和8.0.4-rc)，那么可能没有一个有效的方法来做到这一点。

有时我们需要对超过6000万行的表执行此操作。

对于这些示例，我将使用只有大约150万行的数据，其中查询需要为数据中的所有组找到结果。在我们的实际情况中，我们经常需要返回大约2000个组的数据(假设不需要检查很多数据)。

我将使用以下表格:

CREATE TABLE temperature(
  id INT UNSIGNED NOT NULL AUTO_INCREMENT, 
  groupID INT UNSIGNED NOT NULL, 
  recordedTimestamp TIMESTAMP NOT NULL, 
  recordedValue INT NOT NULL,
  INDEX groupIndex(groupID, recordedTimestamp), 
  PRIMARY KEY (id)
);

CREATE TEMPORARY TABLE selected_group(id INT UNSIGNED NOT NULL, PRIMARY KEY(id)); 

温度表由大约150万条随机记录和100个不同的组填充。 selected_group由这100个组填充(在我们的例子中，所有组通常小于20%)。

由于该数据是随机的，这意味着多行可以具有相同的recordedTimestamps。我们想要的是获得所有选中的组的列表，这些组按groupID顺序排列，每个组的最后一个recordedTimestamp，如果同一个组有多个这样的匹配行，那么这些行的最后一个匹配id。

如果假设MySQL有一个last()函数，它从一个特殊的ORDER BY子句中返回最后一行的值，那么我们可以简单地这样做:

SELECT 
  last(t1.id) AS id, 
  t1.groupID, 
  last(t1.recordedTimestamp) AS recordedTimestamp, 
  last(t1.recordedValue) AS recordedValue
FROM selected_group g
INNER JOIN temperature t1 ON t1.groupID = g.id
ORDER BY t1.recordedTimestamp, t1.id
GROUP BY t1.groupID;

which would only need to examine a few 100 rows in this case as it doesn't use any of the normal GROUP BY functions. This would execute in 0 seconds and hence be highly efficient. Note that normally in MySQL we would see an ORDER BY clause following the GROUP BY clause however this ORDER BY clause is used to determine the ORDER for the last() function, if it was after the GROUP BY then it would be ordering the GROUPS. If no GROUP BY clause is present then the last values will be the same in all of the returned rows.

但是MySQL没有这个，所以让我们看看它有什么不同的想法，并证明这些都不是有效的。

示例1

SELECT t1.id, t1.groupID, t1.recordedTimestamp, t1.recordedValue
FROM selected_group g
INNER JOIN temperature t1 ON t1.id = (
  SELECT t2.id
  FROM temperature t2 
  WHERE t2.groupID = g.id
  ORDER BY t2.recordedTimestamp DESC, t2.id DESC
  LIMIT 1
);

这检查了3,009,254行，在5.7.21上花费了0.859秒，在8.0.4-rc上花费了稍长的时间

示例2

SELECT t1.id, t1.groupID, t1.recordedTimestamp, t1.recordedValue 
FROM temperature t1
INNER JOIN ( 
  SELECT max(t2.id) AS id   
  FROM temperature t2
  INNER JOIN (
    SELECT t3.groupID, max(t3.recordedTimestamp) AS recordedTimestamp
    FROM selected_group g
    INNER JOIN temperature t3 ON t3.groupID = g.id
    GROUP BY t3.groupID
  ) t4 ON t4.groupID = t2.groupID AND t4.recordedTimestamp = t2.recordedTimestamp
  GROUP BY t2.groupID
) t5 ON t5.id = t1.id;

这检查了1,505,331行，在5.7.21上花费了约1.25秒，在8.0.4-rc上花费了稍长时间

示例3

SELECT t1.id, t1.groupID, t1.recordedTimestamp, t1.recordedValue 
FROM temperature t1
WHERE t1.id IN ( 
  SELECT max(t2.id) AS id   
  FROM temperature t2
  INNER JOIN (
    SELECT t3.groupID, max(t3.recordedTimestamp) AS recordedTimestamp
    FROM selected_group g
    INNER JOIN temperature t3 ON t3.groupID = g.id
    GROUP BY t3.groupID
  ) t4 ON t4.groupID = t2.groupID AND t4.recordedTimestamp = t2.recordedTimestamp
  GROUP BY t2.groupID
)
ORDER BY t1.groupID;

这检查了3,009,685行，在5.7.21上花费了约1.95秒，在8.0.4-rc上花费了稍长时间

示例4

SELECT t1.id, t1.groupID, t1.recordedTimestamp, t1.recordedValue
FROM selected_group g
INNER JOIN temperature t1 ON t1.id = (
  SELECT max(t2.id)
  FROM temperature t2 
  WHERE t2.groupID = g.id AND t2.recordedTimestamp = (
      SELECT max(t3.recordedTimestamp)
      FROM temperature t3 
      WHERE t3.groupID = g.id
    )
);

这检查了6,137,810行，在5.7.21上花费了约2.2秒，在8.0.4-rc上花费了稍长时间

示例5

SELECT t1.id, t1.groupID, t1.recordedTimestamp, t1.recordedValue
FROM (
  SELECT 
    t2.id, 
    t2.groupID, 
    t2.recordedTimestamp, 
    t2.recordedValue, 
    row_number() OVER (
      PARTITION BY t2.groupID ORDER BY t2.recordedTimestamp DESC, t2.id DESC
    ) AS rowNumber
  FROM selected_group g 
  INNER JOIN temperature t2 ON t2.groupID = g.id
) t1 WHERE t1.rowNumber = 1;

这检查了6,017,808行，在8.0.4-rc上花费了约4.2秒

例子6

SELECT t1.id, t1.groupID, t1.recordedTimestamp, t1.recordedValue 
FROM (
  SELECT 
    last_value(t2.id) OVER w AS id, 
    t2.groupID, 
    last_value(t2.recordedTimestamp) OVER w AS recordedTimestamp, 
    last_value(t2.recordedValue) OVER w AS recordedValue
  FROM selected_group g
  INNER JOIN temperature t2 ON t2.groupID = g.id
  WINDOW w AS (
    PARTITION BY t2.groupID 
    ORDER BY t2.recordedTimestamp, t2.id 
    RANGE BETWEEN UNBOUNDED PRECEDING AND UNBOUNDED FOLLOWING
  )
) t1
GROUP BY t1.groupID;

这检查了6,017,908行，在8.0.4-rc上花费了约17.5秒

例7

SELECT t1.id, t1.groupID, t1.recordedTimestamp, t1.recordedValue 
FROM selected_group g
INNER JOIN temperature t1 ON t1.groupID = g.id
LEFT JOIN temperature t2 
  ON t2.groupID = g.id 
  AND (
    t2.recordedTimestamp > t1.recordedTimestamp 
    OR (t2.recordedTimestamp = t1.recordedTimestamp AND t2.id > t1.id)
  )
WHERE t2.id IS NULL
ORDER BY t1.groupID;

这只花了很长时间，所以我不得不杀了它。

你也可以从这里看。

http://sqlfiddle.com/ !9 / ef42b / 9

第一个解决方案

SELECT d1.ID,Name,City FROM Demo_User d1
INNER JOIN
(SELECT MAX(ID) AS ID FROM Demo_User GROUP By NAME) AS P ON (d1.ID=P.ID);

第二个解决方案

SELECT * FROM (SELECT * FROM Demo_User ORDER BY ID DESC) AS T GROUP BY NAME ;

你可以通过计数来分组，也可以得到分组的最后一项，比如:

SELECT 
    user,
    COUNT(user) AS count,
    MAX(id) as last
FROM request 
GROUP BY user