**

嗨，这个查询可能会有帮助:

**

SELECT 
  *
FROM 
  message 

WHERE 
  `Id` IN (
    SELECT 
      MAX(`Id`) 
    FROM 
      message 
    GROUP BY 
      `Name`
  ) 
ORDER BY 
   `Id` DESC

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

这个怎么样:

SELECT DISTINCT ON (name) *
FROM messages
ORDER BY name, id DESC;

我也有类似的问题(在postgresql tough上)，在1M的记录表上。这个解决方案需要1.7秒，而使用LEFT JOIN的解决方案需要44秒。 在我的例子中，我必须根据NULL值过滤您的名称字段的对应项，从而使性能更好0.2秒

从MySQL 8.0.14开始，也可以使用横向派生表来实现:

SELECT t.*
FROM messages t
JOIN LATERAL (
  SELECT name, MAX(id) AS id 
  FROM messages t1
  WHERE t.name = t1.name
  GROUP BY name
) trn ON t.name = trn.name AND t.id = trn.id

db < >小提琴

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 ;

我也遇到过类似的问题

子查询和加入救援

SELECT p."Date"
        ,p."Symbol"
        ,p."ratio_roll_qtr_ret"
    FROM PUBLIC."prices_vw" AS p
    JOIN (
        SELECT "Symbol"
            ,max("Date")
        FROM PUBLIC."prices_vw"
        GROUP BY "Symbol"
        ) AS sq ON p."Date" = sq."max"
        AND p."Symbol" = sq."Symbol"
    WHERE p."ratio_roll_qtr_ret" IS NOT NULL
    ORDER BY "ratio_roll_qtr_ret" DESC;