SQL 从 table 中获取每个项目的最新记录

Question

我正在尝试查询数据库以使用 max(Created_at) 并按 stockId 分组获取每只股票的最新记录，但随之而来的是 first id 每个项目及其 last Created_at.

怎样才能把最新的记录集中在一起？请帮忙。我的查询如下：

SELECT a.id as A_id, b.id as B_id, max(b.Created_at) AS Created_at 
  FROM beta b 
 INNER JOIN alpha a ON b.a_id = a.id 
 GROUP BY a.id

Answer 1

可以使用解析函数如ROW_NUMBER() :

SELECT A_id, B_id , Created_at
  FROM
  (
   SELECT a.id as A_id, b.id as B_id , b.Created_at,
          ROW_NUMBER() OVER (PARTITION BY a.id ORDER BY b.Created_at DESC) as rn
     FROM beta b 
     JOIN alpha a ON b.a_id = a.id 
  ) q
  WHERE rn = 1

您的记录按 stockID(a.id) 分组，并在按 created_at 列降序排序后选择 latest。如果关系（每个分组 stockId 的 created_at 值相等）很重要并且所有令人满意的结果都应包含在结果集中，则将 ROW_NUMBER() 替换为 DENSE_RANK().

Answer 2

据我了解你需要这样

declare @item table
(
itemid int
,itemname varchar(100)
)
insert into @item
values(1,'A'),(2,'B'),(3,'C')

declare @stock table
(
itemid int
,stockid int
,updatedate datetime
)
insert into @stock
values(1,1,'2020-08-04 13:11'),(1,1,'2020-08-04 14:11')
      ,(1,2,'2020-08-04 15:11'),(1,2,'2020-08-04 14:11') 


select * from @item a
 inner join
(select stockid,itemid,max(updatedate)lastupdatedate from @stock
group by stockid,itemid
)b on a.itemid=b.itemid

SQL 从 table 中获取每个项目的最新记录

SQL to fetch Latest record of each item from the table

mysql

sql

mysql-8.0