选择表中的最新记录答案

【问题标题】：Selecting the latest record within a table选择表中的最新记录
【发布时间】：2021-05-16 16:53:29
【问题描述】：

我有一个 Oracle v11 数据库，虽然我没有表的架构定义，但我在下面说明了我想要实现的目标。

这就是表格的样子

我正在尝试通过仅选择最新行来转换数据，表格保留更改历史记录，我对更改不感兴趣，只对每个当前问题的最新值感兴趣

这是我目前所拥有的。

select issueno,
  case (when fieldname = 'name' then string_value end) name,
  case (when fieldname = 'point' then string_value end) point
from issues
where issueno = 1234

上面查询的问题是它返回4行，我想只返回一行。

【问题讨论】：

标签： sql oracle oracle11g greatest-n-per-group

【解决方案1】：

您可以通过在MAX() KEEP (..) 值中为transition_date（或load_date 列，取决于您在查询中替换的意思）使用LAST ORDER BY 子句来获取最新日期作为

WITH i AS
(
SELECT CASE WHEN fieldname = 'name' THEN
             MAX(string_value) KEEP (DENSE_RANK LAST ORDER BY transition_date)
                               OVER (PARTITION BY issue_no, fieldname) 
             END AS name,
       CASE WHEN fieldname = 'point' THEN
             MAX(string_value) KEEP (DENSE_RANK LAST ORDER BY transition_date)
                               OVER (PARTITION BY issue_no, fieldname) 
             END AS point                
  FROM issues 
)
SELECT MAX(name) AS name, MAX(point) AS point 
  FROM i

但是，如果相关日期值出现 ties(equal values)，则考虑使用 DENSE_RANK() 函数来计算返回等于 1 和 ROW_NUMBER() 的值能够与主查询中的 JOIN 子句一起使用，例如

WITH i AS
(
SELECT i.*,
       DENSE_RANK() OVER ( PARTITION BY issue_no, fieldname 
                               ORDER BY transition_date DESC) AS dr,
       ROW_NUMBER() OVER ( PARTITION BY issue_no, fieldname 
                               ORDER BY transition_date DESC) AS rn
  FROM issues i
)
SELECT i1.string_value AS name, i2.string_value AS point
  FROM ( SELECT string_value, rn FROM i WHERE dr = 1 AND fieldname = 'name'  ) i1
  FULL JOIN ( SELECT string_value, rn FROM i WHERE dr = 1 AND fieldname = 'point' ) i2
    ON i2.rn = i1.rn

Demo

【讨论】：

【解决方案2】：

假设您希望按列 load_date 获得最新记录

select issueno,
  case (when fieldname = 'name' then string_value end) name,
  case (when fieldname = 'point' then string_value end) point
from issues
where issueno = 1234 and 
(fieldname , load_date) in (select fieldname ,max(load_date) from issues where issueno=1234 group by fieldname)

【讨论】：

【解决方案3】：

我会使用子查询 + 窗口函数来实现您的要求（假设您使用的是基于 load_date 来确定最新记录）

select issueno,
  case (when fieldname = 'name' then string_value end) name,
  case (when fieldname = 'point' then string_value end) point
from 
(
SELECT name, point, ROW_NUMBER() OVER(PARTITION BY ISSUENO, FIELDNAME ORDER BY LOAD_DATE DESC) RN 
FROM issues
)
where issueno = 1234
AND RN = 1

语法ROW_NUMBER() OVER ([query_partition_clause] order_by_clause) 实际上是一个窗口函数，它为每一行分配一个排名，由您在[query_partition_clause] order_by_clause 中声明规则的方式控制

【讨论】：

【解决方案4】：

看看这样的事情是否有帮助；在代码中读取 cmets。

SQL> with issues (issueno, fieldname, string_value,
  2               transition_date, transition_id, load_date)
  3    as
  4    -- sample data; you have it in a table, don't type that
  5    (select 1234, 'name', null , date '2021-01-01', 1, date '2021-01-02' from dual union all
  6     select 1234, 'name', 'Tom', date '2021-02-11', 2, date '2021-02-12' from dual union all
  7     select 1234, 'point', '0' , date '2021-02-04', 3, date '2021-02-05' from dual union all
  8     select 1234, 'point', '5' , date '2021-02-10', 5, date '2021-02-11' from dual
  9    ),
 10  -- query you need begins here
 11  temp as
 12    -- rank values partitioned by ISSUENO and FIELDNAME, sorted by TRANSITION_ID
 13    (select issueno, fieldname, string_value,
 14       row_number() over (partition by issueno, fieldname
 15                          order by transition_id desc) rn
 16     from issues
 17    )
 18  select issueno,
 19    max(case when fieldname = 'name'  then string_value end) name,
 20    max(case when fieldname = 'point' then string_value end) point
 21  from temp
 22  where rn = 1
 23  group by issueno;

   ISSUENO NAME       POINT
---------- ---------- ----------
      1234 Tom        5

SQL>

【讨论】：