选择 table 内的最新记录

Question

我有一个 Oracle v11 数据库，虽然我没有 table 的模式定义，但我在下面说明了我想要实现的目标。

这就是 table 的样子

我正在尝试通过仅选择最新行来转换数据，table 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 return 有 4 行，我想 return 只有一行。

Answer 1

看看这样的事情是否有帮助；阅读代码中的注释。

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>

Answer 2

您可以在 transition_date（或 load_date 列的 MAX() KEEP (..) 值中使用 LAST ORDER BY 子句获取最新日期， 取决于哪个你的意思是在查询中替换) 例如

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平局（相等值），则考虑使用 DENSE_RANK() 函数来计算返回等于 1 以及 ROW_NUMBER() 以便能够在主查询中与 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

Answer 3

假设你想得到最新的记录load_date

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)

Answer 4

我会使用子查询+window函数来实现你的要求（假设你使用的是基于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 window 函数，它根据您在 [query_partition_clause] order_by_clause

中声明规则的方式为每一行分配排名

选择 table 内的最新记录

Selecting the latest record within a table

sql

oracle

oracle11g

greatest-n-per-group