获取最新变化信息

Question

我有一个有很多信息的审计表。我需要找到每个项目的特定列的最新变化是在什么时候和谁做的。

我能够做到返回适当结果的查询，但我发现它是一个巨大的混乱。特别是我需要为很多不同的列做这个工作（我可能会用union来做），并将这些值与最新发布的值进行比较。总之... 有没有比做3级选择更好的方法？

create table Z_AUDIT
(
  v CHAR(1) not null,      -- A value that is part of a project (ex: project manager)
  t DATE not null,         -- Time of change
  w VARCHAR2(10) not null, -- Who did the change
  p VARCHAR2(10)           -- Project
)

INSERT INTO Z_AUDIT (p, v, t, w) VALUES ('project 1', 'a', sysdate, 'bob');
INSERT INTO Z_AUDIT (p, v, t, w) VALUES ('project 1', 'a', sysdate-1, 'judy');
INSERT INTO Z_AUDIT (p, v, t, w) VALUES ('project 1', 'a', sysdate-2, 'bob');
INSERT INTO Z_AUDIT (p, v, t, w) VALUES ('project 1', 'b', sysdate-3, 'judy');
INSERT INTO Z_AUDIT (p, v, t, w) VALUES ('project 1', 'b', sysdate-4, 'bob');
INSERT INTO Z_AUDIT (p, v, t, w) VALUES ('project 1', 'a', sysdate-5, 'judy');
INSERT INTO Z_AUDIT (p, v, t, w) VALUES ('project 1', 'a', sysdate-6, 'bob');
INSERT INTO Z_AUDIT (p, v, t, w) VALUES ('project 1', 'b', sysdate-7, 'judy');
INSERT INTO Z_AUDIT (p, v, t, w) VALUES ('project 1', 'c', sysdate-8, 'bob');

INSERT INTO Z_AUDIT (p, v, t, w) VALUES ('project 2', 'b', sysdate, 'bob');
INSERT INTO Z_AUDIT (p, v, t, w) VALUES ('project 2', 'a', sysdate-1, 'judy');
INSERT INTO Z_AUDIT (p, v, t, w) VALUES ('project 2', 'a', sysdate-2, 'bob');
INSERT INTO Z_AUDIT (p, v, t, w) VALUES ('project 2', 'b', sysdate-3, 'judy');
INSERT INTO Z_AUDIT (p, v, t, w) VALUES ('project 2', 'c', sysdate-4, 'bob');

 -- Get the latest change from the rank 1
 select p, w, t, current_value, previous_value
   from ( -- Get only the changes with the where clause and get the rank
         select i.p,
                 i.w,
                 i.t,
                 i.current_value,
                 i.previous_value,
                 rank() over(partition by i.p order by i.t desc) r
           from ( -- Get the previous value for each audit record
                  select p,
                          w,
                          t,
                          v as current_value,
                          LAG(v, 1) OVER(partition by p ORDER BY t) as previous_value
                    from Z_AUDIT) i
          where nvl(current_value, 'a') <> nvl(previous_value, 'a'))
  where r = 1;

project 1, bob, 592020 7:08:55 AM, a, b project 2, bob, 5112020 7:12:39 AM, b, a

Answer 1

有几种更简单的写法。注意我都是用CTE写的，因为它们更容易阅读（IMO）。你可以使用 FIRST_VALUE:

WITH cp AS (
  SELECT p, w, t, v AS curr,
         LAG(v, 1, '') OVER (PARTITION BY p ORDER BY t) AS prev
  FROM Z_AUDIT
)
SELECT DISTINCT p,
       FIRST_VALUE(w) OVER (PARTITION BY p ORDER BY t DESC) AS w,
       FIRST_VALUE(t) OVER (PARTITION BY p ORDER BY t DESC) AS t,
       FIRST_VALUE(curr) OVER (PARTITION BY p ORDER BY t DESC) AS curr,
       FIRST_VALUE(prev) OVER (PARTITION BY p ORDER BY t DESC) AS prev
FROM cp
WHERE curr != prev

或甲骨文的具体内容 KEEP ... FIRST:

WITH cp AS (
  SELECT p, w, t, v AS curr,
         LAG(v, 1, '') OVER (PARTITION BY p ORDER BY t) AS prev
  FROM Z_AUDIT
)
SELECT p,
       MIN(w) KEEP (DENSE_RANK FIRST ORDER BY t DESC) AS w,
       MIN(t) KEEP (DENSE_RANK FIRST ORDER BY t DESC) AS t,
       MIN(curr) KEEP (DENSE_RANK FIRST ORDER BY t DESC) AS curr,
       MIN(prev) KEEP (DENSE_RANK FIRST ORDER BY t DESC) AS prev
FROM cp
WHERE curr != prev
GROUP BY p

在这两种情况下，输出是。

P           W       T           CURR    PREV
project 1   bob     09-MAY-20   a       b
project 2   bob     11-MAY-20   b       a

演示(还包括你的原始查询的CTE版本)： 哑谜