select 最近的条目

Question

我有以下 table:

LOCATION_ID, PERSON_ID, DATE
3, 65, 2016-06-03
7, 23, 2016-10-28
3, 23, 2016-08-05
5, 65, 2016-07-14

我想在 PL/SQL 中构建一个 select 查询 select 具有最近 location_id 的记录person_id。对于上面的示例，期望的结果应该是：

LOCATION_ID, PERSON_ID, DATE
5, 65, 2016-07-14
7, 23, 2016-10-28

（日期表示为'YYYY-MM-DD'）

谢谢！

Answer 1

您可以先通过 PERSON_ID 对结果进行分组并选择 MAX(DATE).

来提取每个人的最近事件

然后在这两列上将 table 与其自身连接起来以检索 LOCATION_ID

SELECT
  YOUR_TABLE.LOCATION_ID,
  YOUR_TABLE.PERSON_ID,
  YOUR_TABLE.DATE
FROM
  (SELECT
    PERSON_ID, MAX(DATE) AS max_date
  FROM
    YOUR_TABLE
  GROUP BY
    PERSON_ID
  ) AS t1
LEFT JOIN
  YOUR_TABLE
ON
  YOUR_TABLE.PERSON_ID = t1.PERSON_ID
  AND
  YOUR_TABLE.DATE = t1.max_date

顺便说一下，你不应该使用像 DATE 这样的保留字作为列名。

这里 fiddle 展示它的工作原理：http://sqlfiddle.com/#!9/efdcb/2

Answer 2

@quasoft 是正确的。处理这类 GROUP BY 问题的另一种方法（当你想要 return 比你想要分组的列更多时。在你的情况下，你需要 return location_id, person_id。但是你只需要按person_id)分组，就是用analytical functions。

--schema:
CREATE TABLE my_table 
  ( 
     location_id NUMBER, 
     person_id   NUMBER, 
     date_       DATE 
  ); 

INSERT ALL 
INTO my_table 
VALUES (3, 65, To_date('2016-06-03', 'YYYY-MM-DD')) 
INTO my_table 
VALUES (7, 23, To_date('2016-10-28', 'YYYY-MM-DD')) 
INTO my_table 
VALUES (3, 23, To_date('2016-08-05', 'YYYY-MM-DD')) 
INTO my_table 
VALUES (5, 65, To_date('2016-07-14', 'YYYY-MM-DD')) 
SELECT * 
FROM   dual; 

--query:
WITH ordered 
     AS (SELECT location_id, 
                person_id, 
                date_, 
                Row_number() 
                  over ( 
                    PARTITION BY person_id 
                    ORDER BY date_ DESC) RN 
         FROM   my_table) 
SELECT location_id, 
       person_id, 
       date_ 
FROM   ordered 
WHERE  rn = 1;

查询 ordered 按日期对每个组的行进行排序。主查询，returns是每组排序后的第1个。因此，它将 return 在这种情况下是最后一个（我们按 date_desc 排序）。

Answer 3

其他建议是正确的，但是当您使用 FIRST_VALUE and LAST_VALUE Analytic Functions

时最有可能是最紧凑和最快的解决方案

SELECT DISTINCT
   FIRST_VALUE(LOCATION_ID) OVER (PARTITION BY PERSON_ID ORDER BY THE_DATE 
             ROWS BETWEEN UNBOUNDED PRECEDING AND UNBOUNDED FOLLOWING) AS LOCATION_ID, 
   PERSON_ID, 
   MAX(THE_DATE) OVER (PARTITION BY PERSON_ID) AS LAST_DATE
FROM YOUR_TABLE;

其他人更喜欢

SELECT 
   MAX(LOCATION_ID) KEEP (DENSE_RANK FIRST ORDER BY DATE) as LOCATION, 
   PERSON_ID, 
   MAX(DATE) as LAST_DATE
FROM YOUR_TABLE
GROUP BY PERSON_ID;

which也一样，不过我对这个子句不是很熟悉。参见 aggregate_function KEEP

Answer 4

这可能有用！

SELECT * FROM Your_Table A
JOIN (SELECT PERSON_ID,MAX(DATE) as MaxDate FROM Your_Table 
GROUP BY PERSON_ID) B
ON A.PERSON_ID = B.PERSON_ID AND A.DATE = B.MaxDate

select 最近的条目

select the most recent entry

sql

oracle

greatest-n-per-group