Question

我有下表：

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，以选择每人person_id 最近 location_id的记录。对于上述样本，所需的结果应为：

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

（DATE表示为＆＃39; YYYY-MM-DD＆＃39;）

谢谢！

Answer 1

其他提案是正确的，但最简单，最快速的解决方案最有可能在您使用FIRST_VALUE and LAST_VALUE Analytic Functions

时

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;

也是如此，但我对这个条款并不熟悉。见aggregate_function KEEP

Answer 2

您可以先按PERSON_ID对结果进行分组并选择MAX(DATE)，然后为每个人提取最近的事件。

然后在这两列上连接表格以检索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

顺便说一句，你不应该使用像DATE这样的保留字来表示列名。

以下是显示它正常工作的小提琴：http://sqlfiddle.com/#!9/efdcb/2

Answer 3

@quasoft是正确的。处理这些GROUP BY问题的另一种方法（当你想要返回的列多于你想要分组的列时。在你的情况下，你需要返回location_id，person_id。但是你只需要按person_id分组），是使用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。主查询在排序后返回每个组的前1个。因此，在这种情况下，它将返回最后一个（我们按date_desc排序）。

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

选择最近的条目

4 个答案: