Question

我有以下表格：

account

+------------+--------------+
| account_id | account_name |
+------------+--------------+
|        452 | a            |
|        785 | b            |
|        985 | c            |
+------------+--------------+

和

task

+---------+------------+------------+
| task_id | task_date  | account_id |
+---------+------------+------------+
|       2 | 01-01-2013 |        452 |
|       1 | 14-02-2013 |        452 |
|       5 | 03-01-2013 |        452 |
|       1 | 02-02-2013 |        785 |
|       7 | 07-01-2013 |        785 |
|       5 | 01-03-2013 |        785 |
|       1 | 25-03-2013 |        985 |
|       4 | 22-03-2013 |        985 |
+---------+------------+------------+

我需要显示表task和account中的记录，以便只显示最旧的任务并且不显示日期。结果将是：

+--------------+---------+
| account_name | task_id |
+--------------+---------+
| a            |       2 |
| b            |       7 |
| c            |       4 |
+--------------+---------+

我想到了使用min功能。但由于我不想显示日期，我该如何使用该功能。

SELECT A.ACCOUNT_NAME, 
       T.TASK_ID, 
       MIN(T.TASK_DATE)  
  FROM ACCOUNT A  
 INNER JOIN TASK T  
    ON A.ACCOUNT_ID = T.ACCOUNT_ID  
 GROUP BY ACCOUNT_NAME, TASK_ID

Answer 1

另一种方法，有点短：

select max(a.accountname) as account_name
     , max(t.taskid) keep(dense_rank first
                               order by t.taskdate) as task_id
 from account1 a
 join task t
   on (a.accountid = t.accountid)
group by a.accountid

结果：

ACCOUNT_NAME    TASK_ID
------------ ----------
a                     2
b                     7
c                     4

SQLFiddle Demo

Answer 2

您只需使用子查询：

SELECT S.ACCOUNT_NAME, S.TASK_ID FROM (
  SELECT A.ACCOUNT_NAME, T.TASK_ID, MIN(T.TASK_DATE)
  FROM ACCOUNT A
  INNER JOIN TASK T
  ON A.ACCOUNT_ID = T.ACCOUNT_ID
  GROUP BY ACCOUNT_NAME, TASK_ID
) S

Answer 3

也许是这样的：

WITH CTE
AS
(
    SELECT 
        A.ACCOUNT_NAME, 
        T.TASK_ID, 
        ROW_NUMBER() OVER(PARTITION BY T.TASK_ID 
                     ORDER BY T.TASK_DATE DESC) AS RowNbr
    FROM ACCOUNT A
    INNER JOIN TASK T
    ON A.ACCOUNT_ID = T.ACCOUNT_ID
)
SELECT
    CTE.ACCOUNT_NAME,
    CTE.TASK_ID
FROM
    CTE
WHERE 
    CTE.RowNbr=1;

Answer 4

您可以使用分析：

SELECT account_name, task_id
  FROM (SELECT a.account_name,
               t.task_id,
               row_number() 
                  over(PARTITION BY a.account_id ORDER BY t.task_date) task_num
          FROM account a
         INNER JOIN task t ON a.account_id = t.account_id
         GROUP BY account_name, task_id)
 WHERE task_num = 1

或自我加入：

SELECT a.account_name, t.task_id
  FROM account a
 INNER JOIN task t ON a.account_id = t.account_id
 WHERE (t.account_id, t.task_date) = (SELECT t_in.account_id, 
                                             MIN(t_in.task_date) 
                                         FROM task t_in 
                                        GROUP BY t_in.account_id)

Answer 5

您可以使用这种简单的方法：

SELECT a.account_name, t.task_id
FROM   account a,
       task t
WHERE  a.account_id= t.account_id 
AND    t.task_date in (SELECT MIN(t.task_date))
GROUP BY account_name, task_id;

Answer 6

它比看起来更容易：在正常连接表之后，只需按max（tak_date）进行过滤，即可通过简单的子选择进行计算。

SELECT acc.account_name, tsk.task_id
FROM account acc
JOIN task    tsk on (acc.account_id = tsk.account_id)
WHERE tsk.task_date = (
  SELECT MAX(aux.task_date)
  FROM task aux
  WHERE aux.account_id = tsk.account_id 
)

显示最旧或最新的记录而不显示日期

6 个答案: