我需要用表中的数据填充我的事实lds_placement
表。我已经选择了记录,如下所示:
fk1_account_id | fk3_job_role_id | salary | no_of_placements | YEAR
---------------------------------------------------------------------
10 | 3 | 165000 | 5 | 2010
10 | 3 | 132000 | 4 | 2011
10 | 3 | 132000 | 4 | 2012
20 | 2 | 990000 | 3 | 2010
20 | 2 | 132000 | 2 | 2011
20 | 2 | 132000 | 2 | 2012
我想从名为列time_id
的不同表中插入,而不是实际年份本身。time_dim
year
该time_dim
表如下所示:
time_id | year
---------------
5 | 2015
1 | 2013
2 | 2010
3 | 2014
4 | 2012
6 | 2011
我需要插入“年”列实际上是:
year
2
6
4
2
6
4
请给我插入time_id
而不是year
在表格中的方式。这是我用来选择最顶层表格的代码。
SELECT
fk1_account_id,
fk3_job_role_id,
Sum(actual_salary) AS salary,
Count(1) AS no_of_placements,
MAX(EXTRACT(YEAR FROM plt_estimated_end_date)) AS year
FROM lds_placement
GROUP BY fk1_account_id, fk3_job_role_id, EXTRACT(YEAR FROM plt_estimated_end_date)
ORDER BY fk1_account_id;