[英]Join two tables based on date from first
我有两个如下表(日期格式:yyyy-MM-dd):
1) 表 1 - 手术
P_ID | SURGERY_DATE
------------------------------------------------
1 | 2012-04-01
2 | 2012-08-14
1 | 2012-07-22
4 | 2012-10-30
3 | 2012-06-07
2) 表 2 - 访问
P_ID | VISIT_DATE
-----------------------------------------
1 | 2012-03-28
1 | 2012-04-14
1 | 2012-05-17
1 | 2012-09-12
3 | 2012-07-15
4 | 2012-10-10
3 | 2012-06-01
表 SURGERY 和 VISIT 是从其他表连接起来的。 我想查找满足以下条件的所有记录:VISIT_DATE >= SUGERY_DATE
3) 结果表
EMPLOYEE_ID | SUGERY_DATE | NUMBER OF VISIT
-------------------------------------------------------
1 | 2012-04-01 | 4
2 | 2012-08-14 | 0
1 | 2012-07-22 | 2
4 | 2012-10-30 | 1
3 | 2012-06-07 | 1
您可以使用相关子查询:
select s.*,
(select count(*)
from visit v
where v.p_id = s.p_id and v.visit_date > s.surgery_date
) as num_visits_after
from surgery s;
您需要使用 group by 并按如下所示的条件进行计数:
SELECT
S.P_ID,
S.SURGERY_DATE,
SUM(CASE
WHEN V.VISIT_DATE > S.SURGERY_DATE THEN 1
END) AS NUM_VISITS_AFTER
FROM
SURGERY S
LEFT JOIN VISIT V ON ( S.P_ID = V.P_ID )
GROUP BY
S.P_ID,
S.SURGERY_DATE;
干杯!!
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