Question

我需要为service.service_id表格中距离最近service.name和service.date_begin的每一行找到bonus，service.date_begin和service.date_begin <= bonus.date_begin。如果有service个date_begin service，则返回任何service.service_id（例如，最多service.rowid或service）。如果没有此类NULL，请返回bonus。

实施例

bonus_id表（bonus_id date_begin -------------------- 1 2010-04-12 2 2010-04-20是PK）：

service

service_id表（bonus_id service_id name date_begin -------------------------------------- 1 1 'a' 2010-04-10 1 2 'b' 2010-04-11 1 3 'c' 2010-04-11 1 4 'd' 2010-04-15 2 5 'e' 2010-04-22是PK）：

bonus_id  bonus_date_begin  service_id  service_name  service_date_begin
------------------------------------------------------------------------
1         2010-04-12        3           'c'           2010-04-11
2         2010-04-20        NULL        NULL          NULL

期望的输出：

create table bonus (
  bonus_id number primary key,
  date_begin date
);
create table service (
  bonus_id number references bonus(bonus_id),
  service_id number primary key,
  name varchar2(1),
  date_begin date
);
insert into bonus values (1, date '2010-04-12');
insert into bonus values (2, date '2010-04-20');
insert into service values (1, 1, 'a', date '2010-04-10');
insert into service values (1, 2, 'b', date '2010-04-11');
insert into service values (1, 3, 'c', date '2010-04-11');
insert into service values (1, 4, 'd', date '2010-04-15');
insert into service values (2, 5, 'e', date '2010-04-22');
commit;

数据库：Oracle 11.2

人口脚本：

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    mixins: [mixins.reactiveProp],
    props: ["options"],
    mounted () {
        this.renderChart(this.chartData, this.options)
    }
})

Answer 1

SELECT b.bonus_id,
       MAX( b.date_begin ) AS bonus_date_begin,
       MAX( s.service_id ) KEEP ( DENSE_RANK LAST ORDER BY s.date_begin, s.service_id )
         AS service_id,
       MAX( s.name       ) KEEP ( DENSE_RANK LAST ORDER BY s.date_begin, s.service_id )
         AS service_name,
       MAX( s.date_begin ) KEEP ( DENSE_RANK LAST ORDER BY s.date_begin, s.service_id )
         AS service_date_begin
FROM   bonus b
       LEFT OUTER JOIN
       service s
       ON ( b.bonus_id = s.bonus_id AND s.date_begin < b.date_begin )
GROUP BY b.bonus_id;

<强>输出：

BONUS_ID BONUS_DATE_BEGIN SERVICE_ID SERVICE_NAME SERVICE_DATE_BEGIN
-------- ---------------- ---------- ------------ ------------------
1        2010-04-12       3          c            2010-04-11
2        2010-04-20       NULL       NULL         NULL

Answer 2

尝试此代码，这将提供您想要的输出。

select bonus.bonus_id ,bonus.date_begin as bonus_date_begin,max(service.service_id) as service_id,
max(service.date_begin) as service_date_begin 
from bonus left join service
on bonus.bonus_id=service.bonus_id
and service.date_begin <= bonus.date_begin
group by bonus.bonus_id ,bonus.date_begin

Answer 3

select
  t.bonus_id,
  t.bonus_date_begin,
  t.service_id,
  t.service_name,
  t.service_date_begin
from (
  select
    b.bonus_id,
    b.date_begin as bonus_date_begin,
    s.service_id,
    s.name as service_name,
    s.date_begin as service_date_begin,
    row_number() over (
      partition by b.bonus_id
      order by s.date_begin desc, s.service_id desc
    ) as rn
  from avd_bonus b
  left join avd_service s on b.bonus_id = s.bonus_id and s.date_begin <= b.date_begin
) t
where t.rn = 1;

查找相关表格中距离最近日期

3 个答案: