Question

为清晰起见，简单的示例数据：

PERSON_PINCODE

PERSON        START_DATE    END_DATE    PINCODE
10023541700000  01-01-12    31-03-12    6059    
10023541700000  01-01-12    31-03-12    6060  

PINCODE_VALUE

PINCODE START_DATE  END_DATE    VAR     VALUE
6059    01-04-11    30-06-11    3889    28.4
6059    01-07-11    30-09-11    3889    28.2
6059    01-10-11    31-12-11    3890    31.4
6060    01-04-11    30-06-11    3889    29.4
6060    01-07-11    30-09-11    3889    41.2
6060    01-10-11    31-12-11    3890    43.4

Output should be:

PERSON_PINCODE_VALUE

PERSON          START_DATE  END_DATE    PINCODE   VAR   VALUE  DIFF
10023541700000  01-01-12    31-03-12    6059      3889  28.2   90 days
10023541700000  01-01-12    31-03-12    6059      3890  31.4   1 day
10023541700000  01-01-12    31-03-12    6060      3890  41.2   90 days
10023541700000  01-01-12    31-03-12    6060      3890  43.4   1 day

获取PERSON_PINCODE_VALUE，

执行以下操作：

1) Take each row of PERSON_PINCODE and find PINCODE, START_DATE, END_DATE
2) For each PINCODE from step 1, find START_DATE, END_DATE, VAR, VALUE from PINCODE_VALUE
3) Associate all values from step 1 and step 2, on the basis of PINCODE, START_DATE and END_DATE
4) If in step 3, we do NOT get exact PINCODE, START_DATE and END_DATE for each VAR from step 2, find nearest prior START_DATE for remaining VAR
5) Associate values from step 4 with DIFF as  PERSON_PINCODE.START_DATE - PINCODE_VALUE.END_DATE

Answer 1

SELECT   p.person,
         p.start_date,
         p.end_date,
         p.pincode,
         v.var,
         MAX( v.value )
                  KEEP ( DENSE_RANK FIRST ORDER BY ABS( v.start_date - p.start_date ) )
                  AS value,
         DECODE(
           MIN( ABS( v.start_date - p.start_date ) ),
           0, 0,
           MIN( ABS( p.start_date - v.end_date ) )
                  KEEP ( DENSE_RANK FIRST ORDER BY ABS( p.start_date - v.start_date ) )
         ) AS diff
FROM     Person_Pincode p
         INNER JOIN
         Pincode_Value v
         ON ( p.pincode = v.pincode )
GROUP BY p.person,
         p.start_date,
         p.end_date,
         p.pincode,
         v.var;

Answer 2

这是一种“更常规”的方法，可以使用ROW_NUMBER

连接所有行并过滤最佳匹配

with hlpr as (
select pp.PERSON, pp.START_DATE, pp.END_DATE, pp.PINCODE,
pv.START_DATE PV_START_DATE, pv.END_DATE PV_END_DATE,
pv.VALUE, pv.VAR,
pp.START_DATE - pv.END_DATE as diff,
row_number() over(partition by pp.PERSON, pp.START_DATE, pp.END_DATE, pp.PINCODE, pv.VAR 
   order by  pp.START_DATE - pv.END_DATE) as rn 
from PERSON_PINCODE pp
join PINCODE_VALUE pv on pp.PINCODE = pv.PINCODE
where pp.START_DATE >= pv.END_DATE /* prior PV only */
)
select /*+ PARALLEL(5) */ PERSON, START_DATE, END_DATE, PINCODE,  VAR,VALUE, DIFF
from hlpr
where rn = 1
order by 1,2,3,4;

给出

PERSON         START_DATE        END_DATE             PINCODE        VAR      VALUE       DIFF
-------------- ----------------- ----------------- ---------- ---------- ---------- ----------
10023541700000 01.01.12 00:00:00 31.03.12 00:00:00       6059       3889       28,2         93 
10023541700000 01.01.12 00:00:00 31.03.12 00:00:00       6059       3890       31,4          1 
10023541700000 01.01.12 00:00:00 31.03.12 00:00:00       6060       3889       41,2         93 
10023541700000 01.01.12 00:00:00 31.03.12 00:00:00       6060       3890       43,4          1

如果查询具有大量属性（因为排序逻辑集中在一个地方），我更喜欢这个解决方案。

仅对于大型数据集，我会使用@ MT0的GROUP BY方法，因为SORT GROUP BY通常比WINDOW SORT表现得更好。

这里是样本数据

create table PERSON_PINCODE as 
select '10023541700000' PERSON,  to_date('01-01-12','dd-mm-rr') START_DATE, to_date('31-03-12','dd-mm-rr') END_DATE, 6059 PINCODE from dual union all
select '10023541700000' PERSON,  to_date('01-01-12','dd-mm-rr') START_DATE, to_date('31-03-12','dd-mm-rr') END_DATE, 6060 PINCODE from dual
;

create table PINCODE_VALUE as
select 6059 PINCODE,  to_date('01-04-11','dd-mm-rr') START_DATE, to_date('30-06-11','dd-mm-rr') END_DATE, 3889 VAR, 28.4 VALUE from dual union all
select 6059 PINCODE,  to_date('01-07-11','dd-mm-rr') START_DATE, to_date('30-09-11','dd-mm-rr') END_DATE, 3889 VAR, 28.2 VALUE from dual union all
select 6059 PINCODE,  to_date('01-10-11','dd-mm-rr') START_DATE, to_date('31-12-11','dd-mm-rr') END_DATE, 3890 VAR, 31.4 VALUE from dual union all
select 6060 PINCODE,  to_date('01-04-11','dd-mm-rr') START_DATE, to_date('30-06-11','dd-mm-rr') END_DATE, 3889 VAR, 29.4 VALUE from dual union all
select 6060 PINCODE,  to_date('01-07-11','dd-mm-rr') START_DATE, to_date('30-09-11','dd-mm-rr') END_DATE, 3889 VAR, 41.2 VALUE from dual union all
select 6060 PINCODE,  to_date('01-10-11','dd-mm-rr') START_DATE, to_date('31-12-11','dd-mm-rr') END_DATE, 3890 VAR, 43.4 VALUE from dual;

使用带有日期差异

2 个答案: