一、sql利用错位相减的方式得到相同利率下的连续时间的时间区间

说明：环境为oracle

一、需求

二、思路

思路一：

• 需要判断连续发布的两次利率是否一致，如果一致，就只保留最先发布的一次记录
• 需要附带排序后的时间的编号，以便实现错位相减的效果
• 让去重后的数据进行关联，有得取舍

思路二：

• 如果为了能在数据获取后，对其进行序号打标，在mysql等支持procedure的函数中，可以通过如下方式

  begin
      select a.column,(@seq:=@seq+1) seq
      from table_a,(select @seq:=0 from dual) b
  end
  

三、实现

3.1 数据准备

-- 创建表
create table TABLE1
(
  publish_date DATE,
  rate         VARCHAR2(20)
)

-- 插入数据
insert into TABLE1 (PUBLISH_DATE, RATE)
values (to_date('01-01-2010', 'dd-mm-yyyy'), '5.1%');

insert into TABLE1 (PUBLISH_DATE, RATE)
values (to_date('01-10-2010', 'dd-mm-yyyy'), '5.1%');

insert into TABLE1 (PUBLISH_DATE, RATE)
values (to_date('01-01-2011', 'dd-mm-yyyy'), '6.0%');

insert into TABLE1 (PUBLISH_DATE, RATE)
values (to_date('31-10-2012', 'dd-mm-yyyy'), '6.0%');

insert into TABLE1 (PUBLISH_DATE, RATE)
values (to_date('10-11-2012', 'dd-mm-yyyy'), '6.0%');

insert into TABLE1 (PUBLISH_DATE, RATE)
values (to_date('31-12-2012', 'dd-mm-yyyy'), '6.0%');

insert into TABLE1 (PUBLISH_DATE, RATE)
values (to_date('31-03-2013', 'dd-mm-yyyy'), '5.9%');

insert into TABLE1 (PUBLISH_DATE, RATE)
values (to_date('01-09-2013', 'dd-mm-yyyy'), '5.5%');

insert into TABLE1 (PUBLISH_DATE, RATE)
values (to_date('01-05-2014', 'dd-mm-yyyy'), '5.5%');

insert into TABLE1 (PUBLISH_DATE, RATE)
values (to_date('01-01-2015', 'dd-mm-yyyy'), '5.1%');

insert into TABLE1 (PUBLISH_DATE, RATE)
values (to_date('01-06-2016', 'dd-mm-yyyy'), '5.1%');

insert into TABLE1 (PUBLISH_DATE, RATE)
values (to_date('01-09-2017', 'dd-mm-yyyy'), '5.1%');

3.2 求取步骤

1. 获取排序后的时间

   select t1.publish_date,t1.rate,row_number() over(order by t1.publish_date) seq
   from table1 t1
   

   

2. 数据错位实现

   select t3.publish_date,t3.rate,row_number() over(order by t3.publish_date) + 1 seq
   from table1 t3
   

   

3. 关联步骤1和步骤2获取到的数据，得到关联数据

     select t2.*,t4.*
     from (
       select t1.publish_date,t1.rate,row_number() over(order by t1.publish_date) seq
       from table1 t1
     ) t2
     left join (
       select t3.publish_date,t3.rate,row_number() over(order by t3.publish_date) + 1 seq
       from table1 t3
     ) t4
     on t4.seq = t2.seq
     order by t2.publish_date asc
   

   

4. 对步骤3获取到的数据进行处理，排除连续时间下，利率相同的数据

     select t2.publish_date,t2.rate
     from (
       select t1.publish_date,t1.rate,row_number() over(order by t1.publish_date) seq
       from table1 t1
     ) t2
     left join (
       select t3.publish_date,t3.rate,row_number() over(order by t3.publish_date) + 1 seq
       from table1 t3
     ) t4
     on t4.seq = t2.seq
     where t2.rate <> NVL(t4.rate,'-1') -- 对于首次的时间，通过这样让其不被排除
     order  by t2.publish_date asc
   

   

5. 通过步骤4获取到，每次利率变更后的初始时间点，然后通过lead函数进行处理，获取变更节点的前一天作为结束时间

   select a.publish_date start_date,
          nvl(to_char(lead(a.publish_date,1) over(order by a.publish_date)-1,'yyyy-mm-dd'),'9999-12-31') end_date,
          a.rate
   from (
     select t2.publish_date,t2.rate
     from (
       select t1.publish_date,t1.rate,row_number() over(order by t1.publish_date) seq
       from table1 t1
     ) t2
     left join (
       select t3.publish_date,t3.rate,row_number() over(order by t3.publish_date) + 1 seq
       from table1 t3
     ) t4
     on t4.seq = t2.seq
     where t2.rate <> NVL(t4.rate,'-1')
     order  by t2.publish_date asc
   ) a