得到特殊日期的

Question

表結構狀態：（數據庫oracle 12c）

CUSTOMER_ID | STATUS | STATUS_FROM_DATE 
     101 | ABC | 10-01-2015 
     101 | PQR | 27-02-2015 
     101 | LMN | 04-08-2015 
     101 | ABC | 08-09-2015 

問題：如何獲得客戶的status從上面的表格具體日期？

例如：

CUSTOMER_ID | Input Date | Expected Output 
     101 | 15-01-2015 | ABC 
     101 | 27-02-2015 | PQR 
     101 | 28-02-2015 | PQR 
     101 | 10-09-2015 | ABC 

在上面的例子中，
ABC是客戶對15-01-2015的狀態，因爲這是坐落在10-01-2015並沒有改變，直到27-02-2015。
PQR是客戶在28-02-2015上的狀態，因爲這是在27-02-2015上設置的，並且沒有改變，直到04-08-2015。

Answer 1

您可以使用前導分析函數來獲取間隔結束。然後在兩者之間進行搜索。

select * from 
(

    select 
     customer_id, 
     status, 
     status_from_date, 
     nvl(lead(status_from_date) over (partition by customer_id order by status_from_date)-1, 
      to_date('2099','yyyy') 
     )as end_date 
    from your_table 

) 
where your_date_here between status_start_date and end_date 

Answer 2

使用新引進的行限制子句中Oracle 12c，

select <your input date> as date_, expected_status 
from myt 
where status_date <= <your input date> 
order by status_date desc 
fetch first 1 rows only; 

Answer 3

SQL FIDDLE DEMO

with ranges as (
    select t.*, 
    lead(STATUS_FROM_DATE,1, (select sysdate from dual)) 
     over (partition by CUSTOMER_ID order by STATUS_FROM_DATE) as status_change 
    from Table1 t 
) 
select r.status, s."Date", s."Expected Output" 
from ranges r 
inner join TestStatus s 
    on s."Date" < r.status_change 
    and s."Date" >= r.STATUS_FROM_DATE;