在PostgreSQL中查找日期順序

Question

我試圖在我的數據中找到客戶的最大天數序列。我想了解特定客戶所做的最長時間。如果有人在25/8/16和26/08/16和27/08/16和01/09/16和02/09/16輸入我的應用程序 - 最大順序爲3天（25,26， 27）。

最後（輸出）我想得到兩個字段：custid | MaxDaySequence

我在我的數據表中有以下字段：custid |訂購日期（timestemp）

對於exmple：

custid orderdate 
1 25/08/2007 
1 03/10/2007 
1 13/10/2007 
1 15/01/2008 
1 16/03/2008 
1 09/04/2008 
2 18/09/2006 
2 08/08/2007 
2 28/11/2007 
2 04/03/2008 
3 27/11/2006 
3 15/04/2007 
3 13/05/2007 
3 19/06/2007 
3 22/09/2007 
3 25/09/2007 
3 28/01/2008 

我使用PostgreSQL 2014年

感謝

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嘗試：

select custid, max(num_days) as longest 
from ( 
     select custid,rn, count (*) as num_days 
     from (
      select custid, date(orderdate), 
        cast (row_number() over (partition by custid order by date(orderdate)) as varchar(5)) as rn 
      from table_ 
     ) x group by custid, CURRENT_DATE - INTERVAL rn|| ' day' 
) y group by custid 

Answer 1

嘗試：

SELECT custid, max(abc) as max_sequence_of_days 
FROM (
    SELECT custid, yy, count(*) abc 
    FROM (
    SELECT * , 
      SUM(xx) OVER (partition by custid order by orderdate) yy 
    FROM (
     select * , 
      CASE WHEN 
       orderdate - lag(orderdate) over (partition by custid order by orderdate) 
       <= 1 
     THEN 0 ELSE 1 END xx 
     from mytable 
    ) x 
) z 
    GROUP BY custid, yy 
) q 
GROUP BY custid 

演示：http://sqlfiddle.com/#!15/00422/11

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=====編輯===========

---

得到了「操作符不存在：區間< =整數「

這意味着orderdate列的類型爲timestamp，而不是date。
在這種情況下，你需要使用<= interval '1' day條件，而不是<= 1：

請訪問以下鏈接：https://www.postgresql.org/docs/9.0/static/functions-datetime.html瞭解更多關於日期計算PostgreSQL中

請參閱本演示： http://sqlfiddle.com/#!15/7c2200/2

SELECT custid, max(abc) as max_sequence_of_days 
FROM (
    SELECT custid, yy, count(*) abc 
    FROM (
    SELECT * , 
      SUM(xx) OVER (partition by custid order by orderdate) yy 
    FROM (
     select * , 
      CASE WHEN 
       orderdate - lag(orderdate) over (partition by custid order by orderdate) 
       <= interval '1' day 
     THEN 0 ELSE 1 END xx 
     from mytable 
    ) x 
) z 
    GROUP BY custid, yy 
) q 
GROUP BY custid