SQL Oracle计数群集 [英] SQL Oracle Counting Clusters

问题描述

我有一个基于时间戳的数据集.

I have a data set which is based on a timestamp.

     Date                 Value
07-Jul-15 12:05:00          1  
07-Jul-15 12:10:00          1 
07-Jul-15 12:15:00          1 
07-Jul-15 12:20:00          0 
07-Jul-15 12:25:00          0 
07-Jul-15 12:30:00          0 
07-Jul-15 12:35:00          1 
07-Jul-15 12:40:00          1 
07-Jul-15 12:45:00          1 
07-Jul-15 12:50:00          1 
07-Jul-15 12:55:00          0 
07-Jul-15 13:00:00          0 
07-Jul-15 13:05:00          1 
07-Jul-15 13:10:00          1 
07-Jul-15 13:15:00          1 
07-Jul-15 13:20:00          0 
07-Jul-15 13:25:00          0 

我想查询并返回

1. 关闭次数:在这种情况下，关闭次数为3，基于0为ON和1为OFF.

2. 每次关闭之间的时间段

1. Number of shutdowns: The Number of shut down in this case is 3 based on 0 is ON and 1 is OFF.

2. Period Between every shut down

示例:

1. 发件人:15年7月7日12:05:00到:15年7月7日12:15:00时长:15分钟
2. 发件人:15年7月7日12:35:00到:15年7月7日12:50:00时长:20分钟

我正在使用Oracle

I am using Oracle

推荐答案

在ORACLE中使用LEAD和LAG函数，您可以构建以下查询:

Using LEAD and LAG functions in ORACLE you can built these queries:

1.关闭次数:

WITH IntTable AS
( SELECT * FROM
  (
   SELECT dt b_date,value,LEAD(dt) OVER (ORDER BY dt) e_date  FROM
     (
        select "Date" dt,"Value" value,
               LAG("Value") OVER (ORDER BY "Date") pvalue,
               LEAD("Value") OVER (ORDER BY "Date") nvalue
        from T
     ) T1
     WHERE pvalue is NULL or value<>pvalue or nvalue is NULL 
   )
WHERE E_DATE is NOT NULL
)
SELECT COUNT(*) FROM IntTable where value = 0 

SQLFiddle演示

2.每次关闭之间的时间段

2.Period Between every shut down

WITH IntTable AS
( SELECT * FROM
  (
   SELECT dt b_date,value,LEAD(dt) OVER (ORDER BY dt) e_date  FROM
     (
        select "Date" dt,"Value" value,
               LAG("Value") OVER (ORDER BY "Date") pvalue,
               LEAD("Value") OVER (ORDER BY "Date") nvalue
        from T
     ) T1
     WHERE pvalue is NULL or value<>pvalue or nvalue is NULL 
   )
WHERE E_DATE is NOT NULL
)
SELECT b_date,e_date, (e_date-b_date) * 60 * 24 FROM IntTable where value = 1 

SQLFiddle演示

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