TSQL - 带条件的时间总和列

Question

需要创建一个查询，该查询会获取位设置为ON/OFF的时间摘要。

例子：

╔═══════════════════════════╗ 
║   TABLE   ║ 
╠════╦════════╦═════╦═══════╣ 
║ ID ║ TIME ║ BIT ║ VALUE ║ 
╠════╬════════╬═════╬═══════╣ 
║ 1 ║ 13:40 ║ 1 ║ 5 ║ 
║ 2 ║ 13:45 ║ 1 ║ 3 ║ 
║ 3 ║ 13:50 ║ 1 ║ 1 ║ 
║ 4 ║ 13:55 ║ 0 ║ 2 ║ 
║ 5 ║ 14:00 ║ 0 ║ 7 ║ 
║ 6 ║ 14:05 ║ 1 ║ 3 ║ 
║ 7 ║ 14:10 ║ 1 ║ 4 ║ 
║ 8 ║ 14:15 ║ 0 ║ 2 ║ 
║ 9 ║ 14:20 ║ 1 ║ 2 ║ 
╚════╩════════╩═════╩═══════╝ 

我想有TIME（和VALUE - 简单的一个）总总结时BIT是SET ON：

13:40 - 13:50 = 10 mins 
14:05 - 14:10 = 5 mins 
14:20   = no end time, 0 mins 
----------------------------------------- 
       15 mins 

有没有发现：

• How to sum up time field in SQL Server =很好的问题，但有静态的时间开始（0:00:00），这将不会在这种情况下
• Aggregate function over a given time interval工作=有聚集而不是空调，并适用于所有数据

我想，这可能是作为递归函数完成（传递最后处理的日期时间），它将通过处理的最后日期，并且计算自BIT为ON以来的日期时间。

SQL查询求和值（容易的）：

SELECT SUM(Value) 
FROM Table 
WHERE Bit = 1 

我应该如何获得分钟（时间）的总价值，在此期间是BIT设置ON？

---

编辑：查询，可用于测试：

DECLARE @Table TABLE(
    ID INT Identity(1,1) PRIMARY KEY, 
    [TIME] DATETIME NOT NULL, 
    [BIT] BIT NOT NULL, 
    [VALUE] INT NOT NULL 
); 
INSERT INTO @Table([TIME],[BIT],[VALUE]) VALUES('13:40',1,5); 
INSERT INTO @Table([TIME],[BIT],[VALUE]) VALUES('13:45',1,3); 
INSERT INTO @Table([TIME],[BIT],[VALUE]) VALUES('13:50',1,1); 
INSERT INTO @Table([TIME],[BIT],[VALUE]) VALUES('13:55',0,2); 
INSERT INTO @Table([TIME],[BIT],[VALUE]) VALUES('14:00',0,7); 
INSERT INTO @Table([TIME],[BIT],[VALUE]) VALUES('14:05',1,3); 
INSERT INTO @Table([TIME],[BIT],[VALUE]) VALUES('14:10',1,4); 
INSERT INTO @Table([TIME],[BIT],[VALUE]) VALUES('14:15',0,2); 
INSERT INTO @Table([TIME],[BIT],[VALUE]) VALUES('14:20',1,2); 
SELECT * FROM @Table; 

Answer 1

使用LEAD函数获取下一行的时间并计算时间间隔。然后只是组结果由[位]

WITH t AS(
SELECT 
    [time], 
    DATEDIFF(minute, [time], LEAD([time], 1, null) OVER (ORDER BY [time])) AS interval, 
    [bit], 
    [value] 
FROM table1) 
SELECT [bit], CAST(DATEADD(MINUTE, SUM(interval), '00:00') AS TIME), SUM([value]) FROM t 
GROUP BY [bit] 

Answer 2

你有两个问题：总结的时间，并确定相邻值。你可以用行号的方法处理第二个。您可以通过转换为分钟处理前：

select bit, min(time), max(time), 
     sum(datediff(minute, 0, time)) as minutes, 
     sum(value) 
from (select t.*, 
      row_number() over (order by id) as seqnum, 
      row_number() over (partition by bit order by id) as seqnum_b 
     from t 
    ) t 
group by (seqnum - seqnum_b), bit; 

Answer 3

这是一个“空白和孤岛”的问题，有一个非常标准的解决方案。我想出了这个，这与Gordon的几乎相同，但是有一个额外的步骤来计算间隔。这是我发布基本上是重复答案的唯一原因，我不确定在零分钟之内取得差异实际上有效吗？

DECLARE @table TABLE (id int, [time] TIME, [bit] BIT, value INT); 
INSERT INTO @table SELECT 1, '13:40', 1, 5; 
INSERT INTO @table SELECT 2, '13:45', 1, 3; 
INSERT INTO @table SELECT 3, '13:50', 1, 1; 
INSERT INTO @table SELECT 4, '13:55', 0, 2; 
INSERT INTO @table SELECT 5, '14:00', 0, 7; 
INSERT INTO @table SELECT 6, '14:05', 1, 3; 
INSERT INTO @table SELECT 7, '14:10', 1, 4; 
INSERT INTO @table SELECT 8, '14:15', 0, 2; 
INSERT INTO @table SELECT 9, '14:20', 1, 2; 
WITH x AS (
    SELECT *, ROW_NUMBER() OVER (PARTITION BY [bit] ORDER BY id) AS a_id, ROW_NUMBER() OVER (ORDER BY id) AS b_id FROM @table), 
y AS (
    SELECT [bit], MIN([time]) AS min_time, MAX([time]) AS max_time, SUM(value) AS value FROM x GROUP BY a_id - b_id, [bit]) 
SELECT [bit], SUM(value) AS total_value, SUM(DATEDIFF(MINUTE, min_time, max_time)) AS total_minutes FROM y GROUP BY [bit]; 

结果：

bit total_value total_minutes 
0 11   5 
1 18   15 

作为奖金这里是一个解决方案，既解决了实际问题，即有多少经过的时间是在那里当该位被设置为1：

WITH x AS (SELECT id, id - DENSE_RANK() OVER(ORDER BY id) AS grp FROM @table WHERE [bit] = 1), y AS (SELECT MIN(id) AS range_start, MAX(id) AS range_end FROM x GROUP BY grp) 
SELECT SUM(DATEDIFF(MINUTE, t1.[time], t2.[time])) AS minutes_elapsed FROM y INNER JOIN @table t1 ON t1.id = y.range_start INNER JOIN @table t2 ON t2.id = y.range_end;