SQL查询摘要报告groupby状态并计算差异时间

Question

大家好我需要显示一些报告，但我没有这个结果的解决方案。 请帮我指导查询此结果的解决方案。

这是我的表（日志）

我需要编写SQL查询来生成以下结果：

请指导我， 谢谢你的所有答案。

Answer 1

这是一个Island and Gaps问题，可以使用ROW_NUMBER：

来解决

SQL Fiddle

;WITH Cte AS(
    SELECT *,
        grp = ROW_NUMBER() OVER(PARTITION BY NODE, STATUS ORDER BY DATE_TIME)
                - ROW_NUMBER() OVER(PARTITION BY NODE, STATUS, VALUE ORDER BY DATE_TIME)
    FROM [Log]
),
CteMinMax AS(
    SELECT
        NODE,
        STATUS,
        VALUE,
        MIN_DATE_TIME = MIN(DATE_TIME),
        MAX_DATE_TIME = MAX(DATE_TIME)
    FROM Cte 
    GROUP BY 
        NODE, STATUS, VALUE, grp
)
SELECT
    NODE,
    STATUS,
    VALUE,
    [Begin Failed]  = cmm.MIN_DATE_TIME,
    [Last OK]       = x.MIN_DATE_TIME,
    [Time/Minute]   = DATEDIFF(MINUTE, cmm.MIN_DATE_TIME, x.MIN_DATE_TIME)
FROM CteMinMax cmm
CROSS APPLY(
    SELECT TOP 1 MIN_DATE_TIME
    FROM CteMinMax
    WHERE
        NODE = cmm.NODE
        AND STATUS = cmm.STATUS
        AND VALUE = 0
        AND MIN_DATE_TIME > cmm.MIN_DATE_TIME
) x
WHERE cmm.VALUE = 1

<强> RESULT

| NODE |         STATUS | VALUE |               Begin Failed |                    Last OK | Time/Minute |
|------|----------------|-------|----------------------------|----------------------------|-------------|
|    A | SQL Connection |     1 | December, 02 2015 14:02:00 | December, 02 2015 14:04:00 |           2 |
|    A | SQL Connection |     1 | December, 02 2015 14:05:00 | December, 02 2015 14:08:00 |           3 |
|    A | SQL Connection |     1 | December, 02 2015 14:12:00 | December, 02 2015 14:15:00 |           3 |
|    A | SQL Connection |     1 | December, 02 2015 14:17:00 | December, 02 2015 14:18:00 |           1 |