简体   繁体   English

SQL查询摘要报告groupby状态并计算差异时间

[英]SQL query summary report groupby status and calculate difference time

Hi all i need show some report, but i don't have solution for this result. 大家好,我需要显示一些报告,但是我没有针对此结果的解决方案。 Please help me guide solution for Query this result. 请帮助我指导查询此结果的解决方案。

This is my table (log) 这是我的桌子(日志)

这是我的桌子(日志)

I need to write SQL query to generate the following results: 我需要编写SQL查询来生成以下结果: 我需要编写SQL查询以生成以下结果

Please guide me, Thank you all answers. 请指导我,谢谢所有答案。

This is an Island and Gaps problem which can be solved by using ROW_NUMBER : 这是一个孤岛和缺口问题,可以使用ROW_NUMBER解决:

SQL Fiddle SQL小提琴

;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

RESULT 结果

| 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 |

声明:本站的技术帖子网页,遵循CC BY-SA 4.0协议,如果您需要转载,请注明本站网址或者原文地址。任何问题请咨询:yoyou2525@163.com.

 
粤ICP备18138465号  © 2020-2024 STACKOOM.COM