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基本上我想做的是根据 SIP 信令获取通话持续时间。
我有一个包含如下所示记录的表,我正在尝试编写一个返回以下内容的 SELECT 语句:
id callid date micro_ts method duration1 duration2
------------------------------------------------------------------------------------
25 123 2016-04-05 00:00:25 1459814425000320 BYE 00:00:04 3.999876
46 234 2016-04-05 00:01:25 1459814485000000 BYE 00:00:04 4.000000
该 ID 来自给定 CallID 的第一个 BYE。每个单独调用的 CallID 始终相同。 00:00:04 应该是 date 之间的差异和 3.999876 micro_ts 之间的差异。
SQL 应为每个 callid 返回一条记录。这里棘手的部分是我试图找出指示调用开始/停止的消息之间的差异。
我希望它返回基于“第一个方法 = 200 之后方法 = 183 直到第一次 BYE”或“第一个方法 = 200 之后方法 = 180 并且在第一次 BYE 之前找不到方法 = 183”
如何组合这些条件来返回包含多个调用的表所需的内容?
我一直在尝试这样编写 SQL:
select * from (
select id, callid, micro_ts, method, min(date) as StartTime, max(date) as EndTime,
timediff(max(date), min(date)) as duration
from (select t.*,
(select count(*)
from tableA t2
where t2.date <= t.date and
t2.method = 'BYE'
) as grp
from tableA t
) t where t.method = '200'
group by callid, method
order by 3 desc) z;
我相信这是在假设所有调用只有 200 一次然后是 BYE 的情况下有效的。
id callid date micro_ts method
---------------------------------------------------------------
1 123 2016-04-05 00:00:01 1459814401000025 INVITE
2 123 2016-04-05 00:00:02 1459814402000123 407
3 123 2016-04-05 00:00:03 1459814403000941 INVITE
4 123 2016-04-05 00:00:04 1459814404000392 INVITE
5 123 2016-04-05 00:00:05 1459814405000539 INVITE
6 123 2016-04-05 00:00:06 1459814406000101 404
7 123 2016-04-05 00:00:07 1459814407000007 INVITE
8 123 2016-04-05 00:00:08 1459814408000948 404
9 123 2016-04-05 00:00:09 1459814409000784 100
10 123 2016-04-05 00:00:10 1459814410000192 183
11 123 2016-04-05 00:00:11 1459814411000482 183
12 123 2016-04-05 00:00:12 1459814412000561 183
13 123 2016-04-05 00:00:13 1459814413000392 183
14 123 2016-04-05 00:00:14 1459814414000751 180
15 123 2016-04-05 00:00:15 1459814415000012 180
16 123 2016-04-05 00:00:16 1459814416000384 180
17 123 2016-04-05 00:00:17 1459814417000498 180
18 123 2016-04-05 00:00:18 1459814418000533 183
19 123 2016-04-05 00:00:19 1459814419000841 183
20 123 2016-04-05 00:00:20 1459814420000492 183
21 123 2016-04-05 00:00:21 1459814421000444 200 * FIRST 200 after 183
22 123 2016-04-05 00:00:22 1459814422000901 200 |
23 123 2016-04-05 00:00:23 1459814423000294 ACK |
24 123 2016-04-05 00:00:24 1459814424000732 ACK |
25 123 2016-04-05 00:00:25 1459814425000320 BYE * FIRST BYE
26 123 2016-04-05 00:00:26 1459814426000020 BYE
27 123 2016-04-05 00:00:27 1459814427000391 200
28 123 2016-04-05 00:00:28 1459814428000123 200
29 234 2016-04-05 00:01:01 1459814461000000 INVITE
30 234 2016-04-05 00:01:02 1459814462000000 407
31 234 2016-04-05 00:01:03 1459814463000000 INVITE
32 234 2016-04-05 00:01:04 1459814464000000 INVITE
33 234 2016-04-05 00:01:05 1459814465000000 INVITE
34 234 2016-04-05 00:01:06 1459814466000000 404
35 234 2016-04-05 00:01:07 1459814467000000 INVITE
36 234 2016-04-05 00:01:08 1459814468000000 404
37 234 2016-04-05 00:01:09 1459814469000000 100
38 234 2016-04-05 00:01:10 1459814470000000 183
39 234 2016-04-05 00:01:11 1459814471000000 183
40 234 2016-04-05 00:01:12 1459814472000000 183
41 234 2016-04-05 00:01:13 1459814473000000 183
42 234 2016-04-05 00:01:14 1459814474000000 180
43 234 2016-04-05 00:01:15 1459814475000000 180
44 234 2016-04-05 00:01:16 1459814476000000 180
45 234 2016-04-05 00:01:17 1459814477000000 180
46 234 2016-04-05 00:01:21 1459814481000000 200 * FIRST 200 after 180 whern no 183 is present
47 234 2016-04-05 00:01:22 1459814482000000 200 |
48 234 2016-04-05 00:01:23 1459814483000000 ACK |
49 234 2016-04-05 00:01:24 1459814484000000 ACK |
50 234 2016-04-05 00:01:25 1459814485000000 BYE * FIRST BYE
51 234 2016-04-05 00:01:26 1459814486000000 BYE
52 234 2016-04-05 00:01:27 1459814487000000 200
53 234 2016-04-05 00:01:28 1459814488000000 200
最佳答案
我已经尝试过,CASE 表达式可能存在一些转换问题,但这应该可以帮助您完成大部分工作:
SELECT id,
callid
start_date,
start_ms,
end_date,
end_ms,
end_date - start_date AS duration1,
end_ms - start_ms AS duration2
FROM (
SELECT
@same_call := @callid = callid,
@has_ended := @same_call AND (@has_ended OR @has_bye) AS has_ended,
@has_18x := (@has_18x AND @same_call) OR method IN ('180','183'),
@has_200 := (@has_200 AND @same_call) OR (@has_18x AND method = '200'),
@has_bye := (@has_bye AND @same_call) OR (@has_200 AND method = 'BYE') AS has_bye,
@start_date := CASE
WHEN @has_200 THEN COALESCE(@start_date, `date`)
END AS start_date,
@start_ms := CASE
WHEN @has_200 THEN COALESCE(@start_ms, micro_ts)
END AS start_ms,
@end_date := CASE
WHEN @has_bye THEN COALESCE(@end_date, `date`)
END AS end_date,
@end_ms := CASE
WHEN @has_bye THEN COALESCE(@end_ms, micro_ts)
END AS end_ms,
id,
@callid := callid AS callid
FROM sip_signal
JOIN (
SELECT @callid := NULL,
@has_ended := FALSE,
@has_18x := FALSE,
@has_200 := FALSE,
@has_bye := FALSE,
@start_date := NULL,
@start_ms := NULL,
@end_date := NULL,
@end_ms := NULL
) init
ORDER BY callid, micro_ts
) sip_call
WHERE has_bye
AND NOT has_ended
如果您想查看内部工作原理,只需运行内部 SELECT(+ JOIN 和 ORDER BY)即可。
关于表中所选记录的一列上的 MySQL 时间差,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/36426485/
我正在尝试用 Swift 编写这段 JavaScript 代码:k_combinations 到目前为止,我在 Swift 中有这个: import Foundation import Cocoa e
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