mysql选择具有复杂计数+小时范围标准的列答案

【问题标题】：mysql select columns having complex count + hour-range criteriamysql选择具有复杂计数+小时范围标准的列
【发布时间】：2015-08-04 20:33:57
【问题描述】：

问题描述： 我有以下查询来检索过去 15 分钟内的最新警报。

SELECT
   AlmCode,OccurTime,ClearTime....columnN 
FROM 
   TB_ALM 
WHERE 
   AlmCode IN ('3236',....'5978') AND 
   OccurTime >= date_sub(NOW(),interval 15 minute);

表结构：

CREATE TABLE `TB_ALM` (
  `Col1` smallint(2) DEFAULT NULL,
  `Col2` int(4) DEFAULT NULL,
  `Col3` int(2) DEFAULT NULL,
  `Col4` int(10) DEFAULT NULL,
  `Col5` int(10) unsigned DEFAULT NULL,
  `Col6` int(2) DEFAULT NULL,
  `Col7` int(2) DEFAULT NULL,
  `Col8` int(10) DEFAULT NULL,
  `Col9` int(10) unsigned DEFAULT NULL,
  `AlmCode` int(10) unsigned DEFAULT NULL,
  `Col10` int(2) NOT NULL,
  `Col11` int(10) unsigned DEFAULT NULL,
  `Col12` char(12) DEFAULT NULL,
  `Col13` int(2) unsigned DEFAULT NULL,
  `Col14` int(10) unsigned DEFAULT NULL,
  `Col15` int(10) unsigned DEFAULT NULL,
  `Col16` int(10) unsigned DEFAULT NULL,
  `OccurTime` datetime NOT NULL,
  `ClearTime` datetime DEFAULT NULL,
  `AlmDesc` varchar(500) DEFAULT NULL,
  `Col20` int(1) DEFAULT '0',
  `Col21` bigint(20) DEFAULT NULL,
  `Col22` char(120) DEFAULT NULL,
  `Col23` int(10) DEFAULT NULL,
  KEY `TB_ALM_IDX2` (`Col1`,`Col2`,`Col3`,`Col6`,`Col7`,`Col11`,`AlmCode`,`Col9`,`Col4`,`Col8`,`ClearTime`) USING BTREE,
  KEY `TB_ALM_IDX1` (`Col1`,`Col2`,`Col3`,`Col6`,`Col7`,`Col11`,`AlmCode`,`Col5`,`Col21`),
  KEY `TB_ALM_IDX3` (`Col1`,`Col2`,`Col3`,`Col5`) USING BTREE,
  KEY `TB_ALM_IDX4` (`Col1`,`Col2`,`Col3`,`OccurTime`,`ClearTime`,`Col21`) USING BTREE,
  KEY `TB_ALM_IDX5` (`Col23`),
  KEY `TB_ALM_IDX6` (`Col1`,`Col2`,`Col3`,`Col6`,`Col7`,`AlmCode`,`Col11`,`ClearTime`)
) ENGINE=InnoDB DEFAULT CHARSET=utf8

需要什么： 现在我想修改它以检索具有以下条件的警报：

一个。过去 15 分钟内发生了警报（AlmCodes）（原始请求）AND

b.仅当在过去 6 小时内的任何 15 分钟窗口中，每个警报 (AlmCodes) 均未发生超过三次

尝试了什么： 我尝试了以下方法：

在过去 15 分钟内获取 DISTINCT(AlmCodes)。

select distinct(AlmCode) from TB_ALM where AlmCode IN ('3236','4002','4008','4036','4050','4051','4102','4108','4136', '4150','4151','4202','4208','4236','4250','4251','4801','4802','4836','4848','4850','4851 ','4902','4936','4950','4951','5002','5008','5036','5050','5051','5102','5108','5136', '5150','5151','5202','5208','5236','5250','5251','5947','5950','5952','5975','5976','5977 ','5978') AND OccurTime >= date_sub(NOW(),interval 15 分钟) ;
使用 Item-1(above) 作为子查询并获取每个 AlmCode 的出现次数。

select Almcode,concat(date(OccurTime),' ',HOUR(OccurTime)) as HR,count(*) from TB_ALM_HISTORY where AlmCode IN ( select distinct(s.AlmCode) from TB_ALM_HISTORY s where s.AlmCode IN ('3236','4002','4008','4036','4050','4051','4102','4108','4136 ','4150','4151','4202','4208','4236','4250','4251','4801','4802','4836','4848','4850', '4851','4902','4936','4950','4951','5002','5008','5036','5050','5051','5102','5108','5136 ','5150','5151','5202','5208','5236','5250','5251','5947','5950','5952','5975','5976', '5977','5978') AND s.OccurTime >= date_sub(NOW(),interval 15 分钟) ) AND OccurTime >= date_sub(NOW(),interval 15*4*24 minute) group by AlmCode,HR;

问题：

Items-2 查询使用（子查询）一直执行，就好像我将它们作为两个单独的查询运行一样，它立即返回如下。这里缺少什么？

查询 1：获取唯一警报

select distinct(AlmCode)
from TB_ALM_HISTORY 
where AlmCode IN ('3236','4002','4008','4036','4050','4051','4102','4108','4136','4150','4151','4202','4208','4236','4250','4251','4801','4802','4836','4848','4850','4851','4902','4936','4950','4951','5002','5008','5036','5050','5051','5102','5108','5136','5150','5151','5202','5208','5236','5250','5251','5947','5950','5952','5975','5976','5977','5978') 
AND OccurTime >= date_sub(NOW(),interval 15 minute) ;

    +---------+
    | AlmCode |
    +---------+
    |    3236 |
    |    5202 |
    |    5236 |
    +---------+

查询 2：获取过去 6 小时内每个唯一警报的计数

select Almcode,concat(date(OccurTime),' ',LPAD(HOUR(OccurTime),2,'0')) as HR,count(*) from TB_ALM_HISTORY where AlmCode IN ('3236','5202','5236') AND OccurTime >= date_sub(NOW(),interval 15*4*7 minute) group by AlmCode,HR;
+---------+---------------+----------+
| Almcode | HR            | count(*) |
+---------+---------------+----------+
|    3236 | 2015-08-04 11 |        2 |
|    5202 | 2015-08-04 13 |        6 |
|    5202 | 2015-08-04 14 |        4 |
|    5202 | 2015-08-04 15 |        2 |
|    5202 | 2015-08-04 16 |        1 |
|    5202 | 2015-08-04 17 |        2 |
+---------+---------------+----------+

假设此查询在美国东部标准时间下午 6 点运行，AlmCode 5202 已在过去 6 小时内发生（btwn 12-18 小时），因此此 AlmCode 的结果不应包含在最终选择查询中（发生在过去 15 分钟内）。而 AlmCode 3236 在过去 6 小时内没有发生，因此必须包括在过去 15 分钟内针对此特定 AlmCode 发生的所有警报。

如何在一个查询中获得我的最终输出？

一个。获取 OccurTime >= Last 15 Minutes 的唯一 AlmCode

b.对于这些 AlmCode 中的每一个，检查它是否在过去 6 小时内发生了三次

c。 如果否，则使用 OccurTime >= Last 15 Minutes 拉取此 AlmCode 的所有警报 （如果是，则不包括并直接跳过）

【问题讨论】：

能否请您提供创建表语句（包括索引）并分别解释第一次、第二次和复杂查询？
@Vladimir_M 添加了表结构，两步查询的输出具有更详细的预期结果
因此，如果在过去 6 小时内的任何给定 15 分钟内发生三次或更多次警报，是否应该将其包含在结果中？（检查“问题”段落中的描述和“需要什么”）
我已经编辑了我的答案，现在应该适合
在内部连接条件中。 t1 与 t2 具有以下条件：AlmCode 应该相同 (t2.AlmCode = t1.AlmCode ) ，AlmCode_2 应该不迟于 AlmCode_1 后 15 分钟出现 (t2.OccurTime dt1.OccurTime) 之后发生，所以它不可能是相同的发生，因为 OcceTime 是不同的。加入相同的逻辑 t3，在 15 分钟内返回三次

标签： mysql subquery hour in-subquery

【解决方案1】：

过去 15 分钟内创建的所有警报（您的查询）。

select distinct(AlmCode) 
from TB_ALM 
where AlmCode IN ('3236','4002','4008','4036','4050','4051','4102','4108','4136','4150','4151','4202','4208','4236','4250','4251','4801','4802','4836','4848','4850','4851','4902','4936','4950','4951','5002','5008','5036','5050','5051','5102','5108','5136','5150','5151','5202','5208','5236','5250','5251','5947','5950','5952','5975','5976','5977','5978') 
AND OccurTime >= date_sub(NOW(),interval 15 minute)

所有警报，在过去 6 小时内的任意 15 分钟内发生了三次（之后将被排除）

select distinct t1.AlmCode
from TB_ALM t1
inner join TB_ALM t2 on t2.AlmCode = t1.AlmCode 
    and t2.OccurTime <= date_add(t1.OccurTime, interval 15 minute)
    and t2.OccurTime > t1.OccurTime
inner join TB_ALM t3 on t3.AlmCode = t1.AlmCode 
    and t3.OccurTime <= date_add(t1.OccurTime, interval 15 minute)
    and t3.OccurTime > t2.OccurTime
WHERE true
  AND t1.OccurTime >= date_sub(now(), interval 6 hour)
  AND t1.AlmCode IN ('3236','4002','4008','4036','4050','4051','4102','4108','4136','4150','4151','4202','4208','4236','4250','4251','4801','4802','4836','4848','4850','4851','4902','4936','4950','4951','5002','5008','5036','5050','5051','5102','5108','5136','5150','5151','5202','5208','5236','5250','5251','5947','5950','5952','5975','5976','5977','5978')

所以最后的查询是

select distinct(AlmCode) 
from TB_ALM 
where true
  AND OccurTime >= date_sub(NOW(),interval 15 minute) 
  AND AlmCode IN ('3236','4002','4008','4036','4050','4051','4102','4108','4136','4150','4151','4202','4208','4236','4250','4251','4801','4802','4836','4848','4850','4851','4902','4936','4950','4951','5002','5008','5036','5050','5051','5102','5108','5136','5150','5151','5202','5208','5236','5250','5251','5947','5950','5952','5975','5976','5977','5978')
  AND AlmCode NOT IN (select distinct t1.AlmCode
from TB_ALM t1
inner join TB_ALM t2 on t2.AlmCode = t1.AlmCode 
    and t2.OccurTime <= date_add(t1.OccurTime, interval 15 minute)
    and t2.OccurTime > t1.OccurTime
inner join TB_ALM t3 on t3.AlmCode = t1.AlmCode 
    and t3.OccurTime <= date_add(t1.OccurTime, interval 15 minute)
    and t3.OccurTime > t2.OccurTime
WHERE true
      AND t1.OccurTime >= date_sub(now(), interval 6 hour)
      AND t1.AlmCode IN ('3236','4002','4008','4036','4050','4051','4102','4108','4136','4150','4151','4202','4208','4236','4250','4251','4801','4802','4836','4848','4850','4851','4902','4936','4950','4951','5002','5008','5036','5050','5051','5102','5108','5136','5150','5151','5202','5208','5236','5250','5251','5947','5950','5952','5975','5976','5977','5978')
  )

在 AlmCode 列上添加索引，将显着减少执行时间

【讨论】：

能否更正第二个和最后一个查询中的 dt1 拼写错误？
感谢您的关注，现已修复