Question

我有一个名为events的表。它看起来像这样：

id | location_id | type    | date
1  | 123         | success | 2018-01-02
2  | 45          | success | 2018-01-13
3  | 123         | failure | 2018-01-23
4  | 66          | failure | 2018-02-04
5  | 123         | success | 2018-02-06
6  | 66          | failure | 2018-03-04

type列只能有两个值 - “成功”或“失败”。我需要完成的工作如下：查找location_id表中至少有两个连续条目events的每个type=failure。按日期订购条目时，即连续执行。在上面的示例中，只应返回location_id 66，因为type列中有两个连续失败。

显而易见的解决方案是：

iterate through location_ids
    get all entries from events table for each location_id, ordered by date
        iterate through the results and return true if we find two consecutive rows with type=failure

我遇到这种方法的问题：我有几千location_id个，每个events表中都有数百个条目。这意味着每次执行此任务时我们都可以拥有数十万个操作（这通常是因为其结果应显示在我们管理面板的主页上）。

所以我想知道是否有人知道更好的解决方案。我试过搜索一个查询来帮助我解决这个问题，但无济于事。

Answer 1

创建表/插入数据

CREATE TABLE events
    (`id` int, `location_id` int, `type` varchar(7), `date` date)
;

INSERT INTO events
    (`id`, `location_id`, `type`, `date`)
VALUES
    (1, 123, 'success', '2018-01-02'),
    (2, 45, 'success', '2018-01-13'),
    (3, 123, 'failure', '2018-01-23'),
    (4, 66, 'failure', '2018-02-04'),
    (5, 123, 'success', '2018-02-06'),
    (6, 66, 'failure', '2018-03-04')
;

对于这个解决方案，我假设当你连续说出你的意思时。

同一年和同一天的连续月份

所以
2018-02-04
2018-03-04
是一个连续的值
同一年和同月连续一天

所以
2018-02-04
2018-02-05
是一个连续的值

我们只需要以任何方式显示location_id，而不是日期任何事情的最后失败。所以3个或更多的失败不应该成为差

要做的最好的事情是设计一个查询，该查询至少可以根据location_id和type组至少匹配2个或更多不同的日期记录，其中过滤器位于type = 'failure'

<强>查询

SELECT 
   location_id 
 , type
FROM 
 events 
WHERE
 type = 'failure'
GROUP BY
   location_id 
 , type
HAVING 
  COUNT(DISTINCT date) >= 2

<强>结果

| location_id |    type |
|-------------|---------|
|          66 | failure |

参见演示http://sqlfiddle.com/#!9/df4679e/56

现在我们使用INNER JOIN获取所有记录。

<强>查询

SELECT 
 events.*
FROM ( 

  SELECT 
     location_id 
   , type
  FROM 
   events 
  WHERE
   type = 'failure'
  GROUP BY
     location_id 
   , type
  HAVING 
    COUNT(DISTINCT date) >= 2
) AS events_grouped

INNER JOIN
 events
ON
   events_grouped.location_id = events.location_id
 AND
   events_grouped.type = events.type

<强>结果

| id | location_id |    type |       date |
|----|-------------|---------|------------|
|  4 |          66 | failure | 2018-02-04 |
|  6 |          66 | failure | 2018-03-04 |

现在我们需要访问下一条记录。有些数据库支持LEAD。
但目前生产就绪的MySQL版本并不支持因此，我们将通过移动自连接来模拟LEAD。

<强>查询

SELECT 
   events1.*
 , events2.*
FROM ( 

  SELECT 
     location_id 
   , type
  FROM 
   events 
  WHERE
   type = 'failure'
  GROUP BY
     location_id 
   , type
  HAVING 
    COUNT(DISTINCT date) >= 2
) AS events_grouped

INNER JOIN
 events events1
ON
   events_grouped.location_id = events1.location_id
 AND
   events_grouped.type = events1.type

INNER JOIN 
 events events2
ON
   # shift to have acces to the next record.
     events1.id <> events2.id 
   AND
     events1.date <= events2.date

<强>结果

| id | location_id |    type |       date | id | location_id |    type |       date |
|----|-------------|---------|------------|----|-------------|---------|------------|
|  4 |          66 | failure | 2018-02-04 |  5 |         123 | success | 2018-02-06 |
|  4 |          66 | failure | 2018-02-04 |  6 |          66 | failure | 2018-03-04 |

参见演示http://sqlfiddle.com/#!9/df4679e/62

你可以清楚地看到记录在JOIN中移动，所以我们现在可以添加我正在谈论的连续值检查。

最终查询

SELECT 
 events1.location_id
FROM ( 

  SELECT 
     location_id 
   , type
  FROM 
   events 
  WHERE
   type = 'failure'
  GROUP BY
     location_id 
   , type
  HAVING 
    COUNT(DISTINCT date) >= 2
) AS events_grouped

INNER JOIN
 events events1
ON
   events_grouped.location_id = events1.location_id
 AND
   events_grouped.type = events1.type

INNER JOIN 
 events events2
ON
   # shift to have acces to the next record.
     events1.id <> events2.id 
   AND
     events1.date <= events2.date
   AND
   (  
     (
       # check consecutive MONTH, YEAR and DAY need to be the same

       # consecutive month with the same year and same day

       # So <br />
       # 2018-02-04 <br />
       # 2018-03-04 <br />
       # is a consecutive value        
       ABS(YEAR(events1.date) - YEAR(events2.date)) = 0
     AND
       ABS(MONTH(events1.date) - MONTH(events2.date)) = 1
     AND
       ABS(DAY(events1.date) - DAY(events2.date)) = 0   
     )
     OR
     (
       # check consecutive DAY, YEAR and MONTH need to be the same

       # consecutive month with the same year and same day

       # So <br />
       # 2018-02-04 <br />
       # 2018-02-05 <br />
       # is a consecutive value 
         ABS(YEAR(events1.date) - YEAR(events2.date)) = 0
       AND
         ABS(MONTH(events1.date) - MONTH(events2.date)) = 0
       AND
         ABS(DAY(events1.date) - DAY(events2.date)) = 1   
     )   
   )

<强>结果

| location_id |
|-------------|
|          66 |

参见演示http://sqlfiddle.com/#!9/df4679e/65

查询以查找列的两个连续值

1 个答案: