sql - 生成分层数据的复杂递归 SQL

Question

我正在尝试评估商店访客对 COVID-19 传播的影响。

这是一个简单的场景:

访客A 走进商店并遇到Employee1 @ Time = 0。

然后，访问者 A 在时间 = 1 时遇到员工 2。

访客B 走进商店并遇到Employee1 @ Time = 1。

然后访问者 B 在时间 = 2 时遇到了员工 3。

访客A离开商店。

当我收集所有访问者数据以及他们在一段时间内遇到的人时，数据集如下所示:

表 visitorByEmployee :

| VisitorID | EmployeeID | Contact           |
+-----------+------------+-------------------+
| 100       |   X123     | 3/11/2020 1:00    |
| 100       |   X124     | 3/11/2020 1:10    |
| 101       |   X123     | 3/12/2020 1:11    |
| 101       |   X125     | 3/11/2020 1:20    |
| 102       |   X126     | 3/12/2020 10:00   |
| 102       |   X124     | 3/12/2020 10:00   |
| 103       |   X123     | 3/12/2020 11:00   |
| 104       |   X124     | 3/12/2020 12:00   |
| 104       |   X126     | 3/12/2020 12:00   |
| 105       |   X126     | 3/12/2020 12:00   |

我想根据这些数据建立一个层次结构，最终可以表示如下:

每棵树代表访客对病毒传播的影响:

100
  --> X123
    --> 101
      --> X125
    --> 103
  --> X124
    --> 104

102
  --> X126
    --> 104
    --> 105
  --> X124
    --> 104
      --> X126

我试图通过首先找到根节点(不受以前访问者和/或他们看到的员工影响的根访问者)来做到这一点。这些是 100 和 102。

SELECT 
    *,
    ROW_NUMBER() OVER (PARTITION BY EmployeeID ORDER BY Contact) AS SeenOrder
INTO 
    #SeenOrder
FROM
    visitorByEmployee

SELECT *
INTO #RootVisitors
FROM #SeenOrder
WHERE SeenOrder = 1

来自 #RootVisitors和 #SeenOrder ，我想建立一个表格，它可以告诉我影响的层次结构，并可能导致这样的结果:

| InitVisitorID | HLevel     | EmployeeID        |   VisitorID |
+---------------+------------+-------------------+-------------+
| 100           |   0        |  X123             |     100     |
| 100           |   0        |  X124             |     100     |
| 100           |   1        |  X123             |     101     |
| 100           |   1        |  X123             |     103     |
| 100           |   1        |  X124             |     104     |
| 100           |   2        |  X125             |     101     |
| 102           |   0        |  X126             |     102     |
| 102           |   0        |  X124             |     102     |
| 102           |   1        |  X126             |     104     |
| 102           |   1        |  X126             |     105     |
| 102           |   1        |  X124             |     104     |
| 102           |   2        |  X126             |     104     |

这是可以使用递归 CTE 完成的事情吗？我试图这样做，但由于层次结构从访客到员工到访客到员工的转变，我很难创建递归 CTE。

更新
这是我正在研究的递归 CTE。它尚未起作用，但我正在分享的方法是:

; WITH exposure_tree AS (
/* == Anchor with the root visitors == */
/* == You can think of this: The Employees who were exposed by the Visior == */
SELECT re.VisitorID InitVisitor,
    1 as Level, 
    CASE WHEN 1%2=1 THEN 'Visitor' ELSE 'Employee' END ExposerType,
    re.VisitorID Exposer,
    re.EmployeeID Exposee,
    re.SeenOrder,
    re.InitialContact
FROM #SeenOrder re
WHERE re.SeenOrder = 1

/* == Recursive Part #1 ==
Get the visitors who were exposed next by the exposed employees
*/
UNION ALL

SELECT et.VisitorID InitVisitor,
    Level + 1,
    CASE WHEN (Level+1)%2=1 THEN 'Visitor' ELSE 'Employee' END ExposerType,
    re.EmployeeID,
    re.VisitorID, -- These are switched from the anchor.
    re.SeenOrder,
    re.InitialContact
FROM #SeenOrder re
JOIN exposure_tree et ON et.Exposee = re.EmployeeID AND re.SeenOrder > 1 AND re.InitialContact > et.InitialContact

UNION ALL

/* == Recursive Part #2 ==
Get the next employees who were exposed the second level exposed visitors
*/
SELECT et.VisitorID InitVisitor,
    Level + 2,
    CASE WHEN (Level+2)%2=1 THEN 'Visitor' ELSE 'Employee' END ExposerType,
    re.VisitorID,
    re.EmployeeID,
    re.SeenOrder,
    re.InitialContact
FROM #ROOT_EXPOSURES re
JOIN exposure_tree et ON re.VisitorID = et.Exposer and re.SeenOrder > 1 AND re.InitialContact > et.InitialContact
)
select top 1000 * from exposure_tree ORDER BY InitVisitor, Level

Answer 1

您仍然可以使用这些表编写递归 CTE。不过，编码变得棘手。

这是 CTE。您可能需要对其进行调整才能获得您想要的效果。为简单起见，我更改了列名:

with
c as (
  select 'v' as type, vid as id, contact, 0 as lvl, cast(concat('/', vid, '/') as varchar(255)) as path
  from (select *, row_number() over(partition by vid order by contact) as rn from v) x where rn = 1
 union all
  select
   case when type = 'v' then 'e' else 'v' end, -- type
   case when type = 'v' then v.eid else v.vid end, -- id
   v.contact,
   c.lvl + 1,
   cast(concat(c.path, case when type = 'v' then v.eid else v.vid end, '/') as varchar(255))
  from c 
  join v on c.lvl <= 10 and v.contact >= c.contact and (c.type = 'v' and v.vid = c.id or c.type = 'e' and v.eid = c.id)
        and c.path not like concat('%', case when type = 'v' then v.eid else v.vid end, '%')
)
select * from c order by path

结果:

type  id    contact                lvl  path                   
----  ----  ---------------------  ---  -----------------------
v     100   2020-03-11 01:00:00.0    0  /100/                  
e     X123  2020-03-11 01:00:00.0    1  /100/X123/             
v     101   2020-03-12 01:11:00.0    2  /100/X123/101/         
v     103   2020-03-12 11:00:00.0    2  /100/X123/103/         
e     X124  2020-03-11 01:10:00.0    1  /100/X124/             
v     102   2020-03-12 10:00:00.0    2  /100/X124/102/         
e     X126  2020-03-12 10:00:00.0    3  /100/X124/102/X126/    
v     104   2020-03-12 12:00:00.0    4  /100/X124/102/X126/104/
v     105   2020-03-12 12:00:00.0    4  /100/X124/102/X126/105/
v     104   2020-03-12 12:00:00.0    2  /100/X124/104/         
e     X126  2020-03-12 12:00:00.0    3  /100/X124/104/X126/    
v     105   2020-03-12 12:00:00.0    4  /100/X124/104/X126/105/
v     101   2020-03-11 01:20:00.0    0  /101/                  
e     X123  2020-03-12 01:11:00.0    1  /101/X123/             
v     103   2020-03-12 11:00:00.0    2  /101/X123/103/         
e     X125  2020-03-11 01:20:00.0    1  /101/X125/             
v     102   2020-03-12 10:00:00.0    0  /102/                  
e     X124  2020-03-12 10:00:00.0    1  /102/X124/             
v     104   2020-03-12 12:00:00.0    2  /102/X124/104/         
e     X126  2020-03-12 12:00:00.0    3  /102/X124/104/X126/    
v     105   2020-03-12 12:00:00.0    4  /102/X124/104/X126/105/
e     X126  2020-03-12 10:00:00.0    1  /102/X126/             
v     104   2020-03-12 12:00:00.0    2  /102/X126/104/         
e     X124  2020-03-12 12:00:00.0    3  /102/X126/104/X124/    
v     105   2020-03-12 12:00:00.0    2  /102/X126/105/         
v     103   2020-03-12 11:00:00.0    0  /103/                  
e     X123  2020-03-12 11:00:00.0    1  /103/X123/             
v     104   2020-03-12 12:00:00.0    0  /104/                  
e     X124  2020-03-12 12:00:00.0    1  /104/X124/             
e     X126  2020-03-12 12:00:00.0    1  /104/X126/             
v     105   2020-03-12 12:00:00.0    2  /104/X126/105/         
v     105   2020-03-12 12:00:00.0    0  /105/                  
e     X126  2020-03-12 12:00:00.0    1  /105/X126/             
v     104   2020-03-12 12:00:00.0    2  /105/X126/104/         
e     X124  2020-03-12 12:00:00.0    3  /105/X126/104/X124/    

作为引用，这里是我用来测试的数据脚本，如果你需要创建一个 SQL Fiddle 来运行它:

create table v (
  vid varchar(6),
  eid varchar(6),
  contact datetime
);

insert into v (vid, eid, contact) values ('100', 'X123', '2020-03-11 01:00:00');
insert into v (vid, eid, contact) values ('100', 'X124', '2020-03-11 01:10:00');
insert into v (vid, eid, contact) values ('101', 'X123', '2020-03-12 01:11:00');
insert into v (vid, eid, contact) values ('101', 'X125', '2020-03-11 01:20:00');
insert into v (vid, eid, contact) values ('102', 'X126', '2020-03-12 10:00:00');
insert into v (vid, eid, contact) values ('102', 'X124', '2020-03-12 10:00:00');
insert into v (vid, eid, contact) values ('103', 'X123', '2020-03-12 11:00:00');
insert into v (vid, eid, contact) values ('104', 'X124', '2020-03-12 12:00:00');
insert into v (vid, eid, contact) values ('104', 'X126', '2020-03-12 12:00:00');
insert into v (vid, eid, contact) values ('105', 'X126', '2020-03-12 12:00:00');