Question

我正在使用OrientDB的UI /查询工具来分析一些图形数据，我花了几天时间试图解开两个数组失败。

展开子句适用于一个数组，但在尝试展开两个数组时，我似乎无法获得我正在寻找的输出。

以下是我的数据的简化示例：

@class       |  amt  | storeID  | customerID  
transaction     $4        1         1
transaction     $2        1         1
transaction     $6        1         4
transaction     $3        1         4
transaction     $2        2         1
transaction     $7        2         1
transaction     $8        2         2
transaction     $3        2         2
transaction     $4        2         3
transaction     $9        2         3
transaction     $10       3         4
transaction     $3        3         4
transaction     $4        3         5
transaction     $10       3         5

每位客户都是一份包含以下信息的文件：

@class   | customerID | State 
customer     1             NY  
customer     2             NJ  
customer     3             PA
customer     4             NY  
customer     5             NY

每个商店都是一份包含以下信息的文件：

@class   | storeID | State |   Zip 
store         1         NY     1 
store         2         NJ     3    
store         3         NY     2

假设我没有storeID（也不想创建它），我想恢复一个具有以下不同值的扁平表：商店名称，城市，帐号和花费的总和。

该查询有望生成类似下表（对于给定的深度值）。

State | Zip | customerID
 NY      1       4      
 NY      1       5      
 NY      2       1      
 NY      2       4      
 NJ      3       1      
 NJ      3       2      
 NJ      3       3

我尝试了各种扩展/展平/展开操作，但我似乎无法让我的查询工作。

这是我的查询，将State和Zip恢复为两个数组并展平customerID：

SELECT  out().State as State, 
        out().Zip as Zip, 
        customerID 
    FROM ( SELECT EXPAND(IN()) 
            FROM (TRAVERSE * FROM 
                ( SELECT FROM transaction) 
        ) 
    ) ;

哪个收益率，

State            |   Zip     | customerID 
[NY, NY, NJ, NJ]   [1,1,2,2]       1   
[NY, NY, NJ, NJ]   [1,1,2,2]       1   
[NY, NY, PA, PA]   [1,1,3,3]       4
[NY, NY, PA, PA]   [1,1,3,3]       4
...                   ....      ....

这不是我想要的。有人可以提供一些帮助，我可以将这两个数组一起展平/展开吗？

Answer 1

我尝试了这种结构（基于你的例子）：

我使用此查询来检索State，Zip和customerID（而不是数组）：

查询1 ：

SELECT State, Zip, in('transaction').customerID AS customerID FROM Store 
ORDER BY Zip UNWIND customerID

----+------+-----+----+----------
#   |@CLASS|State|Zip |customerID
----+------+-----+----+----------
0   |null  |NY   |1   |1
1   |null  |NY   |1   |1
2   |null  |NY   |1   |4
3   |null  |NY   |1   |4
4   |null  |NY   |2   |4
5   |null  |NY   |2   |4
6   |null  |NY   |2   |5
7   |null  |NY   |2   |5
8   |null  |NJ   |3   |1
9   |null  |NJ   |3   |1
10  |null  |NJ   |3   |2
11  |null  |NJ   |3   |2
12  |null  |NJ   |3   |3
13  |null  |NJ   |3   |3
----+------+-----+----+----------

查询2 ：

SELECT inV('transaction').State AS State, inV('transaction').Zip AS Zip, 
outV('transaction').customerID AS customerID FROM transaction ORDER BY Zip

----+------+-----+----+----------
#   |@CLASS|State|Zip |customerID
----+------+-----+----+----------
0   |null  |NY   |1   |1
1   |null  |NY   |1   |1
2   |null  |NY   |1   |4
3   |null  |NY   |1   |4
4   |null  |NY   |2   |4
5   |null  |NY   |2   |4
6   |null  |NY   |2   |5
7   |null  |NY   |2   |5
8   |null  |NJ   |3   |1
9   |null  |NJ   |3   |1
10  |null  |NJ   |3   |2
11  |null  |NJ   |3   |2
12  |null  |NJ   |3   |3
13  |null  |NJ   |3   |3
----+------+-----+----+----------

<强> EDITED

在以下示例中，通过查询，您将能够检索每个storeID的平均花费和总花费（基于每个customerID）：

SELECT customerID, storeID, avg(amt) AS averagePerStore, sum(amt) AS totalPerStore
FROM transaction GROUP BY customerID,storeID ORDER BY customerID

----+------+----------+-------+---------------+-------------
#   |@CLASS|customerID|storeID|averagePerStore|totalPerStore
----+------+----------+-------+---------------+-------------
0   |null  |1         |1      |3.0            |6.0
1   |null  |1         |2      |4.5            |9.0
2   |null  |2         |2      |5.5            |11.0
3   |null  |3         |2      |6.5            |13.0
4   |null  |4         |1      |4.5            |9.0
5   |null  |4         |3      |6.5            |13.0
6   |null  |5         |3      |7.0            |14.0
----+------+----------+-------+---------------+-------------

希望有所帮助

试图在OrientDB

1 个答案: