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我有一个sql图数据库如下： 在此处输入图像描述

下面是我的 sql 脚本：

DROP TABLE IF EXISTS Person;
CREATE TABLE Person (userName VARCHAR(100)  PRIMARY KEY) AS NODE;

INSERT INTO Person (userName) VALUES ('A'),('B'),('C'),('D'),('E'),('F'); 

DROP TABLE IF EXISTS Follow; 
CREATE TABLE Follow AS EDGE;

INSERT INTO Follow ($from_id, $to_id) VALUES (
   (SELECT $node_id FROM dbo.Person WHERE userName = 'A'),
   (SELECT $node_id FROM dbo.Person WHERE userName = 'E')),

   ((SELECT $node_id FROM dbo.Person WHERE userName = 'E'),
   (SELECT $node_id FROM dbo.Person WHERE userName = 'C')),


   ((SELECT $node_id FROM dbo.Person WHERE userName = 'C'),
   (SELECT $node_id FROM dbo.Person WHERE userName = 'A')),


   ((SELECT $node_id FROM dbo.Person WHERE userName = 'A'),
   (SELECT $node_id FROM dbo.Person WHERE userName = 'F')),

   ((SELECT $node_id FROM dbo.Person WHERE userName = 'F'),
   (SELECT $node_id FROM dbo.Person WHERE userName = 'B')),

   ((SELECT $node_id FROM dbo.Person WHERE userName = 'B'),
   (SELECT $node_id FROM dbo.Person WHERE userName = 'F')),

   ((SELECT $node_id FROM dbo.Person WHERE userName = 'B'),
   (SELECT $node_id FROM dbo.Person WHERE userName = 'E')),

   ((SELECT $node_id FROM dbo.Person WHERE userName = 'E'),
   (SELECT $node_id FROM dbo.Person WHERE userName = 'B'));

我尝试过的：
在我尝试过的查询之后，只能给我以节点 C 结尾的每个节点的最短路径。

SELECT
    username, StartNode, [Edges Path], FinalNode, Levels
FROM (
    SELECT 
        P1.username, 
        P1.username as StartNode, 
        STRING_AGG(P2.userName,'->') WITHIN GROUP (GRAPH PATH) AS [Edges Path],
        LAST_VALUE(P2.userName) WITHIN GROUP (GRAPH PATH) AS FinalNode,
        COUNT(P2.userName) WITHIN GROUP (GRAPH PATH) AS Levels
    FROM
        Person P1,
        Person FOR PATH P2,
        Follow FOR PATH Follow
    WHERE 
        MATCH(SHORTEST_PATH(P1(-(Follow)->P2)+))
 ) AS Q
 WHERE Q.FinalNode = 'C'

我打算实现的目标：

  1. 查找以节点 C 开头的所有路径
  2. 查找以节点 C 结尾的所有路径

这里的关键是我想找到所有以 C 开头并以 C 结尾的路径。我希望我可以通过将 SHORTEST_PATH 更改为 ALL_PATH 来更改上述查询来实现（2），但不存在这样的方法。另一方面，我不知道如何实现(1)。

我怎样才能达到我想要的？

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