SQL Server：具有软删除功能的 Graph-DB 未按预期工作

Posted 2023-02-18

【中文标题】SQL Server：具有软删除功能的 Graph-DB 未按预期工作

【英文标题】：SQL Server : Graph-DB with soft delete isn't working as expected

【发布时间】：2021-05-07 14:06:51

【问题描述】：

我尝试在 SQL Server 中使用图形功能。现在我遇到了软删除问题。

我有以下图表

[1] -> [2] -> [9 (deleted)] -> [4]

当我运行以下脚本时

CREATE TABLE MyNode 
(
    [Id] [bigint] NOT NULL,
    [IsDeleted] [bit] NOT NULL,
) AS NODE;

CREATE TABLE MyEdge 
(
    State [int] NOT NULL
) AS EDGE;

INSERT INTO MyNode (Id, IsDeleted)
VALUES (1, 0), (2, 0), (4, 0), (9, 1);

INSERT INTO MyEdge
VALUES
( (SELECT $node_id FROM MyNode WHERE Id = 1), (SELECT $node_id FROM MyNode WHERE Id = 2), 1),
( (SELECT $node_id FROM MyNode WHERE Id = 2), (SELECT $node_id FROM MyNode WHERE Id = 9), 1),
( (SELECT $node_id FROM MyNode WHERE Id = 9), (SELECT $node_id FROM MyNode WHERE Id = 4), 1)
;

SELECT
    src.Id ID_SOURCE
,   LAST_VALUE(trgt.Id) WITHIN GROUP (GRAPH PATH) AS ID_TARGET
,   STRING_AGG(trgt.Id, '->') WITHIN GROUP (GRAPH PATH) AS ID_CHAIN
--, STRING_AGG(compare.State, '->') WITHIN GROUP (GRAPH PATH) AS STATE_CHAIN
--, STRING_AGG(trgt.IsDeleted, '->') WITHIN GROUP (GRAPH PATH) AS DELETED_CHAIN
FROM
    MyNode AS src
,   (   SELECT
            *
        FROM
            MyEdge
        WHERE
            State = 1
    ) FOR PATH AS compare
,   (   SELECT
            *
        FROM
            MyNode
        WHERE
            IsDeleted = 0
    ) FOR PATH AS trgt
WHERE
    MATCH ( SHORTEST_PATH( src(-(compare)->trgt)+ ) )
    AND src.Id = 1;

SELECT
    src.Id AS SOURCE_ID
,   ed.State AS EDGE_STATE
,   trgt.Id AS TARGET_ID
FROM
    MyNode AS src
,   MyEdge AS ed
,   MyNode AS trgt
WHERE
    MATCH( src-(ed)->trgt )
    AND src.Id = 2;


DROP TABLE MyNode;
DROP TABLE MyEdge;

（提示：这仅适用于 SQL-Server 2019）

我得到以下结果

ID_SOURCE	ID_TARGET	ID_CHAIN
1	2	2
1	4	2->4

没有边缘2->4，而是2->9->4。但是节点 9 被删除了，所以它被用于图的遍历，但在输出中被抑制了。

这是 SQL-Server 中的错误还是我做错了什么？

或者还有其他方法我应该使用 Graph-DB 的软删除吗？

【问题讨论】：

交叉投递到feedback.azure.com/forums/908035-sql-server/suggestions/…

【参考方案1】：

您可以过滤掉来自/到软删除节点的边缘

....
(   
SELECT e.*
FROM MyEdge as e
WHERE
    e.State = 1
    and exists(select * from MyNode as x where x.IsDeleted = 0 and x.$node_id = e.$to_id)
    --and exists(select * from MyNode as x where x.IsDeleted = 0 and x.$node_id = e.$from_id)
) FOR PATH AS compare
....

以下查询不知何故返回错误（执行几秒钟后）：

SELECT
    src.Id ID_SOURCE
,   LAST_VALUE(trgt.Id) WITHIN GROUP (GRAPH PATH) AS ID_TARGET
,   STRING_AGG(trgt.Id, '->') WITHIN GROUP (GRAPH PATH) AS ID_CHAIN
--, STRING_AGG(compare.State, '->') WITHIN GROUP (GRAPH PATH) AS STATE_CHAIN
--, STRING_AGG(trgt.IsDeleted, '->') WITHIN GROUP (GRAPH PATH) AS DELETED_CHAIN
FROM
    MyNode AS src
,   
(   
SELECT e.*
FROM MyEdge as e
WHERE
    e.State = 1
    and exists(select * from MyNode as x where x.IsDeleted = 0 and x.$node_id = e.$to_id)
    --and exists(select * from MyNode as x where x.IsDeleted = 0 and x.$node_id = e.$from_id)
) FOR PATH AS compare

,   (   SELECT
            *
        FROM
            MyNode
        WHERE
            IsDeleted = 0
    ) FOR PATH AS trgt
WHERE
    MATCH ( SHORTEST_PATH( src(-(compare)->trgt)+ ) )

...并且通过向 src 表附加条件/引用来消除错误，这始终是正确的：

AND src.Id = src.Id --or src.IsDeleted = src.IsDeleted

肯定有问题 [Microsoft SQL Server 2019 (RTM-CU8-GDR) (KB4583459)]..

【讨论】：

谢谢，我没有考虑过滤边缘。如果这个解决方案足够快，我会尝试。如果没有，我将在边缘添加一个 IsDeleted-Flag 并在删除和取消删除时更新它们。

【参考方案2】：

将 isDeleted 标志合并到边缘表中，并在相应的（到/从）节点被标记为软删除时更新该标志，并在边缘表子查询上使用 isDeleted 是首选方式。我们不建议直接比较 $node_id 和 $from_id，而是通过 MATCH 函数隐式比较。