Question

我必须计算SQL Server中图表中节点的邻居之间的边缘，而我有GraphNodes和GraphEdges的表。表的结构可在previous question

中找到

关于上一个问题，这里问题的方面是不同的：
我必须执行这些步骤来执行任务，即

从GraphNodes

V

必须有DISTINCT V 的邻居列表（例如，在 TABLE 变量SQL中）
在 V 的邻居之间检查（COUNT DISTINCT）链接 GraphEdges
输出 V ，其邻居之间有DISTINCT链接

我为单个节点尝试的查询工作正常，即

SELECT GN.id, COUNT(DISTINCT(CONCAT(GE.Source_Node,'-', GE.Target_Node))) AS NeighborLinks
FROM GraphEdges GE
JOIN GraphNodes GN ON GN.id = 512
WHERE Source_Node IN (SELECT DISTINCT Target_Node FROM GraphEdges WHERE Source_Node = 512
                      UNION ALL
                      SELECT DISTINCT Source_Node FROM GraphEdges WHERE Target_Node = 512
                     ) 
  AND Target_Node IN (SELECT DISTINCT Target_Node FROM GraphEdges WHERE Source_Node = 512
                       UNION ALL
                       SELECT DISTINCT Source_Node FROM GraphEdges WHERE Target_Node = 512
                     )
GROUP BY GN.id

我已将id = 512作为id中GraphNodes的样本。此查询输出为：

+-------+-----------------+
|  id   |   NeighborLinks |
+-------+-----------------+
|  512  |   6             |
+-------+-----------------+

在UNION ALL子句中使用WHERE的原因是id即512在两列中都存在，即Source_Node和Target_Node为那么，必须从两列中选择DISTINCT邻居。此外，使用GE.Source_Node和GE.Target_Node的相同列表，因为必须仅检查 V 的邻居之间的链接，即512。问题是如何使用我认为TABLE变量或任何其他方法来解决提供长列表值而不是512的问题

我已经提出了关于表变量的这个解决方案，但是在查询中使用表变量得到了错误：
尝试1

DECLARE @ID TABLE(id INT)
DECLARE @S_Neighbor TABLE (id INT)
DECLARE @T_Neighbor TABLE (id INT)

INSERT INTO @ID SELECT id FROM GraphNodes

INSERT INTO @S_Neighbor SELECT DISTINCT Source_Node 
                        FROM GraphEdges 
                        WHERE Target_Node IN (SELECT id FROM @ID)
--UNION ALL
INSERT INTO @T_Neighbor SELECT DISTINCT Target_Node 
                        FROM GraphEdges 
                        WHERE Source_Node IN (SELECT id FROM @ID)

SELECT GN.id,COUNT(DISTINCT(CONCAT(GE.Source_Node,'-', GE.Target_Node))) AS Mutual_Links
FROM GraphEdges GE
JOIN GraphNodes GN ON GN.id = @ID
WHERE Source_Node IN (SELECT DISTINCT Target_Node 
                      FROM GraphEdges 
                      WHERE Source_Node IN @T_Neighbor

                      UNION ALL

                      SELECT DISTINCT Source_Node 
                      FROM GraphEdges 
                      WHERE Target_Node IN @S_Neighbor)
   AND Target_Node IN (SELECT DISTINCT Target_Node 
                       FROM GraphEdges 
                       WHERE Source_Node IN @S_Neighbor

                       UNION ALL

                       SELECT DISTINCT Source_Node 
                       FROM GraphEdges 
                       WHERE Target_Node IN @T_Neighbor)
GROUP BY GN.id

我也试过这个：
尝试2

DECLARE @ID_COUNTER INT
DECLARE @MAX_ID INT

SET @ID_COUNTER = 1
SET @MAX_ID = 148410

WHILE @ID_COUNTER <= @MAX_ID
BEGIN
  (
    SELECT GN.id, 
    COUNT(DISTINCT(CONCAT(GE.Source_Node,'-', GE.Target_Node))) AS Mutual_Links
    FROM GraphEdges GE
    JOIN GraphNodes GN ON GN.id = @ID_COUNTER
    WHERE Source_Node IN (SELECT DISTINCT Target_Node 
                          FROM GraphEdges WHERE Source_Node = @ID_COUNTER
                          UNION ALL
                          SELECT DISTINCT Source_Node 
                          FROM GraphEdges WHERE Target_Node = @ID_COUNTER
                         ) 
       AND Target_Node IN (SELECT DISTINCT Target_Node 
                           FROM GraphEdges WHERE Source_Node = @ID_COUNTER
                           UNION ALL
                           SELECT DISTINCT Source_Node 
                           FROM GraphEdges WHERE Target_Node = @ID_COUNTER
                          )
    GROUP BY GN.id
   )
  SET @ID_COUNTER += 1
END

我使用了@MAX_ID = 3并且返回输出需要56秒，而@MAX_ID最初是= 148410.虽然NeighborLinks的返回值是正确的，但显示的输出位于三个单独的窗口中，如下所示：

id  NeighborLinks
1   53

 id NeighborLinks
 2  318

id  NeighborLinks
3   297

Answer 1

我认为您正在寻找类似下面代码段的内容。

我已经为图形节点（#graph_edges）创建了一个临时表。要查找（不同）邻居数的节点位于临时表#nodes中。

CREATE TABLE #graph_edges(source_node INT NOT NULL,target_node INT NOT NULL);

CREATE TABLE #nodes(id INT NOT NULL PRIMARY KEY);
--INSERT INTO #nodes(id)VALUES(512),(513),(514); -- specific nodes to look up in the graph
INSERT INTO #nodes(id)
SELECT source_node FROM #graph_edges UNION SELECT target_node FROM #graph_edges; -- lookup for all distinct nodes ID's in the graph


SELECT id,neighbor_links=COUNT(*)
FROM
    (
        SELECT n.id,l=ge.source_node,r=ge.target_node
        FROM #nodes AS n
             INNER JOIN #graph_edges AS ge ON
                 ge.source_node=n.id
        UNION -- union of the two sets, this filters duplicate rows (ie no duplicate source_node,target_node row will appear in the derived table)
        SELECT n.id,l=ge.target_node,r=ge.source_node
        FROM #nodes AS n
             INNER JOIN #graph_edges AS ge ON
                 ge.target_node=n.id
    ) AS l
GROUP BY id
ORDER BY id;

DROP TABLE #nodes;
DROP TABLE #graph_edges;

将参数传递给SQL Server

1 个答案: