Question

我想计算接口的数量，然后编写这个代码，从两个表节点和接口中计算它们。这段代码有效，但我想知道这个代码的任何其他方式或优化？计算它们的最佳方式是什么？


       SELECT
    q1.NodeName,
    q1.Nup as up, 
    q2.ndown as down
   FROM (
    SELECT
        Nodes.NodeID AS NodeID,
        Interfaces.NodeID AS InterfaceID,
        Nodes.Caption AS NodeName,
        Interfaces.Status as Status,
        Count(Nodes.Caption) as Nup 
        FROM 
        Nodes INNER JOIN Interfaces ON (Nodes.NodeID = Interfaces.NodeID)
        WHERE  
        (
          (
           (Interfaces.Status = '2'))
        )
        GROUP BY  Nodes.NodeID, Nodes.Caption, Interfaces.Status, Interfaces.NodeID
    ) AS q1
    INNER JOIN (
SELECT   
Interfaces.NodeID AS InterfaceID,
Nodes.Caption AS NodeName,
Interfaces.Status as Status,
Count(Nodes.Caption) as ndown
FROM 
Nodes INNER JOIN Interfaces ON (Nodes.NodeID = Interfaces.NodeID)
WHERE  
(
  (
   (Interfaces.Status = '1'))
)

GROUP BY  Nodes.NodeID, Nodes.Caption, Interfaces.Status, Interfaces.NodeID

   ) AS q2

 ON (q1.NodeID = q2.InterfaceID)
order by down Desc

Answer 1

您可以合并查询：

SELECT
    Nodes.Caption AS NodeName,
    Count(
      CASE WHEN Interfaces.Status = '2'
        THEN 1 
        ELSE NULL 
      END) as up,
    Count(
      CASE WHEN Interfaces.Status = '1'
        THEN 1 
        ELSE NULL 
      END) as down
FROM Nodes 
  INNER JOIN Interfaces ON (Nodes.NodeID = Interfaces.NodeID)
WHERE Interfaces.Status IN('1', '2')
GROUP BY  Nodes.NodeID, Nodes.Caption
order by 3 Desc

注意：在原始查询中，如果某个界面始终处于启用状态或始终处于关闭状态，则由于您使用的是INNER JOIN，因此它不会显示在结果集中。

Answer 2

我会简化接口上的内部查询，因为这是WHERE的完整基础，在结果集完成之前不需要加入节点，然后在单个节点完成后加入节点。

SELECT 
      Nodes.NodeName,
      Q1.CountUp as Up,
      Q1.CountDown as Down
   FROM
      ( SELECT
              Interfaces.NodeID AS InterfaceID,
              SUM( CASE WHEN Interfaces.Status = '1' then 1 else 0 end ) as CountUp,
              SUM( CASE WHEN Interfaces.Status = '2' then 1 else 0 end ) as CountDown
           FROM
              Interfaces 
           WHERE
              Interfaces.Status IN ('1','2')
           GROUP BY
              1 ) AS q1
      INNER JOIN Nodes
         ON q1.NodeID = Nodes.NodeID
   order by 
      3 desc

如何优化查询？

2 个答案: