使用相同条件但不同列并使用聚合函数编写查询

Question

我正在制作一个系统，可以根据客户需求分配分析师。我编写了一个程序来搜索要分配给需求的理想分析师。

主要表格有：

grc_analyst（Table 解决需求的分析师）
grc_requirement（Table 的要求）与 grc_analyst 的多对一关系。这些要求有一个字段“due_date”，表示必须解决要求的截止日期。

寻找分析师的规则如下：

必须计算与每个分析师相关的需求数量。
如果每个人的数字都相同，也就是说，分析师分配的需求数量相同，那么 select 其需求的平均到期日期距当前日期最远的分析师。
如果数量不同，即分析师分配的需求数量不同，则 select 分配的需求量最少的分析师。

代码如下：

    DECLARE @totalRequirementsAnalysts int
    DECLARE @idAnalyst int = null 
    
    SELECT @totalRequirementsAnalysts =count(distinct(a.requirements)) 
    FROM ( 
        SELECT 
            a.id,count(r.id) requirements
        FROM
            grc_analyst a
            INNER JOIN  grc_analyststate ea     ON a.id_analyststate = ea.id AND ea.code = 'A'
            INNER JOIN  grc_analyst_category ac ON a.id = ac.id_analyst 
            INNER JOIN  grc_category c              ON c.id = ac.id_category AND c.code = 'SOAP'        
            LEFT JOIN   grc_requirement r           ON a.id = r.id_analyst AND r.id_requirementstate in (
                SELECT id from grc_requirementstate er where er.code IN ('AS','ER','DL','DC') 
            )
        group by a.id 
    ) a
    
    
    IF  (@totalRequirementsAnalysts = 1)
    BEGIN 
        PRINT 'Analysts have the same amount of requirements assigned'
    
        SET @idAnalyst = (
            SELECT a.id from (
                SELECT TOP(1)
                    a.id,avg (DATEDIFF(DAY,getdate(),r.due_date))average_due_date
                FROM
                    grc_analyst a
                    INNER JOIN  grc_analyststate ea     ON a.id_analyststate = ea.id AND ea.code = 'A'
                    INNER JOIN  grc_analyst_category ac ON a.id = ac.id_analyst 
                    INNER JOIN  grc_category c              ON c.id = ac.id_category AND c.code = 'SOAP'        
                    LEFT JOIN   grc_requirement r           ON a.id = r.id_analyst AND r.id_requirementstate in (
                        SELECT id from grc_requirementstate er where er.code IN ('AS','ER','DL','DC') 
                    )
                group by a.id 
                order by average_due_date DESC
            ) a
        )
    
    END
    ELSE
    BEGIN
        PRINT 'Analysts have different number of requirements assigned'
        SET @idAnalyst = (
            SELECT a.id from (
                SELECT TOP(1)
                    a.id,count(r.id) requirements
                FROM
                    grc_analyst a
                    INNER JOIN  grc_analyststate ea     ON a.id_analyststate = ea.id AND ea.code = 'A'
                    INNER JOIN  grc_analyst_category ac ON a.id = ac.id_analyst 
                    INNER JOIN  grc_category c              ON c.id = ac.id_category AND c.code = 'SOAP'        
                    LEFT JOIN   grc_requirement r           ON a.id = r.id_analyst AND r.id_requirementstate in (
                        SELECT id from grc_requirementstate er where er.code IN ('AS','ER','DL','DC') 
                    )
                group by a.id 
                order by requirements ASC
            ) a
        )
    END
    
    SELECT ga.id from grc_analyst ga where ga.id = @idAnalyst

如您所见，我使用了三个查询，但所有三个查询的“来自”部分都是相同的（具有相同条件的相同连接表）。此过程符合规则并且有效，但我想减少查询次数，因为存在重复的代码。

提前致谢！

Answer 1

这有帮助吗？

WITH base AS
(
  SELECT 
    a.id,
    count(r.id) requirements,
    avg (DATEDIFF(DAY,getdate(),r.due_date)) average_due_date
  FROM grc_analyst a
  JOIN grc_analyststate ea     ON a.id_analyststate = ea.id AND ea.code = 'A'
  JOIN grc_analyst_category ac ON a.id = ac.id_analyst 
  JOIN grc_category c          ON c.id = ac.id_category AND c.code = 'SOAP'        
  LEFT JOIN grc_requirement r  ON a.id = r.id_analyst
  JOIN grc_requirementstate es ON r.id_requirementstate IS NULL OR 
                                  (r.id_requirementstate IS NOT NULL 
                                    AND r.id_requirementstate = er.id 
                                    AND er.code IN ('AS','ER','DL','DC') 
                                  )
  group by a.id 
)
-- etc

如果您不喜欢 CTE，那么您可以使用上面的视图。

此外，当 grc_requirement table 的左连接在日期差异平均值上为空时，我也不知道规则是什么。如果这个结果是错误的，那应该被修复。

Answer 2

为避免重复代码，创建查询公共部分的视图，然后在您的 3 个查询中使用它，例如

CREATE VIEW dbo.grc_analyst_view
AS
SELECT a.id
    , COUNT(r.id) requirements
    , AVG(DATEDIFF(DAY,GETDATE(),r.due_date)) average_due_date
FROM grc_analyst a
INNER JOIN grc_analyststate ea ON a.id_analyststate = ea.id AND ea.code = 'A'
INNER JOIN grc_analyst_category ac ON a.id = ac.id_analyst 
INNER JOIN grc_category c ON c.id = ac.id_category AND c.code = 'SOAP'        
LEFT JOIN grc_requirement r ON a.id = r.id_analyst AND r.id_requirementstate in (
    SELECT id
    FROM grc_requirementstate er
    WHERE er.code IN ('AS','ER','DL','DC') 
)
GROUP BY a.id;
GO

-- QUERY 1
SELECT @totalRequirementsAnalysts = COUNT(DISTINCT(requirements)) 
FROM dbo.grc_analyst_view;

-- QUERY 2
SELECT TOP(1) id
FROM dbo.grc_analyst_view
ORDER BY average_due_date DESC

-- QUERY 3
SELECT TOP(1) id
FROM dbo.grc_analyst_view
ORDER BY requirements ASC;

Answer 3

实际上我相信你可以用一个查询完成你想要的，利用 window 函数 FIRST_VALUE 和一个 sub-query 你可以获得你想要的值而无需完全重复查询。这也应该表现得更好。

原来你不能在COUNT OVER ()中使用DISTINCT所以我们必须使用CTE和CROSS APPLY来代替。

with cte as (
  SELECT a.id
      , COUNT(DISTINCT r.id) Requirements
      , AVG(DATEDIFF(DAY,GETDATE(),r.due_date)) average_due_date
  FROM grc_analyst a
  INNER JOIN grc_analyststate ea ON a.id_analyststate = ea.id AND ea.code = 'A'
  INNER JOIN grc_analyst_category ac ON a.id = ac.id_analyst 
  INNER JOIN grc_category c ON c.id = ac.id_category AND c.code = 'SOAP'        
  LEFT JOIN grc_requirement r ON a.id = r.id_analyst AND r.id_requirementstate in (
      SELECT id
      FROM grc_requirementstate er
      WHERE er.code IN ('AS','ER','DL','DC') 
  )
  GROUP BY a.id
)
SELECT TOP 1 @idAnalyst = CASE WHEN N.DistinctRequirements = 1 THEN FIRST_VALUE(X.id) OVER (ORDER BY average_due_date ASC) ELSE FIRST_VALUE(X.id) OVER (ORDER BY Requirements ASC) END
FROM cte X
CROSS APPLY (
  SELECT COUNT(DISTINCT Requirements) DistinctRequirements
  FROM cte
) N;
-- While one should normally use an 'order by' clause with 'top' it is meaningless in this case as all rows return the same.

如果需要打印选择了哪个选项，则将 requirements 计数分配给一个变量并在您的 IF 语句中使用它。

使用相同条件但不同列并使用聚合函数编写查询

Write query with same conditions but different columns and using aggregate functions

sql

tsql

sql-server

stored-procedures