集團集團在SQL

Question

我有以下查詢：

SELECT Inspections.Pipe_Segment_Reference, 
(COUNT(*) * Conditions.structural_grade) AS Structural_Rating 
FROM (
    SELECT Inspections.Pipe_Segment_Reference, Conditions.structural_grade 
    FROM Conditions 
    INNER JOIN Inspections 
    ON Conditions.InspectionID = Inspections.InspectionID 
    WHERE Conditions.structural_grade IS NOT NULL 
) 
GROUP BY Inspections.Pipe_Segment_Reference, Conditions.structural_grade; 

，輸出

22, 2 
22, 4 
79, 2 
79, 5 
79, 3 

我想通過Pipe_Segment_Reference進一步GROUP獲得的

22, 6 
79, 10 

我試圖輸出在上面的查詢中添加一個額外的選擇，但我無法讓它工作。

Answer 1

您正在正確的軌道上，在您的工作查詢周圍添加額外的SELECT。試試這個...

SELECT 
    DerivedYourInitialQuery.Pipe_Segment_Reference, 
    SUM(DerivedYourInitialQuery.Structural_Rating) AS NewTotal 
FROM 
    (
    SELECT Inspections.Pipe_Segment_Reference, 
    (COUNT(*) * Conditions.structural_grade) AS Structural_Rating 
    FROM (
     SELECT Inspections.Pipe_Segment_Reference, Conditions.structural_grade 
     FROM Conditions 
     INNER JOIN Inspections 
     ON Conditions.InspectionID = Inspections.InspectionID 
     WHERE Conditions.structural_grade IS NOT NULL 
    ) 
    GROUP BY Inspections.Pipe_Segment_Reference, Conditions.structural_grade 
    ) DerivedYourInitialQuery 
GROUP BY 
    DerivedYourInitialQuery.Pipe_Segment_Reference 

Answer 2

如果在附加的select/sub查詢中只對Pipe_Segment_Reference進行分組，

，或者可以是這樣的：

SELECT Inspections.Pipe_Segment_Reference, 
COUNT(*) AS Structural_Rating 
FROM (
    SELECT Inspections.Pipe_Segment_Reference, Conditions.structural_grade 
    FROM Conditions 
    INNER JOIN Inspections 
    ON Conditions.InspectionID = Inspections.InspectionID 
    WHERE Conditions.structural_grade IS NOT NULL 
) 
GROUP BY Inspections.Pipe_Segment_Reference; 

Answer 3

也許：

SELECT

xxx1.Pipe_Segment_Reference，

總和（n字段或selectResult2）作爲的XName

從xtable1 XXX1

GROUP BY 1

有。 （條件）

爲了通過1